Introduction To Aqueous Solutions
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Precipitation Reactions
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Acid-Base Reactions
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Oxidation-Reduction Reactions
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- Oxidation-Reduction Reactions
- Balancing Oxidation-Reduction Reactions in Acidic Solutions
- Balancing Oxidation-Reduction Reactions in Basic Solutions
- Exercise 1
- Exercise 2 - Part a
- Exercise 2 - Part b
- Exercise 2 - Part c
- Exercise 2 - Part d
- Exercise 3 - Part a
- Exercise 3 - Part b
- Exercise 4 - Part a
- Exercise 4 - Part b
- Disproportionation Reactions
- Oxidizing and Reducing Agents

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[{"Name":"Introduction To Aqueous Solutions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Electrolytes and Non-Electrolytes","Duration":"7m 29s","ChapterTopicVideoID":16920,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16920.jpeg","UploadDate":"2019-02-20T23:50:50.4730000","DurationForVideoObject":"PT7M29S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.450","Text":"In this video, we will discuss substances that"},{"Start":"00:03.450 ","End":"00:06.435","Text":"dissolve in water to give aqueous solutions."},{"Start":"00:06.435 ","End":"00:09.000","Text":"We will divide them into 3 categories."},{"Start":"00:09.000 ","End":"00:13.305","Text":"Strong and weak electrolytes and non-electrolytes."},{"Start":"00:13.305 ","End":"00:15.900","Text":"Let\u0027s start with strong electrolytes."},{"Start":"00:15.900 ","End":"00:22.085","Text":"A strong electrolyte is a substance that is completely ionized in an aqueous solution."},{"Start":"00:22.085 ","End":"00:26.510","Text":"Completely ionized means that it consists only of ions."},{"Start":"00:26.510 ","End":"00:29.675","Text":"There\u0027s none of the original substance left."},{"Start":"00:29.675 ","End":"00:31.835","Text":"Let\u0027s take an example."},{"Start":"00:31.835 ","End":"00:34.565","Text":"Sodium chloride dissolved in water."},{"Start":"00:34.565 ","End":"00:36.140","Text":"Now, in a previous video,"},{"Start":"00:36.140 ","End":"00:43.835","Text":"we saw that sodium chloride is an ionic solid consisting of any plus and Cl minus ions."},{"Start":"00:43.835 ","End":"00:46.400","Text":"What happens when it dissolves in water?"},{"Start":"00:46.400 ","End":"00:54.695","Text":"In water, we get solvated or sometimes called hydrated any plus and Cl minus ions."},{"Start":"00:54.695 ","End":"00:56.780","Text":"Let\u0027s look and see how they look."},{"Start":"00:56.780 ","End":"00:58.655","Text":"Here we have a picture of them."},{"Start":"00:58.655 ","End":"01:03.745","Text":"Here\u0027s Na plus ion surrounded by water molecules."},{"Start":"01:03.745 ","End":"01:08.980","Text":"Here\u0027s a Cl minus ion surrounded by water molecules."},{"Start":"01:08.980 ","End":"01:12.130","Text":"Now, you see that the water molecules are oriented"},{"Start":"01:12.130 ","End":"01:16.945","Text":"differently round CNa plus from Cl minus."},{"Start":"01:16.945 ","End":"01:20.440","Text":"In any plus, the O is towards the Na plus,"},{"Start":"01:20.440 ","End":"01:21.805","Text":"and Cl minus,"},{"Start":"01:21.805 ","End":"01:24.625","Text":"the H is nearer to the Cl minus."},{"Start":"01:24.625 ","End":"01:31.300","Text":"The reason for this is that O is more negative than H. We write it as delta minus."},{"Start":"01:31.300 ","End":"01:33.475","Text":"That means a little bit negative."},{"Start":"01:33.475 ","End":"01:36.685","Text":"H is delta plus,"},{"Start":"01:36.685 ","End":"01:38.790","Text":"a little bit more positive."},{"Start":"01:38.790 ","End":"01:44.610","Text":"Any plus is attracted towards the slightly negative O,"},{"Start":"01:44.610 ","End":"01:49.100","Text":"and Cl minus is attracted towards the slightly positive"},{"Start":"01:49.100 ","End":"01:55.470","Text":"H. These are called solvation shells or hydration shells."},{"Start":"01:55.470 ","End":"01:58.740","Text":"They stabilize the ions."},{"Start":"01:58.740 ","End":"02:01.400","Text":"The ions are stable in water."},{"Start":"02:01.400 ","End":"02:06.800","Text":"Now we can see that strong electrolytes conduct electricity."},{"Start":"02:06.800 ","End":"02:08.600","Text":"How can we see this?"},{"Start":"02:08.600 ","End":"02:14.690","Text":"How can we see that the ions in the strong electrolytes conduct electricity?"},{"Start":"02:14.690 ","End":"02:16.970","Text":"Let\u0027s consider this system."},{"Start":"02:16.970 ","End":"02:18.410","Text":"We have a beaker,"},{"Start":"02:18.410 ","End":"02:21.785","Text":"and inside there is a strong electrolyte."},{"Start":"02:21.785 ","End":"02:25.190","Text":"This is a strong electrolyte."},{"Start":"02:25.190 ","End":"02:28.705","Text":"Say sodium chloride in water."},{"Start":"02:28.705 ","End":"02:33.115","Text":"Here we have 2 rods, graphite rods."},{"Start":"02:33.115 ","End":"02:35.690","Text":"Here is a battery."},{"Start":"02:35.690 ","End":"02:40.460","Text":"Electrons flow from the battery to the graphite rod,"},{"Start":"02:40.460 ","End":"02:44.660","Text":"making the graphite rod negatively charged."},{"Start":"02:44.660 ","End":"02:48.650","Text":"The other graphite rod becomes positively"},{"Start":"02:48.650 ","End":"02:52.850","Text":"charged because electrons are taken away from it."},{"Start":"02:52.850 ","End":"02:56.510","Text":"We have a negative charge here on this right-hand side,"},{"Start":"02:56.510 ","End":"02:59.180","Text":"and a positive charge on the left-hand side."},{"Start":"02:59.180 ","End":"03:01.120","Text":"Now what about the ions?"},{"Start":"03:01.120 ","End":"03:05.295","Text":"Here is an Na plus ion rather small one Na plus."},{"Start":"03:05.295 ","End":"03:11.390","Text":"It\u0027s positively charged, so it\u0027s attracted to the negative electrode."},{"Start":"03:11.390 ","End":"03:14.070","Text":"We call this an electrode."},{"Start":"03:14.690 ","End":"03:18.590","Text":"The Cl minus is negatively charged,"},{"Start":"03:18.590 ","End":"03:21.500","Text":"so it attract to the positive electrode."},{"Start":"03:21.500 ","End":"03:27.155","Text":"The negative electrode is called a cathode,"},{"Start":"03:27.155 ","End":"03:31.310","Text":"and the positive electrode is called an anode."},{"Start":"03:31.310 ","End":"03:37.265","Text":"The electrons flow into the graphite rod,"},{"Start":"03:37.265 ","End":"03:41.425","Text":"and the cations and anions move,"},{"Start":"03:41.425 ","End":"03:48.375","Text":"the ions move and carry the charge thereby conducting electricity."},{"Start":"03:48.375 ","End":"03:54.234","Text":"The movement of these ions causes conduction of electricity,"},{"Start":"03:54.234 ","End":"04:01.340","Text":"and then the electrons flow out of the anode back to the battery."},{"Start":"04:01.340 ","End":"04:04.554","Text":"We have a complete movement of charge."},{"Start":"04:04.554 ","End":"04:10.669","Text":"Sodium chloride and other salts are strong electrolytes."},{"Start":"04:10.669 ","End":"04:12.275","Text":"Are there other ones?"},{"Start":"04:12.275 ","End":"04:17.690","Text":"Yes. The strong acids and strong bases like HCl,"},{"Start":"04:17.690 ","End":"04:23.045","Text":"strong acid, and any wage for a strong base,"},{"Start":"04:23.045 ","End":"04:25.370","Text":"these are also strong electrolytes."},{"Start":"04:25.370 ","End":"04:27.485","Text":"We\u0027ll meet these in a few videos."},{"Start":"04:27.485 ","End":"04:30.290","Text":"Now, what about weak electrolytes?"},{"Start":"04:30.290 ","End":"04:34.610","Text":"These are substances that produce some ions in water."},{"Start":"04:34.610 ","End":"04:39.190","Text":"They undergo partial ionization, not complete ionization."},{"Start":"04:39.190 ","End":"04:42.950","Text":"As a result of there only being a few ions,"},{"Start":"04:42.950 ","End":"04:46.160","Text":"they conduct electricity very weakly."},{"Start":"04:46.160 ","End":"04:48.095","Text":"Let\u0027s take an example."},{"Start":"04:48.095 ","End":"04:50.255","Text":"The example is acetic acid."},{"Start":"04:50.255 ","End":"04:51.590","Text":"We\u0027ve met that before."},{"Start":"04:51.590 ","End":"04:53.930","Text":"It\u0027s CH_3CO_2H."},{"Start":"04:53.930 ","End":"04:56.465","Text":"What happens to it in water?"},{"Start":"04:56.465 ","End":"05:05.175","Text":"CH_3CO_2H in water ionizes partially to give CH_3CO_2 minus and H plus."},{"Start":"05:05.175 ","End":"05:07.230","Text":"That\u0027s how we often write it."},{"Start":"05:07.230 ","End":"05:12.965","Text":"Now, these arrows mean that there is an equilibrium."},{"Start":"05:12.965 ","End":"05:16.175","Text":"It means that there\u0027s only a partial process."},{"Start":"05:16.175 ","End":"05:18.590","Text":"That when we come to equilibrium,"},{"Start":"05:18.590 ","End":"05:24.110","Text":"we have some acid and some ions, not totally ions."},{"Start":"05:24.110 ","End":"05:29.360","Text":"In this case, the equilibrium is heavily on the left-hand side,"},{"Start":"05:29.360 ","End":"05:35.075","Text":"so it\u0027s mostly acid and only a small proportion of ions."},{"Start":"05:35.075 ","End":"05:40.950","Text":"Another way of writing this is CH_3CO_2H in"},{"Start":"05:40.950 ","End":"05:48.945","Text":"water plus 1 water molecule giving CH_3CO_2 minus and H_3O plus."},{"Start":"05:48.945 ","End":"05:52.430","Text":"Now this H_3O plus is called hydronium."},{"Start":"05:52.430 ","End":"05:57.425","Text":"It\u0027s meant to be H plus with 1 water molecule."},{"Start":"05:57.425 ","End":"06:03.140","Text":"Of course, there are usually many more than 1 water molecule."},{"Start":"06:03.140 ","End":"06:05.240","Text":"But to simplify matters,"},{"Start":"06:05.240 ","End":"06:08.315","Text":"we write it as if there\u0027s only 1 water molecule."},{"Start":"06:08.315 ","End":"06:11.440","Text":"It\u0027s H. There\u0027s 3 H\u0027s here,"},{"Start":"06:11.440 ","End":"06:13.455","Text":"and 1 O,"},{"Start":"06:13.455 ","End":"06:15.845","Text":"and the whole thing is positively charged."},{"Start":"06:15.845 ","End":"06:19.220","Text":"That\u0027s the hydronium ion."},{"Start":"06:19.220 ","End":"06:22.430","Text":"We can conclude that on a very small proportion of"},{"Start":"06:22.430 ","End":"06:27.410","Text":"the acetic acid molecules dissociate into ions."},{"Start":"06:27.410 ","End":"06:31.230","Text":"Now, what about non-electrolytes?"},{"Start":"06:31.420 ","End":"06:36.980","Text":"So far we\u0027ve had strong electrolytes and weak electrolytes,"},{"Start":"06:36.980 ","End":"06:44.135","Text":"the third category is non-electrolytes."},{"Start":"06:44.135 ","End":"06:48.590","Text":"Now many molecular compounds do not dissociate at all in water."},{"Start":"06:48.590 ","End":"06:52.505","Text":"They therefore do not conduct electricity at all."},{"Start":"06:52.505 ","End":"06:54.210","Text":"What are examples?"},{"Start":"06:54.210 ","End":"06:56.905","Text":"Examples are ethanol,"},{"Start":"06:56.905 ","End":"06:59.240","Text":"that\u0027s alcohol that we drink,"},{"Start":"06:59.240 ","End":"07:02.690","Text":"and glucose, that sugar that we eat."},{"Start":"07:02.690 ","End":"07:06.170","Text":"Note, both of these dissolve in water,"},{"Start":"07:06.170 ","End":"07:08.779","Text":"but don\u0027t form ions."},{"Start":"07:08.779 ","End":"07:11.420","Text":"They do not dissociate at all,"},{"Start":"07:11.420 ","End":"07:14.930","Text":"and they do not conduct electricity at all."},{"Start":"07:14.930 ","End":"07:18.500","Text":"In this video, we classified substances that dissolve in"},{"Start":"07:18.500 ","End":"07:22.700","Text":"water as electrolytes or non-electrolytes."},{"Start":"07:22.700 ","End":"07:29.430","Text":"We classified electrolytes as either strong or weak electrolytes."}],"ID":17667},{"Watched":false,"Name":"Exercise 1 - Part a","Duration":"2m 21s","ChapterTopicVideoID":22888,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22888.jpeg","UploadDate":"2020-12-15T06:22:09.1870000","DurationForVideoObject":"PT2M21S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:03.615","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:03.615 ","End":"00:06.705","Text":"Determine the concentrations of the following ions."},{"Start":"00:06.705 ","End":"00:13.110","Text":"In a, we have the potassium cation in 0.328 molar, potassium chloride."},{"Start":"00:13.110 ","End":"00:18.930","Text":"Now, I just want to remind you that molarity M equals the number of"},{"Start":"00:18.930 ","End":"00:25.629","Text":"moles of the solute divided by the volume of the solution."},{"Start":"00:25.670 ","End":"00:28.920","Text":"Now if we take a look at a,"},{"Start":"00:28.920 ","End":"00:34.395","Text":"you can see then potassium chloride."},{"Start":"00:34.395 ","End":"00:37.965","Text":"Assuming we have 1 mole of potassium chloride,"},{"Start":"00:37.965 ","End":"00:42.330","Text":"there is also 1 mole of the potassium cation."},{"Start":"00:46.880 ","End":"00:50.050","Text":"Now, in order to calculate the concentration of"},{"Start":"00:50.050 ","End":"00:55.025","Text":"the potassium cation in the potassium chloride solution,"},{"Start":"00:55.025 ","End":"01:01.580","Text":"we\u0027re going to take the concentration of the potassium chloride solution which is 0.328."},{"Start":"01:01.580 ","End":"01:05.320","Text":"This is a, we\u0027re starting with the concentration of"},{"Start":"01:05.320 ","End":"01:09.235","Text":"the potassium chloride solution, 0.328 molar."},{"Start":"01:09.235 ","End":"01:12.195","Text":"Again, I want to remind you that this is"},{"Start":"01:12.195 ","End":"01:19.970","Text":"0.328 moles of the solute which in our case is potassium chloride,"},{"Start":"01:19.970 ","End":"01:24.100","Text":"divided by liters of the solution."},{"Start":"01:25.550 ","End":"01:29.825","Text":"Now since we know that for every mole of potassium chloride,"},{"Start":"01:29.825 ","End":"01:32.930","Text":"we have 1 mole of potassium cation,"},{"Start":"01:32.930 ","End":"01:38.190","Text":"we can multiply this by 1 mole of"},{"Start":"01:38.190 ","End":"01:44.470","Text":"the potassium cation for every 1 mole of potassium chloride."},{"Start":"01:45.070 ","End":"01:52.860","Text":"The moles of potassium chloride cancel out and in the end,"},{"Start":"01:52.860 ","End":"01:55.660","Text":"we\u0027re left with 0.328."},{"Start":"01:57.460 ","End":"02:00.800","Text":"Now, the moles of the potassium cation divided by"},{"Start":"02:00.800 ","End":"02:03.905","Text":"the liters of the solution is molar so this equals"},{"Start":"02:03.905 ","End":"02:10.225","Text":"0.328 molar potassium cation."},{"Start":"02:10.225 ","End":"02:15.080","Text":"The concentration for the potassium cation that we found in a equals 0.328"},{"Start":"02:15.080 ","End":"02:20.920","Text":"molar and that is our answer for a. We\u0027re going to go on to b."}],"ID":23709},{"Watched":false,"Name":"Exercise 1 - Part b","Duration":"2m 9s","ChapterTopicVideoID":22881,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22881.jpeg","UploadDate":"2020-12-15T06:20:41.5370000","DurationForVideoObject":"PT2M9S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.400 ","End":"00:04.590","Text":"In b, we need to calculate the concentration of"},{"Start":"00:04.590 ","End":"00:10.875","Text":"the sulfate anion in 0.267 molar aluminum sulfate."},{"Start":"00:10.875 ","End":"00:16.890","Text":"Again, we\u0027re going to start with the concentration of the aluminum sulfate."},{"Start":"00:16.890 ","End":"00:21.610","Text":"We have 0.267 molar,"},{"Start":"00:21.920 ","End":"00:26.170","Text":"which equals 0.267,"},{"Start":"00:26.420 ","End":"00:30.580","Text":"again, moles of aluminum sulfate"},{"Start":"00:36.440 ","End":"00:42.010","Text":"divided by liters of solution."},{"Start":"00:42.350 ","End":"00:45.480","Text":"If we take a look at aluminum sulfate,"},{"Start":"00:45.480 ","End":"00:50.580","Text":"you will see that if we have 1 mole of aluminum sulfate,"},{"Start":"00:52.840 ","End":"00:55.040","Text":"if we look at the sulfate,"},{"Start":"00:55.040 ","End":"00:56.945","Text":"we can see that it\u0027s multiplied by 3,"},{"Start":"00:56.945 ","End":"01:07.060","Text":"meaning we have 3 moles of the sulfate ion in 1 mole of aluminum sulfate."},{"Start":"01:08.090 ","End":"01:10.710","Text":"We\u0027re going to take the concentration of"},{"Start":"01:10.710 ","End":"01:13.840","Text":"the aluminum sulfate and we\u0027re going to multiply this"},{"Start":"01:13.840 ","End":"01:19.990","Text":"by 3 moles of the sulfate anion,"},{"Start":"01:19.990 ","End":"01:26.740","Text":"divided by 1 mole of aluminum sulfate."},{"Start":"01:28.070 ","End":"01:31.995","Text":"The moles of aluminum sulfate cancel out,"},{"Start":"01:31.995 ","End":"01:37.620","Text":"and this equals 0.801 moles of"},{"Start":"01:37.620 ","End":"01:45.750","Text":"the sulfate anion divided by liters of solution,"},{"Start":"01:45.750 ","End":"01:55.745","Text":"and this equals 0.801 molar sulfate anion."},{"Start":"01:55.745 ","End":"02:03.880","Text":"The concentration of the sulfate anion that we found equals 0.801 molar."},{"Start":"02:04.580 ","End":"02:07.080","Text":"That is our answer for b,"},{"Start":"02:07.080 ","End":"02:09.460","Text":"and we\u0027re going to go on to C."}],"ID":23702},{"Watched":false,"Name":"Exercise 1 - Part c","Duration":"1m 56s","ChapterTopicVideoID":22882,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22882.jpeg","UploadDate":"2020-12-15T06:20:48.4470000","DurationForVideoObject":"PT1M56S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.640","Text":"In c, we need to calculate the concentration of"},{"Start":"00:02.640 ","End":"00:08.055","Text":"the aluminum cation in 0.267 molar aluminum sulfate."},{"Start":"00:08.055 ","End":"00:12.030","Text":"Again, we begin with the concentration of the aluminum sulfate,"},{"Start":"00:12.030 ","End":"00:15.930","Text":"which is 0.267 molar aluminum sulfate,"},{"Start":"00:15.930 ","End":"00:20.550","Text":"which equals 0.267 moles of"},{"Start":"00:20.550 ","End":"00:29.050","Text":"aluminum sulfate divided by liters of solution."},{"Start":"00:29.180 ","End":"00:34.690","Text":"Again, if we take a look at the aluminum sulfate, we have Al_2(SO_4)_3."},{"Start":"00:37.250 ","End":"00:40.785","Text":"Now we\u0027re interested in the aluminum cation."},{"Start":"00:40.785 ","End":"00:43.700","Text":"We have 1 mole of aluminum sulfate,"},{"Start":"00:43.700 ","End":"00:47.390","Text":"you can see that we have 2 moles of aluminum,"},{"Start":"00:47.390 ","End":"00:50.810","Text":"since we have 2 aluminums in every 1 aluminum sulfate,"},{"Start":"00:50.810 ","End":"00:55.450","Text":"so 2 moles of the aluminum cation."},{"Start":"00:56.770 ","End":"01:03.560","Text":"In this case, we\u0027re going to multiply by 2 moles of"},{"Start":"01:03.560 ","End":"01:11.910","Text":"the aluminum cation divided by 1 mole of aluminum sulfate."},{"Start":"01:14.750 ","End":"01:21.730","Text":"Aluminum sulfate mole is going to cancel out and this comes to"},{"Start":"01:21.730 ","End":"01:31.960","Text":"0.534 moles of the aluminum cation"},{"Start":"01:32.020 ","End":"01:35.789","Text":"divided by liters of solution."},{"Start":"01:35.900 ","End":"01:44.695","Text":"This equals 0.534 molar and this is the aluminum cation."},{"Start":"01:44.695 ","End":"01:48.905","Text":"The concentration that we found in c for the aluminum cation is"},{"Start":"01:48.905 ","End":"01:56.250","Text":"0.534 molar. That is our answer for c."}],"ID":23703},{"Watched":false,"Name":"Exercise 1 - Part d","Duration":"1m 49s","ChapterTopicVideoID":22883,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22883.jpeg","UploadDate":"2020-12-15T06:20:54.5270000","DurationForVideoObject":"PT1M49S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"In d we need to calculate the concentration of the sodium cation"},{"Start":"00:04.380 ","End":"00:09.640","Text":"in 0.198 molar sodium sulfate."},{"Start":"00:10.100 ","End":"00:13.740","Text":"Again, we start with the concentration of the sodium sulfate,"},{"Start":"00:13.740 ","End":"00:18.600","Text":"which is 0.198 molar sodium sulfate."},{"Start":"00:18.600 ","End":"00:24.240","Text":"This equals 0.198 moles of"},{"Start":"00:24.240 ","End":"00:33.490","Text":"sodium sulfate per liters of solution."},{"Start":"00:33.490 ","End":"00:35.960","Text":"If we take a look at the sodium sulfate,"},{"Start":"00:35.960 ","End":"00:39.900","Text":"for every 1 mole of sodium sulfate,"},{"Start":"00:41.920 ","End":"00:49.320","Text":"there are 2 moles of the sodium cation,"},{"Start":"00:51.190 ","End":"00:56.165","Text":"since we have 2 sodium Na_2."},{"Start":"00:56.165 ","End":"01:01.370","Text":"In this case, we\u0027re going to multiply by 2 moles of"},{"Start":"01:01.370 ","End":"01:11.045","Text":"the sodium cation divided by 1 mole of the sodium sulfate."},{"Start":"01:11.045 ","End":"01:14.555","Text":"The moles of sodium sulfate cancel out."},{"Start":"01:14.555 ","End":"01:26.225","Text":"This equals 0.396 moles of sodium cation"},{"Start":"01:26.225 ","End":"01:32.030","Text":"per liter of solution and this equals"},{"Start":"01:32.030 ","End":"01:39.560","Text":"0.396 molar sodium cation."},{"Start":"01:39.560 ","End":"01:46.535","Text":"Our answer for d is 0.396 molar concentration of the sodium cation."},{"Start":"01:46.535 ","End":"01:48.020","Text":"That is our final answer."},{"Start":"01:48.020 ","End":"01:50.580","Text":"Thank you very much for watching."}],"ID":23704},{"Watched":false,"Name":"Exercise 2 - Part a","Duration":"4m 37s","ChapterTopicVideoID":22884,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22884.jpeg","UploadDate":"2020-12-15T06:21:11.6900000","DurationForVideoObject":"PT4M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.400","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.400 ","End":"00:08.250","Text":"A solution is 0.17 molar sodium sulfate and 0.23 molar copper sulfate."},{"Start":"00:08.250 ","End":"00:11.610","Text":"What are the concentrations of the ions in the solution?"},{"Start":"00:11.610 ","End":"00:16.390","Text":"We\u0027re going to start with the sodium ion,"},{"Start":"00:18.670 ","End":"00:21.260","Text":"and I want to calculate the concentration."},{"Start":"00:21.260 ","End":"00:27.390","Text":"Now we know that the concentration of the sodium sulfate is 0.17 molar,"},{"Start":"00:27.390 ","End":"00:31.365","Text":"so we\u0027re going to start with the concentration of the sodium sulfate."},{"Start":"00:31.365 ","End":"00:33.580","Text":"But just to remind you,"},{"Start":"00:34.610 ","End":"00:38.780","Text":"molarity or molar equals the number of moles of"},{"Start":"00:38.780 ","End":"00:44.460","Text":"the solute divided by the volume of the solution in liters."},{"Start":"00:46.160 ","End":"00:56.720","Text":"We have 0.17 moles of sodium sulfate per liter of solution."},{"Start":"00:56.720 ","End":"01:00.960","Text":"Now if we take a look at the sodium sulfate,"},{"Start":"01:02.260 ","End":"01:08.385","Text":"we have Na_2SO_4,"},{"Start":"01:08.385 ","End":"01:11.760","Text":"so assuming we have 1 mole of sodium sulfate,"},{"Start":"01:11.760 ","End":"01:14.060","Text":"in every 1 molecule of sodium sulfate,"},{"Start":"01:14.060 ","End":"01:16.250","Text":"you can see that we have 2 sodium,"},{"Start":"01:16.250 ","End":"01:18.920","Text":"meaning that if we have 1 mole of sodium sulfate,"},{"Start":"01:18.920 ","End":"01:26.490","Text":"we\u0027re going to have 2 moles of the sodium ion."},{"Start":"01:28.060 ","End":"01:30.380","Text":"We began with the molarity of"},{"Start":"01:30.380 ","End":"01:34.955","Text":"the sodium sulfate and now we want to know the molarity of the sodium ion."},{"Start":"01:34.955 ","End":"01:37.940","Text":"We\u0027re going to multiply this by 2 moles of"},{"Start":"01:37.940 ","End":"01:48.090","Text":"the sodium ion for every 1 mole of sodium sulfate."},{"Start":"01:51.290 ","End":"01:54.800","Text":"Now as you can see, the moles of the sodium sulfate cancel"},{"Start":"01:54.800 ","End":"02:00.080","Text":"out and what is left to do is just multiply the 0.17 by 2,"},{"Start":"02:00.080 ","End":"02:02.825","Text":"so that comes to 0.34."},{"Start":"02:02.825 ","End":"02:11.185","Text":"We\u0027re left with moles of the sodium ion per liter of solution,"},{"Start":"02:11.185 ","End":"02:18.885","Text":"and this equals molarity of the sodium ion."},{"Start":"02:18.885 ","End":"02:25.679","Text":"The answer for is, the concentration of the sodium ion is 0.34 molar."},{"Start":"02:26.830 ","End":"02:30.155","Text":"Now we\u0027re going to go on to b."},{"Start":"02:30.155 ","End":"02:33.060","Text":"We have the concentration of the sodium ion,"},{"Start":"02:33.060 ","End":"02:36.850","Text":"let\u0027s go for the concentration of the copper ion."},{"Start":"02:36.850 ","End":"02:40.270","Text":"The copper has a plus 2 ion"},{"Start":"02:41.540 ","End":"02:46.275","Text":"and we\u0027re going to calculate the concentration in the same way."},{"Start":"02:46.275 ","End":"02:50.110","Text":"We\u0027re going to start with a molarity of the copper sulfate,"},{"Start":"02:50.110 ","End":"02:54.110","Text":"which equals 0.23 molar."},{"Start":"02:54.110 ","End":"02:57.830","Text":"We\u0027re going to start with 0.23 moles of"},{"Start":"02:57.830 ","End":"03:07.570","Text":"copper sulfate per liter of solution and we\u0027re going to multiply this by,"},{"Start":"03:09.800 ","End":"03:12.470","Text":"if we look at our copper sulfate,"},{"Start":"03:12.470 ","End":"03:15.090","Text":"which is CuSO_4,"},{"Start":"03:15.800 ","End":"03:19.950","Text":"for every 1 mole of copper sulfate,"},{"Start":"03:19.950 ","End":"03:29.685","Text":"we have 1 mole of the copper ion since we have 1 copper here,"},{"Start":"03:29.685 ","End":"03:36.555","Text":"Cu_1SO_4, so we\u0027re going to multiply this by 1 mole"},{"Start":"03:36.555 ","End":"03:44.830","Text":"of the copper ion for every 1 mole of copper sulfate."},{"Start":"03:46.750 ","End":"03:57.169","Text":"The moles of copper sulfate are going to cancel out and 0.23"},{"Start":"03:57.169 ","End":"04:01.430","Text":"times 1=0.23 moles of"},{"Start":"04:01.430 ","End":"04:08.409","Text":"the copper cation per liter of solution,"},{"Start":"04:09.410 ","End":"04:16.630","Text":"and this equals 0.23 molar."},{"Start":"04:18.520 ","End":"04:24.150","Text":"The molarity of the copper ion came out to 0.23 molar."},{"Start":"04:27.640 ","End":"04:32.390","Text":"Now we have the molarity of the sodium ion and the copper ion and we\u0027re going to go"},{"Start":"04:32.390 ","End":"04:37.500","Text":"on and measure the molarity of the sulfate anion."}],"ID":23705},{"Watched":false,"Name":"Exercise 2 - Part b","Duration":"3m 35s","ChapterTopicVideoID":22885,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22885.jpeg","UploadDate":"2020-12-15T06:21:25.2930000","DurationForVideoObject":"PT3M35S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:05.500","Text":"Now we\u0027re going to go on to measure the molarity of the sulfate anion."},{"Start":"00:12.980 ","End":"00:16.770","Text":"We\u0027re going to just call this c and we\u0027re going to do it in the same way that we"},{"Start":"00:16.770 ","End":"00:20.970","Text":"measured the sodium ion and the copper ion."},{"Start":"00:20.970 ","End":"00:24.345","Text":"But this time you can see that you have a sulfate ion,"},{"Start":"00:24.345 ","End":"00:28.155","Text":"also in sodium sulfate and also in copper sulfate."},{"Start":"00:28.155 ","End":"00:30.855","Text":"We\u0027re going to start with the sodium sulfate."},{"Start":"00:30.855 ","End":"00:34.680","Text":"It\u0027s 0.17 molar sodium sulfate,"},{"Start":"00:34.680 ","End":"00:42.180","Text":"meaning moles of sodium sulfate divided by liter of solution."},{"Start":"00:42.590 ","End":"00:46.010","Text":"We\u0027re going to multiply this by, again,"},{"Start":"00:46.010 ","End":"00:49.580","Text":"if we take a look at our sodium sulfate,"},{"Start":"00:49.580 ","End":"00:53.480","Text":"you can see that, assuming we have 1 mole of sodium sulfate,"},{"Start":"00:53.480 ","End":"00:58.100","Text":"we\u0027re going to have just 1 mole of"},{"Start":"00:58.100 ","End":"01:05.240","Text":"the sulfate anion since there\u0027s 1 sulfate in every molecule."},{"Start":"01:05.240 ","End":"01:10.760","Text":"We\u0027re going to multiply this by 1 mole of"},{"Start":"01:10.760 ","End":"01:18.750","Text":"the sulfate anion for every 1 mole of sodium sulfate."},{"Start":"01:22.340 ","End":"01:29.630","Text":"Now, we\u0027re going to take a look at the copper sulfate because we have"},{"Start":"01:29.630 ","End":"01:36.500","Text":"to add the molarity from the copper sulfate also."},{"Start":"01:36.500 ","End":"01:41.420","Text":"Here we have 0.23 molar copper sulfate,"},{"Start":"01:41.420 ","End":"01:48.035","Text":"meaning moles of copper sulfate for every liter of solution."},{"Start":"01:48.035 ","End":"01:50.270","Text":"We\u0027re going to multiply this by again,"},{"Start":"01:50.270 ","End":"01:53.975","Text":"if we look at our copper sulfate CuSO_4,"},{"Start":"01:53.975 ","End":"01:56.810","Text":"assuming we have 1 mole of copper sulfate,"},{"Start":"01:56.810 ","End":"02:01.400","Text":"we\u0027re going to have 1 mole of the sulfate anion"},{"Start":"02:01.400 ","End":"02:07.560","Text":"also since there\u0027s 1 sulfate in every molecule of copper sulfate."},{"Start":"02:07.660 ","End":"02:14.109","Text":"We\u0027re going to multiply this by 1 mole of sulfate anion"},{"Start":"02:14.109 ","End":"02:19.980","Text":"for every o1 mole of copper sulfate."},{"Start":"02:19.980 ","End":"02:27.180","Text":"The moles of copper sulfate are going to cancel out. We\u0027re going to take a look."},{"Start":"02:27.180 ","End":"02:37.160","Text":"We have 0.17 moles of the sulfate anion per liter"},{"Start":"02:37.160 ","End":"02:46.835","Text":"of solution plus 0.23"},{"Start":"02:46.835 ","End":"02:55.619","Text":"moles of the sulfate anion per liters of solution."},{"Start":"02:58.720 ","End":"03:03.360","Text":"When we sum this up, it equals 0.40"},{"Start":"03:03.740 ","End":"03:11.099","Text":"moles of the sulfate anion divided by liters of solution."},{"Start":"03:11.099 ","End":"03:21.870","Text":"This equals 0.40 molar sulfate anions."},{"Start":"03:21.870 ","End":"03:23.390","Text":"The molarity of the sulfate ion,"},{"Start":"03:23.390 ","End":"03:28.590","Text":"which we calculated equals 0.40 molar."},{"Start":"03:28.960 ","End":"03:33.410","Text":"That is the concentration that we found for the sulfate anion."},{"Start":"03:33.410 ","End":"03:36.270","Text":"Thank you very much for watching."}],"ID":23706},{"Watched":false,"Name":"Exercise 3","Duration":"5m 24s","ChapterTopicVideoID":22886,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22886.jpeg","UploadDate":"2020-12-15T06:21:47.9230000","DurationForVideoObject":"PT5M24S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:03.854","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.854 ","End":"00:07.485","Text":"What are the molarities of the following ions?"},{"Start":"00:07.485 ","End":"00:13.155","Text":"In a, we have 455 milligrams of the sodium ion per liter of solution."},{"Start":"00:13.155 ","End":"00:19.680","Text":"We\u0027re going to start with a. I want to remind you that molarity, M,"},{"Start":"00:19.680 ","End":"00:28.690","Text":"equals the number of moles of the solute divided by the volume of the solution."},{"Start":"00:30.380 ","End":"00:32.610","Text":"The number of moles,"},{"Start":"00:32.610 ","End":"00:38.435","Text":"n, equals the mass divided by the molar mass."},{"Start":"00:38.435 ","End":"00:43.970","Text":"In a, we\u0027re given the mass of the sodium ion and first of all,"},{"Start":"00:43.970 ","End":"00:47.060","Text":"we\u0027re going to calculate the number of moles."},{"Start":"00:47.060 ","End":"00:50.735","Text":"The number of moles of the sodium ion"},{"Start":"00:50.735 ","End":"00:57.710","Text":"equals the mass of the sodium ion divided by the molar mass of the sodium ion."},{"Start":"00:57.710 ","End":"01:00.230","Text":"The mass is given and it equals"},{"Start":"01:00.230 ","End":"01:09.830","Text":"455 milligrams divided by the molar mass,"},{"Start":"01:09.830 ","End":"01:16.920","Text":"which is given in the periodic table of elements and equals 23 grams per mole."},{"Start":"01:19.190 ","End":"01:23.450","Text":"Now since we have milligrams and grams we\u0027re going to multiply by"},{"Start":"01:23.450 ","End":"01:28.415","Text":"a conversion factor though we have grams in both cases."},{"Start":"01:28.415 ","End":"01:30.365","Text":"We have milligrams,"},{"Start":"01:30.365 ","End":"01:38.952","Text":"we\u0027re going to multiply by 1 gram per every 1,000 milligrams."},{"Start":"01:38.952 ","End":"01:41.580","Text":"This way the milligrams is just going to cancel"},{"Start":"01:41.580 ","End":"01:45.540","Text":"out and we\u0027re going to be left with the grams."},{"Start":"01:45.540 ","End":"01:53.975","Text":"This equals 0.02 mole."},{"Start":"01:53.975 ","End":"01:58.775","Text":"Now I just want you to take a look at the units for 1 second."},{"Start":"01:58.775 ","End":"02:03.450","Text":"You can see that we have grams divided by grams per mole."},{"Start":"02:09.700 ","End":"02:12.890","Text":"I want to remind you that when you\u0027re dividing by"},{"Start":"02:12.890 ","End":"02:18.520","Text":"a fraction is the same as multiplying by the reciprocal of the fraction,"},{"Start":"02:18.520 ","End":"02:22.920","Text":"so it\u0027s the same as grams times mole per gram."},{"Start":"02:22.920 ","End":"02:26.480","Text":"The grams cancel out and we\u0027re left with mole."},{"Start":"02:26.480 ","End":"02:32.100","Text":"The number of moles of the sodium ion equals 0.02 mole."},{"Start":"02:32.290 ","End":"02:36.850","Text":"Now we\u0027ll calculate the molarity of the sodium ion."},{"Start":"02:36.850 ","End":"02:38.030","Text":"The molarity, M,"},{"Start":"02:38.030 ","End":"02:42.335","Text":"equals the number of moles of the solute divided by the volume of the solution."},{"Start":"02:42.335 ","End":"02:51.380","Text":"The number of moles of the solute equals 0.02 mole divided by the volume of the solution,"},{"Start":"02:51.380 ","End":"02:57.748","Text":"which is given in the question, is 1 liter."},{"Start":"02:57.748 ","End":"03:03.590","Text":"Our answer is 0.02 molar."},{"Start":"03:03.590 ","End":"03:07.530","Text":"That\u0027s the molarity of the sodium ion."},{"Start":"03:07.540 ","End":"03:13.090","Text":"Now in b we have the iodide anion."},{"Start":"03:13.090 ","End":"03:18.005","Text":"Again, first we\u0027re going to calculate the number of moles,"},{"Start":"03:18.005 ","End":"03:26.915","Text":"the iodide equals the mass divided by the molar mass."},{"Start":"03:26.915 ","End":"03:35.120","Text":"The mass in this case is 22.8 milligrams divided by the molar mass,"},{"Start":"03:35.120 ","End":"03:42.980","Text":"which is taken from the periodic table of elements and equals 126.9 grams per mole."},{"Start":"03:42.980 ","End":"03:46.454","Text":"Now again, we\u0027re going to multiply by a conversion factor so"},{"Start":"03:46.454 ","End":"03:50.700","Text":"we\u0027re going to multiply it by 1 gram per 1,000 milligrams."},{"Start":"03:51.890 ","End":"03:57.820","Text":"Milligrams is going to cancel out and we\u0027ll be left with grams."},{"Start":"03:59.420 ","End":"04:08.545","Text":"This comes to 1.8 times 10 to the negative 4 mole."},{"Start":"04:08.545 ","End":"04:11.280","Text":"The next step is to calculate the molarity."},{"Start":"04:11.280 ","End":"04:16.504","Text":"The molarity equals the number of moles divided by the volume of the solution."},{"Start":"04:16.504 ","End":"04:26.270","Text":"The number of moles of the solute are 1.8 times 10 to the negative 4 mole divided by,"},{"Start":"04:26.270 ","End":"04:29.345","Text":"and this time we have 100 milliliters of solution,"},{"Start":"04:29.345 ","End":"04:34.235","Text":"so divided by 100 milliliters."},{"Start":"04:34.235 ","End":"04:36.180","Text":"In calculate molarity,"},{"Start":"04:36.180 ","End":"04:38.435","Text":"we need the volume in liters, therefore,"},{"Start":"04:38.435 ","End":"04:43.150","Text":"I\u0027m going to multiply by conversion factor so we\u0027re going to multiply by 1 liter"},{"Start":"04:43.150 ","End":"04:49.050","Text":"divide by 1,000 milliliters,"},{"Start":"04:49.050 ","End":"04:52.140","Text":"the milliliters is going to cancel out."},{"Start":"04:52.910 ","End":"05:03.659","Text":"This equals 1.8 times 10 to the negative 4 mole divided by 0.1"},{"Start":"05:03.659 ","End":"05:09.270","Text":"liters and this equals 1.8 times 10 to"},{"Start":"05:09.270 ","End":"05:15.975","Text":"the negative 3 mole per liter and mole per liter equals molar."},{"Start":"05:15.975 ","End":"05:20.880","Text":"Our answer for b is 1.8 times 10 to the negative 3 molar."},{"Start":"05:20.880 ","End":"05:23.120","Text":"That is our final answer."},{"Start":"05:23.120 ","End":"05:25.650","Text":"Thank you very much for watching."}],"ID":23707},{"Watched":false,"Name":"Exercise 4","Duration":"4m 37s","ChapterTopicVideoID":22887,"CourseChapterTopicPlaylistID":86819,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22887.jpeg","UploadDate":"2020-12-15T06:22:01.7600000","DurationForVideoObject":"PT4M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.999","Text":"We\u0027re going to solve the following exercise."},{"Start":"00:02.999 ","End":"00:06.030","Text":"What molarity of magnesium iodide corresponds to"},{"Start":"00:06.030 ","End":"00:11.055","Text":"an iodine content of 16 milligrams of iodide per 1 liter of solution?"},{"Start":"00:11.055 ","End":"00:15.285","Text":"First, we\u0027re going to calculate the molarity of the iodide ion."},{"Start":"00:15.285 ","End":"00:18.921","Text":"The molarity, M,"},{"Start":"00:18.921 ","End":"00:28.290","Text":"equals the number of moles of solute per volume of solution in liters."},{"Start":"00:28.290 ","End":"00:30.045","Text":"The number of moles,"},{"Start":"00:30.045 ","End":"00:34.245","Text":"N equals the mass divided by the molar mass."},{"Start":"00:34.245 ","End":"00:38.145","Text":"In order to calculate the molarity of the iodide,"},{"Start":"00:38.145 ","End":"00:40.470","Text":"we\u0027re going to begin with calculating the number of moles."},{"Start":"00:40.470 ","End":"00:48.395","Text":"The number of moles of the iodide ion equals the mass divided by the molar mass."},{"Start":"00:48.395 ","End":"00:54.920","Text":"The mass is given, we know it\u0027s 16 milligrams divided by the molar mass,"},{"Start":"00:54.920 ","End":"01:02.735","Text":"which is given in the periodic table of elements N=126.9 grams per mole."},{"Start":"01:02.735 ","End":"01:05.193","Text":"Since we have milligrams and grams here,"},{"Start":"01:05.193 ","End":"01:08.100","Text":"we\u0027re going to multiply by a conversion factor."},{"Start":"01:08.890 ","End":"01:15.120","Text":"You\u0027re going to multiply it by 1 gram per 1,000 milligrams."},{"Start":"01:15.160 ","End":"01:18.305","Text":"Milligrams cancels out,"},{"Start":"01:18.305 ","End":"01:21.065","Text":"and we\u0027re left with grams."},{"Start":"01:21.065 ","End":"01:31.760","Text":"After dividing, we get 1.26 times 10 to the negative 4 mole."},{"Start":"01:31.760 ","End":"01:35.090","Text":"I just want to remind you that if we look at our units,"},{"Start":"01:35.090 ","End":"01:39.335","Text":"we\u0027re dividing grams by grams per mole."},{"Start":"01:39.335 ","End":"01:41.750","Text":"Grams per mole is a fraction."},{"Start":"01:41.750 ","End":"01:43.250","Text":"When you divide by a fraction,"},{"Start":"01:43.250 ","End":"01:46.820","Text":"it\u0027s the same as multiplying by the reciprocal of the fraction."},{"Start":"01:46.820 ","End":"01:52.700","Text":"It\u0027s the same as grams times mole per gram."},{"Start":"01:52.700 ","End":"01:57.770","Text":"The grams cancel out and we\u0027re left with the units mole."},{"Start":"01:57.770 ","End":"02:02.750","Text":"We know the number of moles of the iodide ion."},{"Start":"02:02.750 ","End":"02:06.155","Text":"Next step is to calculate the molarity."},{"Start":"02:06.155 ","End":"02:09.470","Text":"Molarity again equals the number of moles of"},{"Start":"02:09.470 ","End":"02:12.590","Text":"the solute divided by the volume of the solution in liters."},{"Start":"02:12.590 ","End":"02:15.410","Text":"In our case, the number of moles is what we calculated."},{"Start":"02:15.410 ","End":"02:23.780","Text":"It\u0027s 1.26 times 10 to the negative 4 mole divided by the volume of the solution,"},{"Start":"02:23.780 ","End":"02:26.490","Text":"which is given N equals 1 liter."},{"Start":"02:27.730 ","End":"02:35.340","Text":"This comes to 1.26 times 10 to the negative 4 molar."},{"Start":"02:35.340 ","End":"02:40.540","Text":"That\u0027s the molarity of the iodide ion."},{"Start":"02:40.540 ","End":"02:44.930","Text":"Now the next step is to calculate the molarity of the magnesium iodide."},{"Start":"02:45.600 ","End":"02:55.970","Text":"The molarity of the magnesium iodide equals the molarity of the iodide ion"},{"Start":"02:56.460 ","End":"03:01.120","Text":"times 1 mole of"},{"Start":"03:01.120 ","End":"03:11.120","Text":"magnesium iodide per 2 moles of iodide ion."},{"Start":"03:12.570 ","End":"03:16.630","Text":"Because you will see in 1 mole of magnesium iodide,"},{"Start":"03:16.630 ","End":"03:21.680","Text":"we have 2 moles of the iodide ion."},{"Start":"03:21.840 ","End":"03:26.755","Text":"This equals 1.26 times 10 to the negative 4 molar,"},{"Start":"03:26.755 ","End":"03:29.810","Text":"which is the molarity of the iodide."},{"Start":"03:32.610 ","End":"03:36.880","Text":"I want to remind you that molar is mole per liter."},{"Start":"03:36.880 ","End":"03:42.620","Text":"It\u0027s the number of moles of the iodide per liter of solution."},{"Start":"03:44.340 ","End":"03:47.820","Text":"We multiply this by 1 mole of"},{"Start":"03:47.820 ","End":"03:57.390","Text":"magnesium iodide divided by 2 moles iodide."},{"Start":"03:57.390 ","End":"04:01.110","Text":"The moles of iodide cancel out."},{"Start":"04:01.180 ","End":"04:06.500","Text":"This equals 6.3 times 10 to"},{"Start":"04:06.500 ","End":"04:10.520","Text":"the negative 5 moles of"},{"Start":"04:10.520 ","End":"04:18.830","Text":"magnesium iodide per liter of solution."},{"Start":"04:18.830 ","End":"04:27.170","Text":"This equals 6.3 times 10 to the negative 5 molar."},{"Start":"04:27.170 ","End":"04:32.632","Text":"This is the molarity that we calculated,"},{"Start":"04:32.632 ","End":"04:34.955","Text":"and that\u0027s our final answer."},{"Start":"04:34.955 ","End":"04:38.040","Text":"Thank you very much for watching."}],"ID":23708}],"Thumbnail":null,"ID":86819},{"Name":"Precipitation Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Solubility Rules","Duration":"10m 41s","ChapterTopicVideoID":16921,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16921.jpeg","UploadDate":"2019-02-20T23:53:44.5270000","DurationForVideoObject":"PT10M41S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"In this video, we will discuss some rules which allow us to"},{"Start":"00:04.830 ","End":"00:09.675","Text":"predict whether a metal salt is soluble or insoluble."},{"Start":"00:09.675 ","End":"00:12.600","Text":"These are called solubility rules."},{"Start":"00:12.600 ","End":"00:18.180","Text":"The first thing to note is that if there is a conflict between 2 rules,"},{"Start":"00:18.180 ","End":"00:20.580","Text":"the one with the lower number prevails."},{"Start":"00:20.580 ","End":"00:25.860","Text":"For example, if 1 and 3 are at conflict,"},{"Start":"00:25.860 ","End":"00:28.170","Text":"then 1 will prevail."},{"Start":"00:28.170 ","End":"00:34.020","Text":"The first rule is that salts containing group 1 cations, that\u0027s lithium, sodium,"},{"Start":"00:34.020 ","End":"00:38.340","Text":"potassium, cesium, rubidium, and ammonium cation,"},{"Start":"00:38.340 ","End":"00:41.625","Text":"that\u0027s NH_4 plus are soluble."},{"Start":"00:41.625 ","End":"00:45.230","Text":"There are some lithium salts such as slightly soluble,"},{"Start":"00:45.230 ","End":"00:47.105","Text":"so they are exceptions."},{"Start":"00:47.105 ","End":"00:51.650","Text":"Rule 2, salts contain the nitrate anion,"},{"Start":"00:51.650 ","End":"00:54.955","Text":"that\u0027s NO_3 minus are generally soluble."},{"Start":"00:54.955 ","End":"00:58.440","Text":"3, salts containing halide anions,"},{"Start":"00:58.440 ","End":"01:03.950","Text":"that\u0027s Cl minus Br minus and I minus are generally soluble."},{"Start":"01:03.950 ","End":"01:10.085","Text":"Exceptions are halide salts containing silver, lead."},{"Start":"01:10.085 ","End":"01:11.779","Text":"This is a bit difficult."},{"Start":"01:11.779 ","End":"01:17.115","Text":"This is mercury 2 plus and mercury plus,"},{"Start":"01:17.115 ","End":"01:20.430","Text":"all these are insoluble."},{"Start":"01:20.430 ","End":"01:24.510","Text":"4, most salts containing silver are insoluble."},{"Start":"01:24.510 ","End":"01:26.360","Text":"There are exceptions."},{"Start":"01:26.360 ","End":"01:29.495","Text":"Silver nitrate, that\u0027s AgNO_3,"},{"Start":"01:29.495 ","End":"01:32.120","Text":"and silver acetate,"},{"Start":"01:32.120 ","End":"01:37.490","Text":"AgCH_3CO_2 are soluble, so these 2 are soluble."},{"Start":"01:37.490 ","End":"01:42.470","Text":"5, most salts continue the sulfate anion,"},{"Start":"01:42.470 ","End":"01:45.260","Text":"SO_4^2 minus are soluble."},{"Start":"01:45.260 ","End":"01:46.940","Text":"There are exceptions."},{"Start":"01:46.940 ","End":"01:49.879","Text":"Barium sulfate, silver sulfate,"},{"Start":"01:49.879 ","End":"01:53.575","Text":"and strontium sulfate are all insoluble."},{"Start":"01:53.575 ","End":"01:57.780","Text":"6, most salts containing the hydroxide ion,"},{"Start":"01:57.780 ","End":"01:59.245","Text":"that\u0027s OH minus,"},{"Start":"01:59.245 ","End":"02:01.700","Text":"are only slightly soluble."},{"Start":"02:01.700 ","End":"02:05.325","Text":"Hydroxide salts of group 1 are soluble."},{"Start":"02:05.325 ","End":"02:09.340","Text":"Well, that we saw in the very first row."},{"Start":"02:09.340 ","End":"02:12.860","Text":"Hydroxide salts of group 2 cations,"},{"Start":"02:12.860 ","End":"02:17.420","Text":"that\u0027s calcium, strontium and barium, are slightly soluble."},{"Start":"02:17.420 ","End":"02:20.525","Text":"Hydroxides salts of transition elements."},{"Start":"02:20.525 ","End":"02:24.545","Text":"That\u0027s the central block we saw in the periodic table."},{"Start":"02:24.545 ","End":"02:29.525","Text":"Things like iron, copper, and so on."},{"Start":"02:29.525 ","End":"02:34.265","Text":"Also aluminum 3 plus are insoluble."},{"Start":"02:34.265 ","End":"02:37.460","Text":"Aluminum hydroxides are insoluble."},{"Start":"02:37.460 ","End":"02:43.895","Text":"7, most salts contain the sulfide ion are insoluble."},{"Start":"02:43.895 ","End":"02:48.275","Text":"Sulfides of group 2 cations are slightly soluble."},{"Start":"02:48.275 ","End":"02:52.435","Text":"Barium sulfide is slightly soluble."},{"Start":"02:52.435 ","End":"02:58.535","Text":"Most salts containing the carbonate ion are insoluble."},{"Start":"02:58.535 ","End":"03:05.215","Text":"CO_3^2 minus, so things like calcium carbonate, chalk, are insoluble."},{"Start":"03:05.215 ","End":"03:15.085","Text":"9, more salts containing the chromate ion that CRO_4^2 minus are insoluble."},{"Start":"03:15.085 ","End":"03:18.600","Text":"There are 2 more. 10,"},{"Start":"03:18.600 ","End":"03:21.620","Text":"most salts contain the phosphate ion,"},{"Start":"03:21.620 ","End":"03:25.490","Text":"that\u0027s PO_4^3 minus are insoluble."},{"Start":"03:25.490 ","End":"03:28.215","Text":"The final 11,"},{"Start":"03:28.215 ","End":"03:34.370","Text":"most salts containing the fluoride ion F minus are insoluble."},{"Start":"03:34.370 ","End":"03:37.130","Text":"Here are 11 rules."},{"Start":"03:37.130 ","End":"03:42.949","Text":"We can use them to predict where the precipitation reactions will occur."},{"Start":"03:42.949 ","End":"03:45.520","Text":"We\u0027re going to take 2 examples."},{"Start":"03:45.520 ","End":"03:47.185","Text":"Here\u0027s the first 1."},{"Start":"03:47.185 ","End":"03:54.020","Text":"Sodium phosphate and aluminum chloride are put in water and will be dissolved?"},{"Start":"03:54.020 ","End":"03:58.330","Text":"That\u0027s the first thing to decide whether these substances will dissolve."},{"Start":"03:58.330 ","End":"04:04.370","Text":"Rule number 1 told us that something continue sodium would dissolve."},{"Start":"04:04.370 ","End":"04:06.065","Text":"That was rule number 1."},{"Start":"04:06.065 ","End":"04:09.260","Text":"That is more important than rule number 10,"},{"Start":"04:09.260 ","End":"04:17.100","Text":"which told us that salts containing phosphate would not dissolve, would be insoluble."},{"Start":"04:17.100 ","End":"04:20.915","Text":"Rule number 1 overcomes rule number 10,"},{"Start":"04:20.915 ","End":"04:24.035","Text":"so sodium phosphate is soluble."},{"Start":"04:24.035 ","End":"04:27.250","Text":"What about aluminum chloride?"},{"Start":"04:27.250 ","End":"04:35.390","Text":"Well rule number 3 told us that salts containing chloride ion where soluble."},{"Start":"04:35.390 ","End":"04:37.970","Text":"Aluminum chloride is soluble."},{"Start":"04:37.970 ","End":"04:40.780","Text":"Let\u0027s write out all the ions."},{"Start":"04:40.780 ","End":"04:51.175","Text":"Sodium phosphate dissociates to sodium and 3-Na plus and PO_4^3 minus."},{"Start":"04:51.175 ","End":"04:58.190","Text":"Aluminum chloride dissociates to Al^3 plus and 3Cl minus."},{"Start":"04:58.190 ","End":"05:03.530","Text":"Now we have to look and see what happens when sodium,"},{"Start":"05:03.530 ","End":"05:07.570","Text":"instead of going with phosphate, goes with chloride."},{"Start":"05:07.570 ","End":"05:13.445","Text":"Aluminum instead of going with chloride goes with phosphate."},{"Start":"05:13.445 ","End":"05:15.980","Text":"Let\u0027s look at these 2 things."},{"Start":"05:15.980 ","End":"05:19.565","Text":"First of all, sodium and chloride."},{"Start":"05:19.565 ","End":"05:22.190","Text":"Well, we all know that that is soluble,"},{"Start":"05:22.190 ","End":"05:28.965","Text":"but the rules 1 and 3 tell us that this is true."},{"Start":"05:28.965 ","End":"05:32.095","Text":"Rule 1 is related to sodium."},{"Start":"05:32.095 ","End":"05:34.780","Text":"Salts containing sodium are soluble."},{"Start":"05:34.780 ","End":"05:40.495","Text":"Rule 3 tells us that salts containing chloride are soluble."},{"Start":"05:40.495 ","End":"05:44.245","Text":"Sodium and chloride stay in solution."},{"Start":"05:44.245 ","End":"05:47.439","Text":"Here they are staying in solution."},{"Start":"05:47.439 ","End":"05:50.140","Text":"What about the other combination?"},{"Start":"05:50.140 ","End":"05:56.270","Text":"The other combination is aluminum and phosphate."},{"Start":"05:56.970 ","End":"06:00.640","Text":"Is that soluble or not?"},{"Start":"06:00.640 ","End":"06:02.770","Text":"Well, if we look at rule number 10,"},{"Start":"06:02.770 ","End":"06:08.120","Text":"it says more salts containing the phosphate ion are insoluble."},{"Start":"06:08.120 ","End":"06:16.390","Text":"Rule number 10 tells us that aluminum phosphate is insoluble, AlPO_4."},{"Start":"06:16.390 ","End":"06:19.920","Text":"Here it is, AlPO_4."},{"Start":"06:19.920 ","End":"06:23.480","Text":"AlPO_4 is insoluble."},{"Start":"06:23.480 ","End":"06:25.565","Text":"It precipitate."},{"Start":"06:25.565 ","End":"06:28.920","Text":"Here only 1 substance precipitates,"},{"Start":"06:28.920 ","End":"06:31.775","Text":"and 1 substance is still in solution."},{"Start":"06:31.775 ","End":"06:37.130","Text":"Now let\u0027s look and see whether we have any spectator ions."},{"Start":"06:37.130 ","End":"06:40.310","Text":"We see sodium on the left-hand side,"},{"Start":"06:40.310 ","End":"06:42.350","Text":"sodium the left-hand side,"},{"Start":"06:42.350 ","End":"06:44.030","Text":"and sodium on the right-hand side."},{"Start":"06:44.030 ","End":"06:47.420","Text":"These cancel, these are spectator ions."},{"Start":"06:47.420 ","End":"06:51.975","Text":"We see chloride on the left and chloride on the right."},{"Start":"06:51.975 ","End":"06:54.860","Text":"The chloride is also spectator."},{"Start":"06:54.860 ","End":"07:03.009","Text":"Now we can write the net ionic equation."},{"Start":"07:03.009 ","End":"07:05.625","Text":"It\u0027s Al^3 plus,"},{"Start":"07:05.625 ","End":"07:09.150","Text":"plus PO_4^3 minus that\u0027s phosphate,"},{"Start":"07:09.150 ","End":"07:12.820","Text":"giving us AlPO_4 solid,"},{"Start":"07:12.820 ","End":"07:15.305","Text":"that\u0027s something that precipitate."},{"Start":"07:15.305 ","End":"07:18.380","Text":"That\u0027s our net ionic equation."},{"Start":"07:18.380 ","End":"07:20.660","Text":"Let\u0027s look at another example."},{"Start":"07:20.660 ","End":"07:25.775","Text":"This is barium sulfide and copper sulfate."},{"Start":"07:25.775 ","End":"07:31.295","Text":"The first thing to decide is whether barium sulfide is soluble."},{"Start":"07:31.295 ","End":"07:33.049","Text":"Now if we look at the rules,"},{"Start":"07:33.049 ","End":"07:36.905","Text":"we\u0027ll see that rule number 7 says"},{"Start":"07:36.905 ","End":"07:41.170","Text":"that most salts containing the sulfide ion are insoluble."},{"Start":"07:41.170 ","End":"07:48.800","Text":"However, sulfides of group 2 cations and bariums in group 2 are slightly soluble."},{"Start":"07:48.800 ","End":"07:52.160","Text":"Barium sulfide is slightly soluble."},{"Start":"07:52.160 ","End":"07:56.060","Text":"The rule that told us that was rule number 7."},{"Start":"07:56.060 ","End":"07:58.025","Text":"What about copper sulfate?"},{"Start":"07:58.025 ","End":"08:00.540","Text":"Which rule tells us about that?"},{"Start":"08:00.540 ","End":"08:02.870","Text":"Well, if we look carefully,"},{"Start":"08:02.870 ","End":"08:11.705","Text":"we\u0027ll see that rule number 5 tells us most salts continued to sulfate ion are soluble."},{"Start":"08:11.705 ","End":"08:15.485","Text":"This is rule number 5 tells us soluble."},{"Start":"08:15.485 ","End":"08:17.554","Text":"Let\u0027s look at the ions."},{"Start":"08:17.554 ","End":"08:22.805","Text":"Barium sulfide goes to barium ion and sulfide ion."},{"Start":"08:22.805 ","End":"08:27.845","Text":"Copper sulfate goes to copper ion and sulfate ion."},{"Start":"08:27.845 ","End":"08:30.799","Text":"These are all our ions."},{"Start":"08:30.799 ","End":"08:36.845","Text":"Now we have to decide whether barium and sulfate,"},{"Start":"08:36.845 ","End":"08:39.745","Text":"that\u0027s barium sulfate,"},{"Start":"08:39.745 ","End":"08:43.050","Text":"whether that is soluble or insoluble."},{"Start":"08:43.050 ","End":"08:48.350","Text":"The other combinations copper and sulfide, giving a CuS."},{"Start":"08:48.350 ","End":"08:51.365","Text":"Is that soluble or insoluble?"},{"Start":"08:51.365 ","End":"08:56.705","Text":"Let\u0027s look at this, barium and sulfate."},{"Start":"08:56.705 ","End":"09:00.900","Text":"What rule tells us about sulfates?"},{"Start":"09:00.900 ","End":"09:04.220","Text":"We have rule number 5."},{"Start":"09:04.220 ","End":"09:06.470","Text":"Rule number 5 says,"},{"Start":"09:06.470 ","End":"09:09.545","Text":"the more salts containing sulfates are soluble,"},{"Start":"09:09.545 ","End":"09:13.055","Text":"but an exception is barium sulfate."},{"Start":"09:13.055 ","End":"09:16.850","Text":"Barium and sulfate will be insoluble."},{"Start":"09:16.850 ","End":"09:21.290","Text":"Here we have barium sulfate forming a solid."},{"Start":"09:21.290 ","End":"09:25.920","Text":"Now we need copper and sulfide."},{"Start":"09:26.800 ","End":"09:29.980","Text":"What about copper and sulfide?"},{"Start":"09:29.980 ","End":"09:32.285","Text":"If we look at the rules,"},{"Start":"09:32.285 ","End":"09:40.865","Text":"we will see that rule number 7 says more salts containing the sulfide ion are insoluble."},{"Start":"09:40.865 ","End":"09:43.405","Text":"That\u0027s rule number 7,"},{"Start":"09:43.405 ","End":"09:51.030","Text":"telling us that copper sulfide is insoluble."},{"Start":"09:51.030 ","End":"09:58.285","Text":"Here we have not 1 substance precipitating, but 2 precipitating."},{"Start":"09:58.285 ","End":"10:02.075","Text":"We have barium sulfate and copper sulfide."},{"Start":"10:02.075 ","End":"10:05.000","Text":"We have no spectator ions."},{"Start":"10:05.000 ","End":"10:10.370","Text":"There are no ions that appear on the left and on the right."},{"Start":"10:10.370 ","End":"10:17.865","Text":"This equation is not only the full ionic equation,"},{"Start":"10:17.865 ","End":"10:20.805","Text":"it\u0027s also the net ionic equation."},{"Start":"10:20.805 ","End":"10:24.685","Text":"This is a net ionic equation."},{"Start":"10:24.685 ","End":"10:28.295","Text":"Here, I\u0027ve written it out, no spectator ions."},{"Start":"10:28.295 ","End":"10:32.090","Text":"Final equation is also net ionic equation."},{"Start":"10:32.090 ","End":"10:35.930","Text":"In this video, we discussed the solubility rules and used"},{"Start":"10:35.930 ","End":"10:40.830","Text":"them to predict where the precipitation reactions would occur."}],"ID":17674},{"Watched":false,"Name":"Precipitation Reactions and Net Ionic Equations","Duration":"4m 22s","ChapterTopicVideoID":16922,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16922.jpeg","UploadDate":"2019-02-20T23:54:27.3970000","DurationForVideoObject":"PT4M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"In this video, we will discuss"},{"Start":"00:03.030 ","End":"00:07.740","Text":"precipitation reactions between metal salts dissolved in water."},{"Start":"00:07.740 ","End":"00:10.425","Text":"Some metal salts are soluble in water."},{"Start":"00:10.425 ","End":"00:13.530","Text":"They undergo complete ionization of water."},{"Start":"00:13.530 ","End":"00:16.935","Text":"In other words, they turn totally into ions."},{"Start":"00:16.935 ","End":"00:21.990","Text":"Here\u0027s some examples, sodium chloride and silver nitrate."},{"Start":"00:21.990 ","End":"00:24.405","Text":"Both of these are soluble in water."},{"Start":"00:24.405 ","End":"00:26.070","Text":"Sodium chloride is of course,"},{"Start":"00:26.070 ","End":"00:29.295","Text":"ordinary salt, and we all know that soluble in water."},{"Start":"00:29.295 ","End":"00:30.900","Text":"Let\u0027s write the equation."},{"Start":"00:30.900 ","End":"00:33.215","Text":"Sodium chloride put in water,"},{"Start":"00:33.215 ","End":"00:38.055","Text":"gives us Na^+ ions and Cl minus ions."},{"Start":"00:38.055 ","End":"00:46.505","Text":"Silver nitrate and water gives us Ag^+ ions and NO_3 minus ions."},{"Start":"00:46.505 ","End":"00:49.820","Text":"However, not all metal salts are soluble in water."},{"Start":"00:49.820 ","End":"00:52.295","Text":"Some are insoluble."},{"Start":"00:52.295 ","End":"00:57.485","Text":"Some metal salts are insoluble or almost insoluble in water."},{"Start":"00:57.485 ","End":"01:01.225","Text":"The stay as solids in the bottom of the beaker."},{"Start":"01:01.225 ","End":"01:06.195","Text":"Examples of silver iodide and calcium carbonate."},{"Start":"01:06.195 ","End":"01:08.330","Text":"Calcium carbonate is chalk."},{"Start":"01:08.330 ","End":"01:11.420","Text":"We all know chalk is insoluble in water."},{"Start":"01:11.420 ","End":"01:14.630","Text":"Now let\u0027s talk about precipitation reactions."},{"Start":"01:14.630 ","End":"01:20.810","Text":"These are reactions in which 1 or more solids dropped to the bottom of the beaker."},{"Start":"01:20.810 ","End":"01:22.459","Text":"Let\u0027s take an example."},{"Start":"01:22.459 ","End":"01:25.070","Text":"I\u0027ve deliberately taken the example of"},{"Start":"01:25.070 ","End":"01:29.480","Text":"the 2 substances we talked about before that were soluble in water,"},{"Start":"01:29.480 ","End":"01:33.335","Text":"sodium chloride, and silver nitrate."},{"Start":"01:33.335 ","End":"01:39.635","Text":"Here we have sodium chloride and silver nitrate in water giving 4 ions;"},{"Start":"01:39.635 ","End":"01:43.520","Text":"sodium plus, Cl minus,"},{"Start":"01:43.520 ","End":"01:46.700","Text":"Ag^+ and NO_3 minus, 4 ions."},{"Start":"01:46.700 ","End":"01:53.810","Text":"Now supposing, instead of sodium going with chlorine it went with nitrate,"},{"Start":"01:53.810 ","End":"01:58.625","Text":"and silver instead of going with nitrate, went with chlorine."},{"Start":"01:58.625 ","End":"02:00.665","Text":"What would happen?"},{"Start":"02:00.665 ","End":"02:05.562","Text":"If we take sodium with nitrate ion,"},{"Start":"02:05.562 ","End":"02:12.515","Text":"nothing would happen because sodium nitrate is soluble in water."},{"Start":"02:12.515 ","End":"02:14.390","Text":"So nothing would happen."},{"Start":"02:14.390 ","End":"02:19.805","Text":"However, if we have chloride with Ag^+,"},{"Start":"02:19.805 ","End":"02:22.445","Text":"if we have these 2 ions together,"},{"Start":"02:22.445 ","End":"02:29.425","Text":"we will get a solid because silver chloride is insoluble in water."},{"Start":"02:29.425 ","End":"02:31.735","Text":"We can write that again."},{"Start":"02:31.735 ","End":"02:38.005","Text":"There are 4 ions and we get silver chloride as a solid precipitate,"},{"Start":"02:38.005 ","End":"02:40.405","Text":"dropping to the bottom of the beaker."},{"Start":"02:40.405 ","End":"02:45.880","Text":"Sodium plus and nitrate ions left in solution."},{"Start":"02:45.880 ","End":"02:50.680","Text":"These stay in solution and this precipitate."},{"Start":"02:50.680 ","End":"02:55.870","Text":"We can say that AgCl will precipitate."},{"Start":"02:55.870 ","End":"02:59.095","Text":"Here\u0027s a picture illustrating it."},{"Start":"02:59.095 ","End":"03:02.350","Text":"We have 1 substance that\u0027s soluble in water."},{"Start":"03:02.350 ","End":"03:05.200","Text":"We add another substance soluble in water."},{"Start":"03:05.200 ","End":"03:07.090","Text":"When the 2 are mixed,"},{"Start":"03:07.090 ","End":"03:10.160","Text":"we get a solid precipitated."},{"Start":"03:10.160 ","End":"03:14.320","Text":"We get some liquid in which they are still ions."},{"Start":"03:14.320 ","End":"03:17.020","Text":"Now let\u0027s look at the equation again."},{"Start":"03:17.020 ","End":"03:22.895","Text":"We see that sodium plus appears on the left-hand side and also on the right hand side,"},{"Start":"03:22.895 ","End":"03:24.900","Text":"so we can cancel."},{"Start":"03:24.900 ","End":"03:28.300","Text":"We\u0027ll see that NO_3 minus appears in"},{"Start":"03:28.300 ","End":"03:32.740","Text":"the left hand side and also on the right-hand side so we can cancel them."},{"Start":"03:32.740 ","End":"03:36.370","Text":"These are called spectator ions."},{"Start":"03:36.370 ","End":"03:38.125","Text":"They have to be there,"},{"Start":"03:38.125 ","End":"03:42.760","Text":"but they don\u0027t actively precipitate in the reaction."},{"Start":"03:42.760 ","End":"03:44.500","Text":"They just spectators."},{"Start":"03:44.500 ","End":"03:46.285","Text":"They just look at what\u0027s happening."},{"Start":"03:46.285 ","End":"03:50.405","Text":"Now we can write the net ionic equation."},{"Start":"03:50.405 ","End":"03:55.310","Text":"That\u0027s the equation without the spectator ions."},{"Start":"03:55.310 ","End":"04:00.155","Text":"It\u0027s Ag^+, Cl minus giving AgCl,"},{"Start":"04:00.155 ","End":"04:01.625","Text":"which is a solid."},{"Start":"04:01.625 ","End":"04:04.235","Text":"Sometimes we write it like this."},{"Start":"04:04.235 ","End":"04:08.660","Text":"An arrow down the way indicating precipitation."},{"Start":"04:08.660 ","End":"04:10.340","Text":"In this video, we talked about"},{"Start":"04:10.340 ","End":"04:15.485","Text":"precipitation reactions and how to write them as net ionic equations."},{"Start":"04:15.485 ","End":"04:16.880","Text":"In the next video,"},{"Start":"04:16.880 ","End":"04:22.020","Text":"we will learn to predict whether such reactions will occur."}],"ID":17675},{"Watched":false,"Name":"Exercise 5","Duration":"6m 53s","ChapterTopicVideoID":22889,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22889.jpeg","UploadDate":"2020-12-15T06:37:23.0600000","DurationForVideoObject":"PT6M53S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23710},{"Watched":false,"Name":"Exercise 5 - Part a","Duration":"3m 29s","ChapterTopicVideoID":22890,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22890.jpeg","UploadDate":"2020-12-15T06:37:32.3500000","DurationForVideoObject":"PT3M29S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23711},{"Watched":false,"Name":"Exercise 5 - Part b","Duration":"3m 7s","ChapterTopicVideoID":22891,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22891.jpeg","UploadDate":"2020-12-15T06:37:41.3100000","DurationForVideoObject":"PT3M7S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23712},{"Watched":false,"Name":"Exercise 5 - Part c","Duration":"2m 38s","ChapterTopicVideoID":22892,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22892.jpeg","UploadDate":"2020-12-15T06:37:48.6500000","DurationForVideoObject":"PT2M38S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23713},{"Watched":false,"Name":"Exercise 6 - Part a","Duration":"3m 28s","ChapterTopicVideoID":22893,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22893.jpeg","UploadDate":"2020-12-15T06:37:58.1630000","DurationForVideoObject":"PT3M28S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23714},{"Watched":false,"Name":"Exercise 6 - Part b","Duration":"3m 1s","ChapterTopicVideoID":22894,"CourseChapterTopicPlaylistID":86820,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22894.jpeg","UploadDate":"2020-12-15T06:38:07.1830000","DurationForVideoObject":"PT3M1S","Description":null,"VideoComments":[],"Subtitles":[],"ID":23715}],"Thumbnail":null,"ID":86820},{"Name":"Acid-Base Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Acids and Bases","Duration":"8m 38s","ChapterTopicVideoID":16929,"CourseChapterTopicPlaylistID":86821,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/16929.jpeg","UploadDate":"2019-02-20T23:58:46.7600000","DurationForVideoObject":"PT8M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.035","Text":"In this video, we\u0027ll talk about acids and bases."},{"Start":"00:04.035 ","End":"00:09.015","Text":"Let\u0027s first consider some examples of acids and bases that you\u0027re familiar with."},{"Start":"00:09.015 ","End":"00:11.715","Text":"Here are some examples of acidic solutions."},{"Start":"00:11.715 ","End":"00:12.990","Text":"You can have vinegar,"},{"Start":"00:12.990 ","End":"00:16.140","Text":"lemon juice, that\u0027s pretty obvious, urine,"},{"Start":"00:16.140 ","End":"00:22.020","Text":"if you do a test at the doctors and measure how acidic your urine is,"},{"Start":"00:22.020 ","End":"00:24.085","Text":"beer, and milk."},{"Start":"00:24.085 ","End":"00:28.130","Text":"Bases; soap, seawater, and blood."},{"Start":"00:28.130 ","End":"00:29.825","Text":"Of course, there are many more,"},{"Start":"00:29.825 ","End":"00:33.155","Text":"but these are examples we\u0027ll take at the moment."},{"Start":"00:33.155 ","End":"00:36.275","Text":"Now, there are a few definitions of an acid,"},{"Start":"00:36.275 ","End":"00:38.585","Text":"and in this video, we\u0027ll talk about 2 of them."},{"Start":"00:38.585 ","End":"00:40.745","Text":"The first one is the oldest one."},{"Start":"00:40.745 ","End":"00:43.190","Text":"The Arrhenius definition of an acid."},{"Start":"00:43.190 ","End":"00:45.480","Text":"It\u0027s really quite old, 1884."},{"Start":"00:45.580 ","End":"00:49.430","Text":"Arrhenius said that an acid is a substance that"},{"Start":"00:49.430 ","End":"00:53.555","Text":"provides H plus ions in an aqueous solution."},{"Start":"00:53.555 ","End":"00:57.184","Text":"There are strong acids and weak acids."},{"Start":"00:57.184 ","End":"01:01.760","Text":"Before I go on, I want to talk about H plus aqueous."},{"Start":"01:01.760 ","End":"01:04.055","Text":"H plus ion in water."},{"Start":"01:04.055 ","End":"01:07.250","Text":"This ion is solvated or hydrated by water."},{"Start":"01:07.250 ","End":"01:11.045","Text":"In other words, it\u0027s surrounded by a shell of water molecules."},{"Start":"01:11.045 ","End":"01:17.345","Text":"Let\u0027s draw it. Here\u0027s H plus and it\u0027s surrounded by water molecules."},{"Start":"01:17.345 ","End":"01:19.840","Text":"I\u0027ll just draw 2 of them."},{"Start":"01:19.840 ","End":"01:24.425","Text":"We often write it as if we\u0027re surrounded just by 1."},{"Start":"01:24.425 ","End":"01:29.270","Text":"We take these H plus and 1 water molecule and"},{"Start":"01:29.270 ","End":"01:37.440","Text":"write H_3O plus and that\u0027s called a hydronium ion."},{"Start":"01:37.440 ","End":"01:41.490","Text":"We\u0027ll often use that instead of H plus."},{"Start":"01:41.490 ","End":"01:44.420","Text":"Now, let\u0027s first consider strong acids."},{"Start":"01:44.420 ","End":"01:48.785","Text":"A strong acid completely dissociates into ions."},{"Start":"01:48.785 ","End":"01:51.485","Text":"It\u0027s also a strong electrolyte."},{"Start":"01:51.485 ","End":"01:55.160","Text":"Remember we talked about electrolytes in the previous video."},{"Start":"01:55.160 ","End":"01:58.420","Text":"Here\u0027s an example, hydrochloric acid."},{"Start":"01:58.420 ","End":"02:02.150","Text":"HCl, that\u0027s hydrochloric acid, in water,"},{"Start":"02:02.150 ","End":"02:11.360","Text":"dissociates into H plus and Cl minus and H plus is solvated or hydrated."},{"Start":"02:11.360 ","End":"02:16.370","Text":"Another way of writing the same thing and one will need later on in this video is"},{"Start":"02:16.370 ","End":"02:23.120","Text":"HCl plus a water molecule giving us H_3O plus and Cl minus."},{"Start":"02:23.120 ","End":"02:29.215","Text":"HCl in water dissociates into H_3O plus and Cl minus."},{"Start":"02:29.215 ","End":"02:31.365","Text":"There are quite a few strong acids,"},{"Start":"02:31.365 ","End":"02:33.480","Text":"for example, sulfuric acid,"},{"Start":"02:33.480 ","End":"02:36.240","Text":"H_2SO_4 and nitric acid,"},{"Start":"02:36.240 ","End":"02:39.170","Text":"HNO_3, and there are others."},{"Start":"02:39.170 ","End":"02:41.119","Text":"What about weak acids?"},{"Start":"02:41.119 ","End":"02:45.335","Text":"A weak acid only partially dissociates into ions,"},{"Start":"02:45.335 ","End":"02:48.275","Text":"so it is also a weak electrolyte."},{"Start":"02:48.275 ","End":"02:53.750","Text":"Here\u0027s our favorite example, acetic acid, CH_3CO_2H."},{"Start":"02:53.750 ","End":"03:03.635","Text":"It dissociates into acetate and H plus."},{"Start":"03:03.635 ","End":"03:12.489","Text":"We use this double arrow to indicate that the process is only partial."},{"Start":"03:12.489 ","End":"03:17.080","Text":"In fact, there are very few acetate ions at H plus ions."},{"Start":"03:17.080 ","End":"03:19.945","Text":"It mostly stays as acetic acid."},{"Start":"03:19.945 ","End":"03:24.330","Text":"Another way of writing it again is with the H_2O so that we get,"},{"Start":"03:24.330 ","End":"03:25.765","Text":"again, acetate,"},{"Start":"03:25.765 ","End":"03:28.855","Text":"but an H_3O plus ions."},{"Start":"03:28.855 ","End":"03:33.049","Text":"Now, what about Arrhenius definition of a base?"},{"Start":"03:33.590 ","End":"03:38.500","Text":"A base is a substance that provides hydroxide ions,"},{"Start":"03:38.500 ","End":"03:41.635","Text":"OH minus, in an aqueous solution."},{"Start":"03:41.635 ","End":"03:46.165","Text":"Once again, there are strong bases and weak bases."},{"Start":"03:46.165 ","End":"03:48.590","Text":"Here\u0027s an example of a strong base."},{"Start":"03:48.590 ","End":"03:52.413","Text":"A strong base completely dissociates into ions,"},{"Start":"03:52.413 ","End":"03:55.355","Text":"so it is also a strong electrolyte."},{"Start":"03:55.355 ","End":"03:58.665","Text":"Here\u0027s an example, NaOH."},{"Start":"03:58.665 ","End":"04:03.710","Text":"NaOH dissociates to give us Na plus and OH minus."},{"Start":"04:03.710 ","End":"04:06.230","Text":"Here\u0027s our hydroxide ion."},{"Start":"04:06.230 ","End":"04:11.525","Text":"The other strong bases from group 1; LiOH, KOH,"},{"Start":"04:11.525 ","End":"04:14.780","Text":"and group 2 of the periodic table;"},{"Start":"04:14.780 ","End":"04:17.395","Text":"Ca(OH)_2 and Ba(OH)_2,"},{"Start":"04:17.395 ","End":"04:20.645","Text":"calcium hydroxide, barium hydroxide."},{"Start":"04:20.645 ","End":"04:22.650","Text":"What about weak bases?"},{"Start":"04:22.650 ","End":"04:26.600","Text":"Now, weak base only partially dissociates into ions."},{"Start":"04:26.600 ","End":"04:29.053","Text":"It\u0027s also weak electrolyte."},{"Start":"04:29.053 ","End":"04:32.360","Text":"The example that\u0027s always given is ammonia."},{"Start":"04:32.360 ","End":"04:35.300","Text":"Ammonia doesn\u0027t contain OH minus,"},{"Start":"04:35.300 ","End":"04:37.370","Text":"but when it reacts with water,"},{"Start":"04:37.370 ","End":"04:39.140","Text":"we get NH_4 plus,"},{"Start":"04:39.140 ","End":"04:43.555","Text":"which is called ammonium and OH minus."},{"Start":"04:43.555 ","End":"04:47.300","Text":"It doesn\u0027t directly contain OH minus,"},{"Start":"04:47.300 ","End":"04:49.805","Text":"but it gives us OH minus,"},{"Start":"04:49.805 ","End":"04:52.400","Text":"so not exactly the definition."},{"Start":"04:52.400 ","End":"04:56.795","Text":"In order to make this definition more accurate,"},{"Start":"04:56.795 ","End":"05:01.940","Text":"came 2 people, a Bronsted and Lowry, 2 different people."},{"Start":"05:01.940 ","End":"05:11.270","Text":"In 1923, they gave a slightly different definition of an acid and a base."},{"Start":"05:11.270 ","End":"05:14.825","Text":"Now, their definition includes Arrhenius."},{"Start":"05:14.825 ","End":"05:17.105","Text":"Everything that\u0027s an acid and base according to"},{"Start":"05:17.105 ","End":"05:21.401","Text":"Arrhenius will also be an acid and a base according to Bronsted and Lowry,"},{"Start":"05:21.401 ","End":"05:24.855","Text":"but it\u0027s a much wider definition."},{"Start":"05:24.855 ","End":"05:32.570","Text":"What\u0027s the definition? An acid is a proton donor and a base is a proton acceptor."},{"Start":"05:32.570 ","End":"05:37.235","Text":"Now, by proton, they mean H because H is a proton and an electron,"},{"Start":"05:37.235 ","End":"05:39.955","Text":"so H plus is just like a proton."},{"Start":"05:39.955 ","End":"05:42.785","Text":"Let\u0027s consider some examples."},{"Start":"05:42.785 ","End":"05:49.070","Text":"The examples I\u0027ve taken are the same examples as I used for the Arrhenius acid and base."},{"Start":"05:49.070 ","End":"05:51.805","Text":"Here we have hydrochloric acid,"},{"Start":"05:51.805 ","End":"05:54.590","Text":"HCl, reacting with water."},{"Start":"05:54.590 ","End":"05:56.120","Text":"According to this definition,"},{"Start":"05:56.120 ","End":"06:02.315","Text":"we need to have the water here to get H_3O plus and Cl minus."},{"Start":"06:02.315 ","End":"06:04.445","Text":"What is happening here."},{"Start":"06:04.445 ","End":"06:13.620","Text":"The HCl is an acid because it\u0027s donating it\u0027s proton to H_2O."},{"Start":"06:13.620 ","End":"06:20.505","Text":"It\u0027s giving a proton to H_2O to get H_3O plus and H_2O,"},{"Start":"06:20.505 ","End":"06:24.515","Text":"water, is accepting that proton, so it\u0027s a base."},{"Start":"06:24.515 ","End":"06:28.280","Text":"This is an acid and this is a base."},{"Start":"06:28.280 ","End":"06:30.215","Text":"In more advanced courses,"},{"Start":"06:30.215 ","End":"06:34.145","Text":"we talk about conjugate acids and bases on the other side."},{"Start":"06:34.145 ","End":"06:38.735","Text":"An acid, the HCl goes to Cl minus."},{"Start":"06:38.735 ","End":"06:41.923","Text":"We call that a conjugate base,"},{"Start":"06:41.923 ","End":"06:46.200","Text":"conj is just short for conjugate,"},{"Start":"06:46.200 ","End":"06:49.160","Text":"and H_2O goes to H_3O plus,"},{"Start":"06:49.160 ","End":"06:52.490","Text":"so we call that a conjugate acid."},{"Start":"06:52.490 ","End":"06:54.530","Text":"Here\u0027s our second example,"},{"Start":"06:54.530 ","End":"07:01.835","Text":"acetic acid plus water gives us acetate and H_3O plus."},{"Start":"07:01.835 ","End":"07:07.590","Text":"Acetic acid is donating its proton to water."},{"Start":"07:09.290 ","End":"07:12.450","Text":"This proton is going to water,"},{"Start":"07:12.450 ","End":"07:16.570","Text":"so it is an acid."},{"Start":"07:17.750 ","End":"07:20.700","Text":"Water is accepting the proton,"},{"Start":"07:20.700 ","End":"07:22.240","Text":"so it\u0027s a base."},{"Start":"07:22.240 ","End":"07:24.700","Text":"If we want to do the conjugates,"},{"Start":"07:24.700 ","End":"07:29.510","Text":"then the conjugate acid is H_3O plus,"},{"Start":"07:29.690 ","End":"07:34.205","Text":"and the conjugate base is the acetate."},{"Start":"07:34.205 ","End":"07:38.860","Text":"The last example is the one that we started from,"},{"Start":"07:38.860 ","End":"07:44.605","Text":"the one that made us realize that a better definition was acquired, and that\u0027s ammonia."},{"Start":"07:44.605 ","End":"07:50.465","Text":"We see that ammonia acquires a proton."},{"Start":"07:50.465 ","End":"07:55.735","Text":"It accepts a proton from the water to give us NH_4 plus."},{"Start":"07:55.735 ","End":"07:59.345","Text":"It must be a base because it accepts a proton"},{"Start":"07:59.345 ","End":"08:04.535","Text":"and water loses a proton to give us OH minus."},{"Start":"08:04.535 ","End":"08:08.065","Text":"Water, in this case, is an acid."},{"Start":"08:08.065 ","End":"08:15.890","Text":"We see that water sometimes behaves like a base and water sometimes behaves like an acid."},{"Start":"08:15.890 ","End":"08:18.365","Text":"If we want the conjugate business,"},{"Start":"08:18.365 ","End":"08:28.025","Text":"then OH minus will be a conjugate base and NH_4 plus will be a conjugate acid."},{"Start":"08:28.025 ","End":"08:32.270","Text":"In this video, we learned about acids and bases according to"},{"Start":"08:32.270 ","End":"08:38.550","Text":"2 different definitions; Arrhenius and Bronsted-Lowry."}],"ID":18112},{"Watched":false,"Name":"Acidic and Basic Solutions","Duration":"8m 23s","ChapterTopicVideoID":17366,"CourseChapterTopicPlaylistID":86821,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/17366.jpeg","UploadDate":"2019-03-03T01:05:40.3600000","DurationForVideoObject":"PT8M23S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.255","Text":"In this video, you\u0027ll learn about pH."},{"Start":"00:03.255 ","End":"00:09.450","Text":"The pH value tells us whether a solution is acidic or basic and to what extent."},{"Start":"00:09.450 ","End":"00:10.905","Text":"In the previous video,"},{"Start":"00:10.905 ","End":"00:14.940","Text":"we learnt about the Arrhenius definition of an acid or base."},{"Start":"00:14.940 ","End":"00:18.435","Text":"We saw that an acidic solution, according to Arrhenius,"},{"Start":"00:18.435 ","End":"00:23.160","Text":"contains H plus or H_3O plus ions."},{"Start":"00:23.160 ","End":"00:27.390","Text":"A basic solution contains OH minus ions."},{"Start":"00:27.390 ","End":"00:33.315","Text":"Now in water, really pure water there are also a few ions."},{"Start":"00:33.315 ","End":"00:38.145","Text":"There are some H_3O plus ions and some OH minus ions,"},{"Start":"00:38.145 ","End":"00:39.435","Text":"a very small amount,"},{"Start":"00:39.435 ","End":"00:41.325","Text":"but never less they\u0027re there."},{"Start":"00:41.325 ","End":"00:44.480","Text":"We can write the self ionization or"},{"Start":"00:44.480 ","End":"00:48.620","Text":"self dissociation of water equation in the following way."},{"Start":"00:48.620 ","End":"00:55.430","Text":"H_2O gives us a few H plus irons and a few OH minus ions."},{"Start":"00:55.430 ","End":"00:58.120","Text":"Notice that the same amount."},{"Start":"00:58.120 ","End":"01:04.680","Text":"For every H plus is 1 OH minus or we can write it using H_3O"},{"Start":"01:04.680 ","End":"01:11.494","Text":"plus 2 water molecules gives us H_3O plus an OH minus ions."},{"Start":"01:11.494 ","End":"01:19.330","Text":"Again, we see that the same number of H_3O plus ions has OH minus so 1 to 1 ratio."},{"Start":"01:19.330 ","End":"01:25.360","Text":"Now it can be shown experimentally that the product of the concentration of H_3O plus,"},{"Start":"01:25.360 ","End":"01:28.565","Text":"remember this square brackets means concentration."},{"Start":"01:28.565 ","End":"01:33.020","Text":"The concentration of OH minus ions is precisely equal to"},{"Start":"01:33.020 ","End":"01:37.580","Text":"10 to the minus 14 at 25 degrees Celsius."},{"Start":"01:37.580 ","End":"01:39.740","Text":"This is true in water,"},{"Start":"01:39.740 ","End":"01:42.065","Text":"in an acid or in a base."},{"Start":"01:42.065 ","End":"01:45.470","Text":"It\u0027s the product is always 10 to the minus 14."},{"Start":"01:45.470 ","End":"01:48.500","Text":"If 1 goes up, the other must come down,"},{"Start":"01:48.500 ","End":"01:52.565","Text":"H_3O plus comes down OH minus will go up."},{"Start":"01:52.565 ","End":"01:54.920","Text":"Now in pure water,"},{"Start":"01:54.920 ","End":"02:01.115","Text":"we see there\u0027s precisely the same number of H_3O plus and OH minus ions."},{"Start":"02:01.115 ","End":"02:08.525","Text":"Let\u0027s write x is equal to H_3O plus the concentration of H_3O plus,"},{"Start":"02:08.525 ","End":"02:13.110","Text":"which is the same as the concentration of OH minus."},{"Start":"02:13.690 ","End":"02:17.945","Text":"10 to the minus 14 is the same as x^2."},{"Start":"02:17.945 ","End":"02:24.225","Text":"We can work out that x is equal to 10 to the minus 7."},{"Start":"02:24.225 ","End":"02:30.555","Text":"In pure water H_3O plus is equal OH minus is equal to 10 to the minus 7."},{"Start":"02:30.555 ","End":"02:35.915","Text":"Now if we\u0027ve got a little bit more concentration of H_3O plus,"},{"Start":"02:35.915 ","End":"02:37.484","Text":"then 10 to the minus 7,"},{"Start":"02:37.484 ","End":"02:41.960","Text":"if H_3O plus concentration is greater than 10 to the minus 7,"},{"Start":"02:41.960 ","End":"02:48.050","Text":"the solution is acidic because then we\u0027ll have more H_3O plus and we\u0027ll have OH minus."},{"Start":"02:48.050 ","End":"02:49.895","Text":"If the opposite is true,"},{"Start":"02:49.895 ","End":"02:52.880","Text":"the concentration of OH minus is greater than 10 to the minus"},{"Start":"02:52.880 ","End":"02:56.195","Text":"7 which implies that the concentration of"},{"Start":"02:56.195 ","End":"02:58.715","Text":"H_3O plus will be less than 10 to minus 7"},{"Start":"02:58.715 ","End":"03:03.095","Text":"because the product is always 10 to the minus 14."},{"Start":"03:03.095 ","End":"03:06.100","Text":"Then we say the solution is basic."},{"Start":"03:06.100 ","End":"03:12.234","Text":"Now, chemists have devised a logarithmic scale for measuring acidity."},{"Start":"03:12.234 ","End":"03:14.960","Text":"They call this the pH."},{"Start":"03:14.960 ","End":"03:17.845","Text":"I\u0027m sure you\u0027ve heard that name many times."},{"Start":"03:17.845 ","End":"03:27.530","Text":"The definition of the pH is negative minus the log_10 of the concentration of H_3O plus."},{"Start":"03:27.530 ","End":"03:29.605","Text":"Supposing we have pure water,"},{"Start":"03:29.605 ","End":"03:34.120","Text":"the concentration of H_3O plus is 10 to the minus 7 as we saw before."},{"Start":"03:34.120 ","End":"03:43.625","Text":"The log of H_3O plus is this little number here, the exponent."},{"Start":"03:43.625 ","End":"03:45.570","Text":"It\u0027s minus 7."},{"Start":"03:45.570 ","End":"03:48.185","Text":"The pH is the negative of that."},{"Start":"03:48.185 ","End":"03:50.735","Text":"We saw, it\u0027s the negative of the log."},{"Start":"03:50.735 ","End":"03:53.185","Text":"The pH will be 7."},{"Start":"03:53.185 ","End":"03:57.560","Text":"We find that the pH is 7 for a neutral solution."},{"Start":"03:57.560 ","End":"04:03.740","Text":"If the concentration of H_3O plus is greater than 10 to the minus 7 that means the log,"},{"Start":"04:03.740 ","End":"04:08.465","Text":"this little number here will be greater than minus 7."},{"Start":"04:08.465 ","End":"04:13.310","Text":"Because the pH has a negative this the pH,"},{"Start":"04:13.310 ","End":"04:20.600","Text":"which has a negative of log_10 of H_3O plus will be less than 7 for an acidic solution."},{"Start":"04:20.600 ","End":"04:22.475","Text":"Now how did I get to this?"},{"Start":"04:22.475 ","End":"04:24.005","Text":"We know, for example,"},{"Start":"04:24.005 ","End":"04:29.880","Text":"that minus 6 is greater than minus 7 but if we change"},{"Start":"04:29.880 ","End":"04:36.475","Text":"the signs we see that 6 is less than 7."},{"Start":"04:36.475 ","End":"04:40.580","Text":"We go from greater than to less than."},{"Start":"04:40.580 ","End":"04:42.815","Text":"For an acidic solution,"},{"Start":"04:42.815 ","End":"04:45.515","Text":"the pH is less than 7."},{"Start":"04:45.515 ","End":"04:47.990","Text":"Now, if we have a basic solution,"},{"Start":"04:47.990 ","End":"04:53.165","Text":"we said the H_3O plus concentration was less than 10 to the minus 7."},{"Start":"04:53.165 ","End":"04:56.770","Text":"The log will be less than minus 7."},{"Start":"04:56.770 ","End":"05:00.515","Text":"Once again, we have to change the order, the pH,"},{"Start":"05:00.515 ","End":"05:07.360","Text":"which is the negative of the log_10 of H_3O plus concentration will be greater than 7."},{"Start":"05:07.360 ","End":"05:10.470","Text":"It\u0027s greater than 7 for a basic solution."},{"Start":"05:10.470 ","End":"05:14.835","Text":"Now we can draw a little chart, 7 means neutral."},{"Start":"05:14.835 ","End":"05:17.895","Text":"Less than 7 is acidic,"},{"Start":"05:17.895 ","End":"05:20.629","Text":"and greater than 7 means basic."},{"Start":"05:20.629 ","End":"05:22.429","Text":"Let\u0027s take an example."},{"Start":"05:22.429 ","End":"05:25.854","Text":"Supposing we\u0027re told that the pH of a solution is 6.4,"},{"Start":"05:25.854 ","End":"05:28.690","Text":"is the solution acidic or basic?"},{"Start":"05:28.690 ","End":"05:34.220","Text":"Then calculate H_3O plus concentration and OH minus concentration."},{"Start":"05:34.220 ","End":"05:36.350","Text":"Well, the first part is easy."},{"Start":"05:36.350 ","End":"05:39.380","Text":"We see that 6.4 is less than 7."},{"Start":"05:39.380 ","End":"05:41.329","Text":"The solution is acidic."},{"Start":"05:41.329 ","End":"05:45.920","Text":"Now we want to calculate the concentration of H_3O plus."},{"Start":"05:45.920 ","End":"05:48.380","Text":"We\u0027re given that the pH of 6.4,"},{"Start":"05:48.380 ","End":"05:52.310","Text":"which implies the log_10 of H_3O"},{"Start":"05:52.310 ","End":"05:57.109","Text":"plus is minus 6.4 because that\u0027s the definition of the pH."},{"Start":"05:57.109 ","End":"06:00.815","Text":"Now in order to find the concentration of H_3O plus,"},{"Start":"06:00.815 ","End":"06:05.865","Text":"we have to do take 10 to the power minus 6.4."},{"Start":"06:05.865 ","End":"06:08.225","Text":"If we use our calculators,"},{"Start":"06:08.225 ","End":"06:13.820","Text":"will see that that\u0027s equal to 3.98 times 10 to the minus 7."},{"Start":"06:13.820 ","End":"06:16.045","Text":"What about OH minus?"},{"Start":"06:16.045 ","End":"06:22.275","Text":"The concentration of OH minus times the concentration of H_3O plus, remember,"},{"Start":"06:22.275 ","End":"06:30.210","Text":"H_3O plus times OH minus is equal to 10 to the power minus 14."},{"Start":"06:30.210 ","End":"06:39.700","Text":"OH minus is 10 to the minus 14 divided by the concentration of H_3O plus."},{"Start":"06:39.700 ","End":"06:43.600","Text":"We found that was 3.98 times 10 to the minus 7."},{"Start":"06:43.600 ","End":"06:45.460","Text":"If we calculate that,"},{"Start":"06:45.460 ","End":"06:52.180","Text":"it\u0027s 0.251 times 10 to the minus 7 which is the same as 2.51 times 10 to the minus 8."},{"Start":"06:52.180 ","End":"06:58.690","Text":"The H_3O plus concentration is greater than 10 to the minus"},{"Start":"06:58.690 ","End":"07:05.650","Text":"7 whereas the OH minus concentration is less than 10 to the minus 7."},{"Start":"07:05.650 ","End":"07:09.355","Text":"Now, often we use something called an indicator."},{"Start":"07:09.355 ","End":"07:14.560","Text":"Indicators are substances that change the color according to the pH of the solution."},{"Start":"07:14.560 ","End":"07:17.080","Text":"They are often spread on filter paper."},{"Start":"07:17.080 ","End":"07:21.880","Text":"Now, there\u0027s a whole lot of indicators or different colors,"},{"Start":"07:21.880 ","End":"07:26.800","Text":"and they each cover a different range of pH values."},{"Start":"07:26.800 ","End":"07:28.219","Text":"Let\u0027s take an example,"},{"Start":"07:28.219 ","End":"07:32.290","Text":"litmus paper, you\u0027ve probably seen this in labs."},{"Start":"07:32.290 ","End":"07:34.600","Text":"The pH range is 4.5,"},{"Start":"07:34.600 ","End":"07:38.935","Text":"which is acidic, to 8.3, which is basic."},{"Start":"07:38.935 ","End":"07:40.765","Text":"Here\u0027s a picture of it."},{"Start":"07:40.765 ","End":"07:44.680","Text":"The neutral paper is some light purple."},{"Start":"07:44.680 ","End":"07:46.540","Text":"When you put it into base,"},{"Start":"07:46.540 ","End":"07:48.590","Text":"it goes blue, that\u0027s a base."},{"Start":"07:49.040 ","End":"07:52.390","Text":"Blue is a base."},{"Start":"07:52.390 ","End":"07:55.420","Text":"You put it into an acid, it goes red."},{"Start":"07:55.420 ","End":"07:57.410","Text":"Red is for acid."},{"Start":"07:57.410 ","End":"08:03.860","Text":"This is very easy way of determining whether we have a base or an acid."},{"Start":"08:03.860 ","End":"08:07.730","Text":"Of course nowadays there\u0027re much more sophisticated ways of doing these things."},{"Start":"08:07.730 ","End":"08:10.610","Text":"We have something called a pH meter,"},{"Start":"08:10.610 ","End":"08:15.695","Text":"which we can dip into our solution and it will give us exactly the pH."},{"Start":"08:15.695 ","End":"08:17.510","Text":"In this video, we talked about"},{"Start":"08:17.510 ","End":"08:23.640","Text":"useful concept pH for measuring whether solution is acidic or basic."}],"ID":18113},{"Watched":false,"Name":"Reactions between Acids and Bases","Duration":"6m 55s","ChapterTopicVideoID":17367,"CourseChapterTopicPlaylistID":86821,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/17367.jpeg","UploadDate":"2019-03-03T01:07:31.7370000","DurationForVideoObject":"PT6M55S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.890 ","End":"00:05.579","Text":"In previous videos, we learned about acids and bases."},{"Start":"00:05.579 ","End":"00:10.065","Text":"We also learned about pH and net ionic equations."},{"Start":"00:10.065 ","End":"00:16.170","Text":"In this video, we\u0027ll discuss reactions between a strong acid and a strong base."},{"Start":"00:16.170 ","End":"00:19.035","Text":"When a strong acid and strong base react,"},{"Start":"00:19.035 ","End":"00:23.265","Text":"they neutralize each other to give salt and water."},{"Start":"00:23.265 ","End":"00:24.989","Text":"Let us take an example."},{"Start":"00:24.989 ","End":"00:28.380","Text":"Hydrochloric acid reacts with sodium hydroxide,"},{"Start":"00:28.380 ","End":"00:30.570","Text":"hydrochloric acid is of course, an acid,"},{"Start":"00:30.570 ","End":"00:32.565","Text":"sodium hydroxide is a base,"},{"Start":"00:32.565 ","End":"00:34.200","Text":"to give us sodium chloride,"},{"Start":"00:34.200 ","End":"00:36.915","Text":"which is common salt and water."},{"Start":"00:36.915 ","End":"00:40.905","Text":"We see that 1 mole of acid and 1 mole of base,"},{"Start":"00:40.905 ","End":"00:44.855","Text":"gives us 1 mole of sodium chloride and 1 mole of water."},{"Start":"00:44.855 ","End":"00:49.219","Text":"Let\u0027s write the net ionic equation for this reaction."},{"Start":"00:49.219 ","End":"00:53.480","Text":"We saw in previous videos that hydrochloric acid reacts"},{"Start":"00:53.480 ","End":"00:57.800","Text":"with water to give us H_3O plus and Cl minus."},{"Start":"00:57.800 ","End":"01:06.605","Text":"We also saw that sodium hydroxide reacts in water to give us sodium plus and OH minus."},{"Start":"01:06.605 ","End":"01:12.335","Text":"These HCl ionizes and NaOH also ionizes."},{"Start":"01:12.335 ","End":"01:14.455","Text":"Let\u0027s now take the reaction."},{"Start":"01:14.455 ","End":"01:17.115","Text":"Here we have the HCl,"},{"Start":"01:17.115 ","End":"01:19.580","Text":"and here we have the sodium hydroxide."},{"Start":"01:19.580 ","End":"01:24.170","Text":"When they react, the sodium plus stays in solution and"},{"Start":"01:24.170 ","End":"01:29.305","Text":"the CO minus stays in solution because sodium chloride is soluble in water."},{"Start":"01:29.305 ","End":"01:35.920","Text":"The H_3O plus reacts with the OH minus to give us water 2H_2O."},{"Start":"01:35.920 ","End":"01:38.175","Text":"Now when we look at this reaction,"},{"Start":"01:38.175 ","End":"01:40.730","Text":"we see that there are spectator ions."},{"Start":"01:40.730 ","End":"01:44.600","Text":"We see this sodium plus on the left and sodium plus on the right."},{"Start":"01:44.600 ","End":"01:47.280","Text":"We can cancel, Cl minus on the left,"},{"Start":"01:47.280 ","End":"01:48.930","Text":"Cl minus on the right."},{"Start":"01:48.930 ","End":"01:54.530","Text":"All we\u0027ve got leftover is H_3O plus and the OH minus,"},{"Start":"01:54.530 ","End":"01:59.345","Text":"which react to give us 2 water molecules, 2H2O."},{"Start":"01:59.345 ","End":"02:02.210","Text":"This is our net ionic equation."},{"Start":"02:02.210 ","End":"02:06.350","Text":"Now it\u0027s important to know that the same net equation will be reached"},{"Start":"02:06.350 ","End":"02:10.820","Text":"for all the reactions involving a strong acid and strong base."},{"Start":"02:10.820 ","End":"02:13.175","Text":"Whatever the acid or base are,"},{"Start":"02:13.175 ","End":"02:17.030","Text":"we will always get the same net ionic equation."},{"Start":"02:17.030 ","End":"02:19.370","Text":"What\u0027s the equivalence point?"},{"Start":"02:19.370 ","End":"02:24.635","Text":"Supposing we are adding a strong base to a strong acid."},{"Start":"02:24.635 ","End":"02:26.555","Text":"We\u0027re adding it gradually."},{"Start":"02:26.555 ","End":"02:33.260","Text":"When we get to the point where we have a 1:1 ratio between the HCl,"},{"Start":"02:33.260 ","End":"02:35.915","Text":"for example, and the NaOH,"},{"Start":"02:35.915 ","End":"02:39.490","Text":"then this is called the equivalence point."},{"Start":"02:39.490 ","End":"02:44.225","Text":"Point at which the acid to base ratio matches the stoichiometric ratio."},{"Start":"02:44.225 ","End":"02:49.630","Text":"In the case of HCl and NaOH, it\u0027s 1:1."},{"Start":"02:49.630 ","End":"02:55.920","Text":"Now we\u0027re going to talk about the titration between HCl and NaOH."},{"Start":"02:55.920 ","End":"02:58.940","Text":"We take a solution of unknown concentration, say,"},{"Start":"02:58.940 ","End":"03:01.295","Text":"an acid and place it in a flask."},{"Start":"03:01.295 ","End":"03:02.944","Text":"Here\u0027s our flask."},{"Start":"03:02.944 ","End":"03:09.680","Text":"We add a base of known concentration to the flask using a buret. What\u0027s a buret?"},{"Start":"03:09.680 ","End":"03:14.450","Text":"It\u0027s a long glass tube with a stopcock"},{"Start":"03:14.450 ","End":"03:19.750","Text":"on it so we can control the amount that goes into the flask."},{"Start":"03:19.750 ","End":"03:22.720","Text":"Usually, it has graduated here."},{"Start":"03:22.720 ","End":"03:26.794","Text":"Tells us how much we\u0027ve put into the flask."},{"Start":"03:26.794 ","End":"03:30.050","Text":"We also add to this an indicator."},{"Start":"03:30.050 ","End":"03:33.805","Text":"In addition, we add a few drops of an indicator."},{"Start":"03:33.805 ","End":"03:37.610","Text":"When the acid and base are in stoichiometric ratio,"},{"Start":"03:37.610 ","End":"03:39.665","Text":"then this case it\u0027s 1:1."},{"Start":"03:39.665 ","End":"03:41.405","Text":"The pH will be 7,"},{"Start":"03:41.405 ","End":"03:45.860","Text":"it will be neutral and the indicator will change color."},{"Start":"03:45.860 ","End":"03:52.710","Text":"The point at which the color of indicator changes is called the end point."},{"Start":"03:52.820 ","End":"03:55.985","Text":"If we\u0027ve done the experiment carefully,"},{"Start":"03:55.985 ","End":"03:58.430","Text":"this should happen at the equivalence point."},{"Start":"03:58.430 ","End":"04:04.150","Text":"It should happen when we\u0027ve got the stoichiometric ratio between the acid and base."},{"Start":"04:04.150 ","End":"04:09.770","Text":"What might be a suitable indicator for this reaction between HCl and NaOH?"},{"Start":"04:09.770 ","End":"04:13.510","Text":"A suitable indicator is phenol red,"},{"Start":"04:13.510 ","End":"04:16.635","Text":"which is yellow at pH 6,"},{"Start":"04:16.635 ","End":"04:18.675","Text":"orange at pH 7,"},{"Start":"04:18.675 ","End":"04:20.850","Text":"and red at pH 8."},{"Start":"04:20.850 ","End":"04:22.915","Text":"When we do our titration,"},{"Start":"04:22.915 ","End":"04:26.620","Text":"we very slowly add the base to the acid,"},{"Start":"04:26.620 ","End":"04:29.065","Text":"and when we get to pH 7,"},{"Start":"04:29.065 ","End":"04:33.400","Text":"we should get a change in color of the indicator."},{"Start":"04:33.400 ","End":"04:37.360","Text":"Let\u0027s take an example of how we could use the titration."},{"Start":"04:37.360 ","End":"04:41.949","Text":"100 milliliters of hydrochloric acid of unknown concentration."},{"Start":"04:41.949 ","End":"04:43.750","Text":"We don\u0027t know it\u0027s concentration."},{"Start":"04:43.750 ","End":"04:45.580","Text":"Is placed in a flask."},{"Start":"04:45.580 ","End":"04:49.390","Text":"A few drops of phenol red are added, that\u0027s our indicator."},{"Start":"04:49.390 ","End":"04:52.575","Text":"Then slowly we add the NaOH."},{"Start":"04:52.575 ","End":"05:00.470","Text":"We add 200 milliliters of 0.1 molar NaOH until the solution changes color."},{"Start":"05:00.470 ","End":"05:02.105","Text":"That\u0027s our end point,"},{"Start":"05:02.105 ","End":"05:05.690","Text":"which would be equivalent to the equivalence point."},{"Start":"05:05.690 ","End":"05:09.799","Text":"We ask, what is the concentration of HCl?"},{"Start":"05:09.799 ","End":"05:13.415","Text":"Let\u0027s go back to our equation."},{"Start":"05:13.415 ","End":"05:21.210","Text":"1 mole of HCl reacts with 1 mole of NaOH to give us 1 mole of NaCL and water."},{"Start":"05:21.210 ","End":"05:25.535","Text":"The moles of NaOH should be equal to the moles of HCl."},{"Start":"05:25.535 ","End":"05:32.780","Text":"We recall our equation that the number of moles is equal to the molarity types of volume."},{"Start":"05:32.780 ","End":"05:41.930","Text":"A molarity of NaOH is 0.1 molar and the volume of NaOH is 0.2 liters."},{"Start":"05:41.930 ","End":"05:45.050","Text":"Remember they told us it was 200 milliliters,"},{"Start":"05:45.050 ","End":"05:48.025","Text":"200 milliliters is 0.2 liters."},{"Start":"05:48.025 ","End":"05:50.040","Text":"When we multiply these 2,"},{"Start":"05:50.040 ","End":"05:51.585","Text":"we get the number of moles,"},{"Start":"05:51.585 ","End":"05:55.740","Text":"so that\u0027s 0.02 moles of NaOH."},{"Start":"05:55.740 ","End":"06:00.190","Text":"Now we know that the number moles of NaOH must be equal to the number of moles of HCl,"},{"Start":"06:00.190 ","End":"06:05.615","Text":"so that must also be 0.02 moles of HCl."},{"Start":"06:05.615 ","End":"06:08.450","Text":"From that, we need to find the molarity."},{"Start":"06:08.450 ","End":"06:15.265","Text":"We recall the molarity is number of moles divided by the volume in liters."},{"Start":"06:15.265 ","End":"06:19.790","Text":"We have 0.02 moles of HCl divided"},{"Start":"06:19.790 ","End":"06:24.349","Text":"by the volume given to us in the question of a 100 milliliters,"},{"Start":"06:24.349 ","End":"06:26.465","Text":"which is 0.1 liter."},{"Start":"06:26.465 ","End":"06:32.765","Text":"When we divide the 2, we get 0.2M."},{"Start":"06:32.765 ","End":"06:37.130","Text":"Because moles divided by volume gives us molarity."},{"Start":"06:37.130 ","End":"06:44.065","Text":"We can conclude that the concentration of HCl is 0.2 molar."},{"Start":"06:44.065 ","End":"06:49.820","Text":"In this video, we learned about reactions between strong acids and bases,"},{"Start":"06:49.820 ","End":"06:55.620","Text":"and how to use a titration to determine an unknown concentration."}],"ID":18114}],"Thumbnail":null,"ID":86821},{"Name":"Oxidation-Reduction Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Oxidation-Reduction Reactions","Duration":"8m 33s","ChapterTopicVideoID":17368,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/17368.jpeg","UploadDate":"2019-03-03T01:04:12.2770000","DurationForVideoObject":"PT8M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"In a previous video,"},{"Start":"00:02.145 ","End":"00:03.840","Text":"we gave rules for determining"},{"Start":"00:03.840 ","End":"00:09.480","Text":"the oxidation states of atoms and molecules, ionic solids, etc."},{"Start":"00:09.480 ","End":"00:14.160","Text":"Now we\u0027re going to talk about oxidation-reduction reactions."},{"Start":"00:14.160 ","End":"00:16.440","Text":"Let\u0027s take an example."},{"Start":"00:16.440 ","End":"00:21.315","Text":"Here\u0027s a reaction 2FeO, solid,"},{"Start":"00:21.315 ","End":"00:26.400","Text":"that\u0027s iron oxide plus 3O_2,"},{"Start":"00:26.400 ","End":"00:32.130","Text":"oxygen gas giving 2Fe_2O_3 solid."},{"Start":"00:32.130 ","End":"00:34.410","Text":"That\u0027s another oxide of iron."},{"Start":"00:34.410 ","End":"00:38.085","Text":"We\u0027re going from 1 oxide of iron FeO reacting it"},{"Start":"00:38.085 ","End":"00:42.020","Text":"with oxygen to give another oxide of iron."},{"Start":"00:42.020 ","End":"00:46.205","Text":"Let\u0027s look at the ratios of iron to oxygen."},{"Start":"00:46.205 ","End":"00:51.040","Text":"In FeO, the ratio of Fe to O is 1 to 1."},{"Start":"00:51.040 ","End":"00:53.899","Text":"In Fe_2O_3 which is a product,"},{"Start":"00:53.899 ","End":"01:00.650","Text":"it increases to 2-3, 2-3 that\u0027s 1-1.5."},{"Start":"01:00.650 ","End":"01:05.525","Text":"The ratio goes from 1-1 to 1-1.5."},{"Start":"01:05.525 ","End":"01:07.760","Text":"We say it is oxidized,"},{"Start":"01:07.760 ","End":"01:10.070","Text":"the iron has been oxidized."},{"Start":"01:10.070 ","End":"01:17.270","Text":"Now that definition, that the ratio of iron to oxygen is increasing is very,"},{"Start":"01:17.270 ","End":"01:21.910","Text":"very limited because we don\u0027t always have compounds with oxygen in them."},{"Start":"01:21.910 ","End":"01:25.600","Text":"Let\u0027s discuss something much more general."},{"Start":"01:25.600 ","End":"01:30.310","Text":"We\u0027re going to relate oxidation-reduction reactions to"},{"Start":"01:30.310 ","End":"01:35.600","Text":"the change in the oxidation state which we abbreviate as OS."},{"Start":"01:35.600 ","End":"01:41.190","Text":"In oxidation-reduction reactions, which we call redox from re,"},{"Start":"01:41.190 ","End":"01:44.110","Text":"from reduction and ox from oxidation."},{"Start":"01:44.110 ","End":"01:46.450","Text":"We call it redox reactions."},{"Start":"01:46.450 ","End":"01:50.139","Text":"The oxidation states of some of the atoms change."},{"Start":"01:50.139 ","End":"01:53.395","Text":"Let\u0027s look at the reaction we had before."},{"Start":"01:53.395 ","End":"01:55.660","Text":"Let\u0027s start with FeO."},{"Start":"01:55.660 ","End":"01:57.940","Text":"On the left-hand side, we have FeO."},{"Start":"01:57.940 ","End":"02:04.490","Text":"Remember from our rules that the oxidation state of oxygen is minus 2."},{"Start":"02:04.880 ","End":"02:09.600","Text":"FeO if it\u0027s minus 2,"},{"Start":"02:09.600 ","End":"02:18.310","Text":"Fe must be plus 2 because FeO has to be completely without any oxidation states."},{"Start":"02:18.310 ","End":"02:20.320","Text":"The total has to be 0."},{"Start":"02:20.320 ","End":"02:23.475","Text":"Fe has oxidation state plus 2,"},{"Start":"02:23.475 ","End":"02:27.025","Text":"that\u0027s why we called it iron 2 oxide."},{"Start":"02:27.025 ","End":"02:29.065","Text":"Let\u0027s look at the oxygen."},{"Start":"02:29.065 ","End":"02:32.830","Text":"From our rules, we know that each atom of the element"},{"Start":"02:32.830 ","End":"02:36.755","Text":"has an oxidation state of 0 so that\u0027s our reactants."},{"Start":"02:36.755 ","End":"02:38.850","Text":"Let\u0027s look at the products."},{"Start":"02:38.850 ","End":"02:41.700","Text":"Our product\u0027s Fe_2O_3."},{"Start":"02:41.700 ","End":"02:46.240","Text":"Now once again, we know from the rules that oxygen has oxidation state minus"},{"Start":"02:46.240 ","End":"02:51.670","Text":"2 so the oxidation state of iron must be plus 3."},{"Start":"02:51.670 ","End":"02:59.130","Text":"Why? Because we have 3 oxygens that\u0027s 3 times minus 2 equals minus 6."},{"Start":"02:59.130 ","End":"03:06.150","Text":"So to balance that we need plus 6 and we have 2 irons so it must be 2 times 3."},{"Start":"03:06.150 ","End":"03:10.710","Text":"The oxidation state must be 3."},{"Start":"03:10.710 ","End":"03:12.658","Text":"Here we have it written,"},{"Start":"03:12.658 ","End":"03:17.900","Text":"the oxidation state of oxygen is minus 2 and iron is plus 3."},{"Start":"03:17.900 ","End":"03:21.595","Text":"That\u0027s why we call it iron 3 oxide."},{"Start":"03:21.595 ","End":"03:27.500","Text":"What\u0027s happened? The oxidation state of iron has gone from plus 2 to plus 3."},{"Start":"03:27.500 ","End":"03:29.630","Text":"It has increased."},{"Start":"03:29.630 ","End":"03:32.405","Text":"When the oxidation state increases,"},{"Start":"03:32.405 ","End":"03:35.765","Text":"we say that iron is oxidized,"},{"Start":"03:35.765 ","End":"03:39.155","Text":"oxidized from plus 2 to plus 3."},{"Start":"03:39.155 ","End":"03:42.035","Text":"But every time we have something oxidized,"},{"Start":"03:42.035 ","End":"03:45.425","Text":"we must also have something reduced."},{"Start":"03:45.425 ","End":"03:49.910","Text":"The oxidation state of oxygen goes from"},{"Start":"03:49.910 ","End":"03:57.030","Text":"0 when it\u0027s an O_2 to minus 2 when it\u0027s an, Fe_2O_3."},{"Start":"03:57.030 ","End":"03:59.180","Text":"It has decreased."},{"Start":"03:59.180 ","End":"04:02.025","Text":"We say that oxygen is reduced."},{"Start":"04:02.025 ","End":"04:03.950","Text":"Every time we have oxidation,"},{"Start":"04:03.950 ","End":"04:06.110","Text":"we must also have reduction."},{"Start":"04:06.110 ","End":"04:08.555","Text":"The 2 always go together."},{"Start":"04:08.555 ","End":"04:14.810","Text":"Now we can write oxidation-reduction reactions as 2 half-reactions."},{"Start":"04:14.810 ","End":"04:16.490","Text":"That\u0027s very very useful."},{"Start":"04:16.490 ","End":"04:21.485","Text":"We\u0027ll see it\u0027s useful when we want to balance oxidation-reduction reactions."},{"Start":"04:21.485 ","End":"04:25.925","Text":"It also has important implications in electrochemistry."},{"Start":"04:25.925 ","End":"04:29.660","Text":"We can divide redox reactions into 2 half-reactions,"},{"Start":"04:29.660 ","End":"04:32.740","Text":"1 for oxidation and 1 for reduction."},{"Start":"04:32.740 ","End":"04:35.175","Text":"Let\u0027s take an example."},{"Start":"04:35.175 ","End":"04:36.680","Text":"Here we have an example."},{"Start":"04:36.680 ","End":"04:40.460","Text":"We have bar of zinc reacting with"},{"Start":"04:40.460 ","End":"04:45.395","Text":"a solution containing copper 2 plus ions like copper sulfate."},{"Start":"04:45.395 ","End":"04:55.210","Text":"The result is we get zinc going into zinc 2 plus ions and copper going into a solid."},{"Start":"04:55.210 ","End":"05:00.645","Text":"Now, you can see that this is a net ionic equation."},{"Start":"05:00.645 ","End":"05:03.100","Text":"Here\u0027s a picture of it."},{"Start":"05:03.100 ","End":"05:04.730","Text":"On the left-hand side,"},{"Start":"05:04.730 ","End":"05:10.055","Text":"we have a zinc bar in a copper sulfate solution,"},{"Start":"05:10.055 ","End":"05:12.845","Text":"that\u0027s a solution containing copper 2 plus."},{"Start":"05:12.845 ","End":"05:19.785","Text":"Afterwards, we have copper solid deposited on the zinc rod."},{"Start":"05:19.785 ","End":"05:24.710","Text":"Here\u0027s copper solid and inside the solution,"},{"Start":"05:24.710 ","End":"05:26.900","Text":"in addition to the copper 2 plus,"},{"Start":"05:26.900 ","End":"05:29.905","Text":"there will be zinc 2 plus."},{"Start":"05:29.905 ","End":"05:31.560","Text":"Inside the solution,"},{"Start":"05:31.560 ","End":"05:34.875","Text":"we will have zinc 2 plus."},{"Start":"05:34.875 ","End":"05:39.555","Text":"We can write this as 2 half-reactions."},{"Start":"05:39.555 ","End":"05:45.780","Text":"Zinc solid going to zinc 2 plus in solution."},{"Start":"05:45.780 ","End":"05:50.770","Text":"But we see we have to balance not only the number of atoms,"},{"Start":"05:50.770 ","End":"05:56.265","Text":"1 zinc, and 1 zinc but also the number of charges."},{"Start":"05:56.265 ","End":"05:58.065","Text":"Here we have 2 plus,"},{"Start":"05:58.065 ","End":"06:01.025","Text":"and we balance it by 2 electrons."},{"Start":"06:01.025 ","End":"06:03.745","Text":"This is called oxidation."},{"Start":"06:03.745 ","End":"06:05.670","Text":"Here\u0027s the reduction,"},{"Start":"06:05.670 ","End":"06:14.690","Text":"copper 2 plus in solution reacts with 2 electrons to give copper solid."},{"Start":"06:14.690 ","End":"06:16.400","Text":"Here\u0027s the copper solid."},{"Start":"06:16.400 ","End":"06:18.590","Text":"We call that reduction."},{"Start":"06:18.590 ","End":"06:21.460","Text":"Now if we add up these 2 equations,"},{"Start":"06:21.460 ","End":"06:25.790","Text":"we have 2 electrons on the left-hand side and 2 electrons on the right-hand side,"},{"Start":"06:25.790 ","End":"06:33.095","Text":"so they cancel and the result is the full equation,"},{"Start":"06:33.095 ","End":"06:39.220","Text":"zinc solid plus copper 2 plus give you a zinc 2 plus and copper."},{"Start":"06:39.220 ","End":"06:42.270","Text":"We have 2 features of oxidation."},{"Start":"06:42.270 ","End":"06:45.095","Text":"The oxidation state of an element increases."},{"Start":"06:45.095 ","End":"06:46.865","Text":"In this case, it was zinc,"},{"Start":"06:46.865 ","End":"06:49.130","Text":"and electrons are lost."},{"Start":"06:49.130 ","End":"06:53.240","Text":"Zinc lost electrons when it became zinc 2 plus."},{"Start":"06:53.240 ","End":"06:58.255","Text":"The electrons always appear on the right-hand side of the equation here."},{"Start":"06:58.255 ","End":"07:02.995","Text":"Then reduction, the oxidation state of an element decreases."},{"Start":"07:02.995 ","End":"07:05.810","Text":"In this case, it was copper."},{"Start":"07:05.810 ","End":"07:08.045","Text":"Electrons are gained,"},{"Start":"07:08.045 ","End":"07:12.604","Text":"electrons always appear on the left-hand side of the equation."},{"Start":"07:12.604 ","End":"07:17.420","Text":"Here\u0027s a picture that tries to illustrate the process."},{"Start":"07:17.420 ","End":"07:20.240","Text":"It was the illustration of a redox reaction."},{"Start":"07:20.240 ","End":"07:23.854","Text":"Here\u0027s our zinc bar."},{"Start":"07:23.854 ","End":"07:27.200","Text":"Zinc atoms go into the solution,"},{"Start":"07:27.200 ","End":"07:29.500","Text":"become zinc 2 plus."},{"Start":"07:29.500 ","End":"07:33.050","Text":"An atom goes into the solution becomes zinc 2 plus."},{"Start":"07:33.050 ","End":"07:36.830","Text":"Here\u0027s an atom going into the solution, zinc 2 plus."},{"Start":"07:36.830 ","End":"07:38.855","Text":"Now, what about the electrons?"},{"Start":"07:38.855 ","End":"07:41.920","Text":"When it goes from zinc to zinc 2 plus,"},{"Start":"07:41.920 ","End":"07:44.579","Text":"2 electrons are released."},{"Start":"07:44.579 ","End":"07:46.905","Text":"Here are the 2 electrons."},{"Start":"07:46.905 ","End":"07:50.585","Text":"Now copper is in solution,"},{"Start":"07:50.585 ","End":"07:52.045","Text":"copper 2 plus,"},{"Start":"07:52.045 ","End":"07:54.045","Text":"here is copper 2 plus."},{"Start":"07:54.045 ","End":"07:55.920","Text":"When it becomes copper 2 plus,"},{"Start":"07:55.920 ","End":"08:00.565","Text":"it has to absorb 2 electrons to gain 2 electrons."},{"Start":"08:00.565 ","End":"08:02.495","Text":"At the end of the day,"},{"Start":"08:02.495 ","End":"08:09.520","Text":"we have zinc 2 plus in the solution and copper deposited on the zinc bar."},{"Start":"08:09.520 ","End":"08:14.375","Text":"That\u0027s an approximate illustration of the process."},{"Start":"08:14.375 ","End":"08:15.950","Text":"In this video, we defined"},{"Start":"08:15.950 ","End":"08:20.825","Text":"a redox reaction and showed how it can be written as 2 half-reactions."},{"Start":"08:20.825 ","End":"08:23.570","Text":"These half-reactions can be used as the basis of"},{"Start":"08:23.570 ","End":"08:27.770","Text":"electrochemical cell, which produces electricity."},{"Start":"08:27.770 ","End":"08:33.510","Text":"So it\u0027s very important to understand how to write half-reactions."}],"ID":18115},{"Watched":false,"Name":"Balancing Oxidation-Reduction Reactions in Acidic Solutions","Duration":"10m 10s","ChapterTopicVideoID":17369,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/17369.jpeg","UploadDate":"2019-03-03T01:06:50.2870000","DurationForVideoObject":"PT10M10S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.205","Text":"In a previous video,"},{"Start":"00:02.205 ","End":"00:06.975","Text":"we learned about oxidation-reduction reactions and half-reactions."},{"Start":"00:06.975 ","End":"00:12.720","Text":"In this video, we\u0027ll learn how to balance oxidation-reduction reactions."},{"Start":"00:12.720 ","End":"00:18.810","Text":"We\u0027re going to balance oxidation-reduction-reactions in acidic solutions."},{"Start":"00:18.810 ","End":"00:25.020","Text":"We always assume that the solution is acidic when we balance the equation the first time."},{"Start":"00:25.020 ","End":"00:27.525","Text":"If the medium is in fact basic,"},{"Start":"00:27.525 ","End":"00:30.705","Text":"we correct for this after balancing the equation."},{"Start":"00:30.705 ","End":"00:35.745","Text":"There are several steps which are best illustrated by considering an example."},{"Start":"00:35.745 ","End":"00:37.980","Text":"So here\u0027s our example."},{"Start":"00:37.980 ","End":"00:42.060","Text":"Balance the following equation in an acidic solution."},{"Start":"00:42.060 ","End":"00:47.030","Text":"HCO_2H, that\u0027s formic acid plus MnO_4 minus,"},{"Start":"00:47.030 ","End":"00:53.180","Text":"that\u0027s permanganate to give carbon dioxide and Mn^2 plus."},{"Start":"00:53.180 ","End":"00:57.740","Text":"The first step is to write half equations for this reaction."},{"Start":"00:57.740 ","End":"00:59.270","Text":"In order to do that,"},{"Start":"00:59.270 ","End":"01:05.585","Text":"we need to know the oxidation states of the atoms that change in the reaction."},{"Start":"01:05.585 ","End":"01:07.490","Text":"Here\u0027s our reaction."},{"Start":"01:07.490 ","End":"01:09.780","Text":"Let\u0027s first look at carbon."},{"Start":"01:09.780 ","End":"01:14.365","Text":"On the left-hand side, we have HCO_2H."},{"Start":"01:14.365 ","End":"01:19.130","Text":"Now, we know that the oxidation state of hydrogen is plus 1."},{"Start":"01:19.130 ","End":"01:25.080","Text":"We have 2 times 1 and that of oxygen is minus 2."},{"Start":"01:25.080 ","End":"01:29.260","Text":"So we have 2 times minus 2."},{"Start":"01:29.260 ","End":"01:33.350","Text":"That gives us a total of minus 2."},{"Start":"01:33.350 ","End":"01:41.060","Text":"The oxidation state of carbon must be plus 2 in order that the total will be 0."},{"Start":"01:41.060 ","End":"01:43.940","Text":"Carbon here is plus 2."},{"Start":"01:43.940 ","End":"01:47.060","Text":"Let\u0027s look at carbon dioxide."},{"Start":"01:47.060 ","End":"01:51.095","Text":"We know the oxidation state of oxygen is minus 2,"},{"Start":"01:51.095 ","End":"01:54.565","Text":"so we have 2 times minus 2,"},{"Start":"01:54.565 ","End":"01:56.910","Text":"giving us minus 4."},{"Start":"01:56.910 ","End":"01:59.570","Text":"In order that the total will be 0,"},{"Start":"01:59.570 ","End":"02:04.590","Text":"we need the carbon will have an oxidation state of plus 4."},{"Start":"02:05.330 ","End":"02:12.200","Text":"Carbon goes from plus 2 to plus 4 and is therefore oxidized."},{"Start":"02:12.200 ","End":"02:17.225","Text":"Remember, oxidation is when the oxidation state increases."},{"Start":"02:17.225 ","End":"02:22.585","Text":"Here we have the beginning of the half equation is not yet balanced,"},{"Start":"02:22.585 ","End":"02:25.455","Text":"each going to CO_2."},{"Start":"02:25.455 ","End":"02:28.290","Text":"Now look at the manganese."},{"Start":"02:28.290 ","End":"02:30.780","Text":"We have MnO_4 minus,"},{"Start":"02:30.780 ","End":"02:33.270","Text":"so the total must be minus 1."},{"Start":"02:33.270 ","End":"02:38.730","Text":"We know that 4 oxygens gives us minus 8."},{"Start":"02:38.730 ","End":"02:41.645","Text":"In order to get total of minus 1,"},{"Start":"02:41.645 ","End":"02:45.965","Text":"the oxidation states of Mn must be 7."},{"Start":"02:45.965 ","End":"02:48.045","Text":"We write plus 7 here."},{"Start":"02:48.045 ","End":"02:52.700","Text":"And of course for Mn 2 plus the oxidation state is plus 2."},{"Start":"02:52.700 ","End":"02:54.020","Text":"So here\u0027s the summary."},{"Start":"02:54.020 ","End":"02:58.780","Text":"Manganese goes from plus 7 to plus 2 and is reduced."},{"Start":"02:58.780 ","End":"03:01.640","Text":"We know that something is reduced when it\u0027s"},{"Start":"03:01.640 ","End":"03:06.695","Text":"oxidation state is less and the product and the reactant."},{"Start":"03:06.695 ","End":"03:11.950","Text":"We can write the half-reaction MnO_4 minus going to Mn_2 plus."},{"Start":"03:11.950 ","End":"03:14.399","Text":"This is not yet balanced."},{"Start":"03:14.399 ","End":"03:19.940","Text":"Step 2 is to balance all atoms except hydrogen and oxygen."},{"Start":"03:19.940 ","End":"03:21.875","Text":"If we look at the equations,"},{"Start":"03:21.875 ","End":"03:25.741","Text":"we see that we have 1 carbon here on the left and 1 on the right,"},{"Start":"03:25.741 ","End":"03:28.490","Text":"so that balances with respect to carbon."},{"Start":"03:28.490 ","End":"03:30.710","Text":"We have 1 manganese on the left and 1 on"},{"Start":"03:30.710 ","End":"03:34.580","Text":"the right so it\u0027s balanced with respect to manganese."},{"Start":"03:34.580 ","End":"03:38.870","Text":"In this case, both carbon and manganese are balanced."},{"Start":"03:38.870 ","End":"03:45.215","Text":"Step 3, is to balance oxygen and hydrogen in each half equation using water,"},{"Start":"03:45.215 ","End":"03:47.810","Text":"H_2O for the oxygen,"},{"Start":"03:47.810 ","End":"03:50.000","Text":"and H plus for the hydrogen."},{"Start":"03:50.000 ","End":"03:55.355","Text":"The way to do this is first adding water and then hydrogen plus."},{"Start":"03:55.355 ","End":"04:02.850","Text":"Let\u0027s look at our equation and HCO_2H going to CO_2."},{"Start":"04:02.850 ","End":"04:07.775","Text":"We have in fact 2 hydrogens on the left and none of the right."},{"Start":"04:07.775 ","End":"04:11.945","Text":"But we have 2 oxygens on the left and 2 oxygens on the right."},{"Start":"04:11.945 ","End":"04:14.054","Text":"Oxygen is already balanced,"},{"Start":"04:14.054 ","End":"04:16.865","Text":"all we have to balance is the hydrogen."},{"Start":"04:16.865 ","End":"04:22.680","Text":"We can do that by adding 2 hydrogen, 2H plus."},{"Start":"04:22.680 ","End":"04:24.390","Text":"Here it\u0027s written out."},{"Start":"04:24.390 ","End":"04:27.230","Text":"Just 2H plus on the right-hand side."},{"Start":"04:27.230 ","End":"04:29.640","Text":"Let\u0027s look at the second equation."},{"Start":"04:29.640 ","End":"04:32.710","Text":"MnO4 minus Mn_2 plus."},{"Start":"04:32.710 ","End":"04:35.000","Text":"We have 4 oxygens on the left,"},{"Start":"04:35.000 ","End":"04:37.910","Text":"so we need to add 4 oxygens on the right."},{"Start":"04:37.910 ","End":"04:40.595","Text":"We can add 4 water molecules."},{"Start":"04:40.595 ","End":"04:42.270","Text":"But after doing that,"},{"Start":"04:42.270 ","End":"04:46.775","Text":"we now have 8 hydrogens on the right and none on the left."},{"Start":"04:46.775 ","End":"04:51.095","Text":"On the left we have to add 8 hydrogens."},{"Start":"04:51.095 ","End":"04:52.910","Text":"Here it\u0027s written out,"},{"Start":"04:52.910 ","End":"04:58.995","Text":"8 hydrogens plus MnO_4 minus giving Mn_2 plus plus 4H_2O."},{"Start":"04:58.995 ","End":"05:02.645","Text":"Now, these are all balanced with respect to the atoms."},{"Start":"05:02.645 ","End":"05:09.545","Text":"The next step is to balance the charges in each half equation using electrons."},{"Start":"05:09.545 ","End":"05:11.885","Text":"Now, how are we going to do this?"},{"Start":"05:11.885 ","End":"05:14.930","Text":"If we look at our equation,"},{"Start":"05:14.930 ","End":"05:20.765","Text":"we see we have no charges on the left and plus 2 on the right,"},{"Start":"05:20.765 ","End":"05:23.195","Text":"2 times 1 for the hydrogens."},{"Start":"05:23.195 ","End":"05:26.780","Text":"In order to make the whole thing equal,"},{"Start":"05:26.780 ","End":"05:33.485","Text":"we have equal to 0 because I have 0 on the left and plus 2 on the right."},{"Start":"05:33.485 ","End":"05:35.555","Text":"In order to meet them both 0,"},{"Start":"05:35.555 ","End":"05:39.610","Text":"we have to add 2 electrons minus 2."},{"Start":"05:39.610 ","End":"05:41.645","Text":"Here are 2 electrons,"},{"Start":"05:41.645 ","End":"05:44.090","Text":"each with a charge of minus 1."},{"Start":"05:44.090 ","End":"05:46.625","Text":"Now let\u0027s look at the second reaction."},{"Start":"05:46.625 ","End":"05:49.115","Text":"We have 8 hydrogens,"},{"Start":"05:49.115 ","End":"05:50.570","Text":"that\u0027s plus 8,"},{"Start":"05:50.570 ","End":"05:54.305","Text":"1 minus from the MnO4 minus."},{"Start":"05:54.305 ","End":"05:56.995","Text":"On the left-hand side, that\u0027s plus 7."},{"Start":"05:56.995 ","End":"05:58.310","Text":"On the right-hand side,"},{"Start":"05:58.310 ","End":"05:59.585","Text":"we just have plus 2,"},{"Start":"05:59.585 ","End":"06:02.120","Text":"so in order for the 2 of them to be equal,"},{"Start":"06:02.120 ","End":"06:06.140","Text":"we can add 5 electrons each with a charge of minus"},{"Start":"06:06.140 ","End":"06:10.850","Text":"1 on the left-hand side and that will give plus 2."},{"Start":"06:10.850 ","End":"06:15.275","Text":"Now we have the same charge on the left-hand side and the right-hand side."},{"Start":"06:15.275 ","End":"06:17.215","Text":"Here\u0027s written out."},{"Start":"06:17.215 ","End":"06:20.540","Text":"We see that in the first equation,"},{"Start":"06:20.540 ","End":"06:23.825","Text":"there are 2 electrons on the right-hand side."},{"Start":"06:23.825 ","End":"06:29.540","Text":"Our reactants have lost electrons, that\u0027s called oxidation."},{"Start":"06:29.540 ","End":"06:31.294","Text":"In the second equation,"},{"Start":"06:31.294 ","End":"06:35.070","Text":"we have gained electrons on the left-hand side."},{"Start":"06:35.070 ","End":"06:39.470","Text":"The reactant has gained electrons that\u0027s called reduction."},{"Start":"06:39.470 ","End":"06:45.695","Text":"Now we\u0027d like to add these equations that have no electrons left at all."},{"Start":"06:45.695 ","End":"06:48.760","Text":"We don\u0027t want to have any electrons leftover."},{"Start":"06:48.760 ","End":"06:52.290","Text":"To do this, we can do Step 5."},{"Start":"06:52.290 ","End":"06:55.150","Text":"If necessary, multiply each equation by"},{"Start":"06:55.150 ","End":"06:59.379","Text":"a factor so the same number of electrons appears in each equation."},{"Start":"06:59.379 ","End":"07:03.355","Text":"Then when we add them they\u0027ll be no electrons left at all."},{"Start":"07:03.355 ","End":"07:05.080","Text":"So if we look at the equations,"},{"Start":"07:05.080 ","End":"07:12.090","Text":"we see that we get 2 electrons in the first step and 5 in a second."},{"Start":"07:12.090 ","End":"07:17.028","Text":"We need to multiply the first equation by 5 and the second equation by 2,"},{"Start":"07:17.028 ","End":"07:20.155","Text":"then we\u0027ll have 10 electrons in each of them."},{"Start":"07:20.155 ","End":"07:22.885","Text":"Here\u0027s multiplying the first equation by 5,"},{"Start":"07:22.885 ","End":"07:26.485","Text":"that gives us 5 HCO_H,"},{"Start":"07:26.485 ","End":"07:31.680","Text":"giving 5CO_2 plus 10 hydrogens, plus 10 electrons."},{"Start":"07:31.680 ","End":"07:35.460","Text":"We can multiply the second equation by 2."},{"Start":"07:35.460 ","End":"07:42.090","Text":"Here, 2 times this equation gives 10 electrons plus 16 hydrogens plus"},{"Start":"07:42.090 ","End":"07:48.540","Text":"2MnO_4 minus 2Mn_2 plus, plus 8H_2O."},{"Start":"07:48.540 ","End":"07:51.195","Text":"We have 10 electrons left here,"},{"Start":"07:51.195 ","End":"07:53.465","Text":"10 electrons on the right."},{"Start":"07:53.465 ","End":"07:57.500","Text":"Now, we can add the half equations to get final equation."},{"Start":"07:57.500 ","End":"07:59.385","Text":"Here are 2 equations,"},{"Start":"07:59.385 ","End":"08:00.830","Text":"now we\u0027re going to add to them,"},{"Start":"08:00.830 ","End":"08:03.875","Text":"remembering that the 10 electrons cancels."},{"Start":"08:03.875 ","End":"08:06.005","Text":"Is there anything else that cancels?"},{"Start":"08:06.005 ","End":"08:11.695","Text":"Yes, we have 10 hydrogens in the right here and 16 on the left."},{"Start":"08:11.695 ","End":"08:17.885","Text":"I can cancel the 10 hydrogens here and our left here with just 6 hydrogens."},{"Start":"08:17.885 ","End":"08:19.745","Text":"Now, we can add them up."},{"Start":"08:19.745 ","End":"08:26.450","Text":"We get 5HCO_2H plus 2MnO_4 minus, plus 6 hydrogens."},{"Start":"08:26.450 ","End":"08:29.100","Text":"That\u0027s indicative of a acid."},{"Start":"08:29.100 ","End":"08:35.115","Text":"An acidic solution have H plus give me 5CO_2 plus 2Mn2 plus,"},{"Start":"08:35.115 ","End":"08:37.205","Text":"plus 8 water molecules."},{"Start":"08:37.205 ","End":"08:38.660","Text":"Now once we\u0027ve done that,"},{"Start":"08:38.660 ","End":"08:42.815","Text":"we should really check the final equation just to make sure it\u0027s correct."},{"Start":"08:42.815 ","End":"08:45.020","Text":"We have to check 2 possible things."},{"Start":"08:45.020 ","End":"08:49.774","Text":"We have to check the number of atoms and also the charge."},{"Start":"08:49.774 ","End":"08:53.090","Text":"We have to check that atoms and charges are balanced."},{"Start":"08:53.090 ","End":"08:55.369","Text":"Let\u0027s look first at the atoms."},{"Start":"08:55.369 ","End":"08:58.670","Text":"Look at carbon, 5 carbons here,"},{"Start":"08:58.670 ","End":"09:01.210","Text":"5 carbons on the right-hand side."},{"Start":"09:01.210 ","End":"09:03.000","Text":"Look at oxygen,"},{"Start":"09:03.000 ","End":"09:04.500","Text":"5 times 2,"},{"Start":"09:04.500 ","End":"09:07.470","Text":"that\u0027s 10, and another 2 times 4,"},{"Start":"09:07.470 ","End":"09:09.750","Text":"that\u0027s 18 all together."},{"Start":"09:09.750 ","End":"09:14.910","Text":"The right-hand side, 5 times 2 is 10 and 8 times 1."},{"Start":"09:14.910 ","End":"09:17.685","Text":"There\u0027s another 8, giving us 18."},{"Start":"09:17.685 ","End":"09:20.595","Text":"Now 18 oxygens all together."},{"Start":"09:20.595 ","End":"09:25.095","Text":"Manganese, we have 2 manganese the left and 2 on the right."},{"Start":"09:25.095 ","End":"09:30.150","Text":"Hydrogens, we have 5 times 2 that\u0027s 10 here,"},{"Start":"09:30.150 ","End":"09:32.360","Text":"another 6 that\u0027s 16."},{"Start":"09:32.360 ","End":"09:36.230","Text":"And the right-hand side we have 8 times 2 also 16."},{"Start":"09:36.230 ","End":"09:37.759","Text":"That\u0027s balanced."},{"Start":"09:37.759 ","End":"09:39.241","Text":"The atoms are all balanced,"},{"Start":"09:39.241 ","End":"09:41.170","Text":"what about the charges?"},{"Start":"09:41.170 ","End":"09:44.355","Text":"Here we have 2 times minus 1,"},{"Start":"09:44.355 ","End":"09:48.630","Text":"giving us minus 2 and another 6 that\u0027s plus 4."},{"Start":"09:48.630 ","End":"09:51.015","Text":"On the right-hand side we have 2 times 2,"},{"Start":"09:51.015 ","End":"09:52.560","Text":"that\u0027s also plus 4."},{"Start":"09:52.560 ","End":"09:54.365","Text":"This equation is correct."},{"Start":"09:54.365 ","End":"09:58.970","Text":"It\u0027s balanced with respect to both atoms and charges."},{"Start":"09:58.970 ","End":"10:04.895","Text":"In this video, we learned how to balance redox equations in acidic solutions."},{"Start":"10:04.895 ","End":"10:06.365","Text":"In the next video,"},{"Start":"10:06.365 ","End":"10:10.290","Text":"we\u0027ll learn to do it in basic solutions."}],"ID":18116},{"Watched":false,"Name":"Balancing Oxidation-Reduction Reactions in Basic Solutions","Duration":"9m 6s","ChapterTopicVideoID":17370,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/17370.jpeg","UploadDate":"2019-03-03T01:08:37.3400000","DurationForVideoObject":"PT9M6S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"In the previous video,"},{"Start":"00:01.980 ","End":"00:06.675","Text":"we learned how to balance oxidation-reduction reactions in acidic solutions."},{"Start":"00:06.675 ","End":"00:11.865","Text":"In this video, we will learn how to balance them in basic solutions."},{"Start":"00:11.865 ","End":"00:16.590","Text":"So we\u0027re going to balance oxidation-reduction reactions in basic solutions."},{"Start":"00:16.590 ","End":"00:18.540","Text":"First, we\u0027ll balance them as if they\u0027re"},{"Start":"00:18.540 ","End":"00:22.650","Text":"acidic and then afterwards we\u0027ll correct for the base."},{"Start":"00:22.650 ","End":"00:24.975","Text":"Let\u0027s take an example."},{"Start":"00:24.975 ","End":"00:30.525","Text":"Balance the following reaction in a basic solution: chromium hydroxide,"},{"Start":"00:30.525 ","End":"00:32.790","Text":"Cr(OH)_3 plus Br_2,"},{"Start":"00:32.790 ","End":"00:34.275","Text":"that\u0027s bromine gas,"},{"Start":"00:34.275 ","End":"00:39.865","Text":"to give chromate, CrO_4^2 minus plus bromide Br^minus."},{"Start":"00:39.865 ","End":"00:44.210","Text":"The steps are just the same steps as we had before."},{"Start":"00:44.210 ","End":"00:48.280","Text":"First of all, we write half equations."},{"Start":"00:48.280 ","End":"00:50.990","Text":"Here\u0027s the equation we want to balance."},{"Start":"00:50.990 ","End":"00:53.390","Text":"We want to write half equations for it."},{"Start":"00:53.390 ","End":"00:57.815","Text":"The first thing to do is to calculate the oxidation states"},{"Start":"00:57.815 ","End":"01:02.915","Text":"of the atoms that change from reactants to products."},{"Start":"01:02.915 ","End":"01:04.850","Text":"Let\u0027s begin with chromium."},{"Start":"01:04.850 ","End":"01:07.070","Text":"We have chromium hydroxide."},{"Start":"01:07.070 ","End":"01:08.885","Text":"There are 3 oxygens,"},{"Start":"01:08.885 ","End":"01:11.645","Text":"that\u0027s 3 times minus 2."},{"Start":"01:11.645 ","End":"01:19.055","Text":"That\u0027s the oxidation state of oxygen is minus 2 and 3 hydrogens, 3 times 1."},{"Start":"01:19.055 ","End":"01:22.730","Text":"That gives us a total of minus 3."},{"Start":"01:22.730 ","End":"01:25.325","Text":"In order to get 0,"},{"Start":"01:25.325 ","End":"01:29.885","Text":"which is the oxidation state of Cr(OH)_3,"},{"Start":"01:29.885 ","End":"01:37.530","Text":"we need to add plus 3 so the oxidation state of chromium is plus 3."},{"Start":"01:37.960 ","End":"01:40.640","Text":"Now look at the right-hand side."},{"Start":"01:40.640 ","End":"01:42.245","Text":"We have chromate."},{"Start":"01:42.245 ","End":"01:44.300","Text":"We have 4 oxygens,"},{"Start":"01:44.300 ","End":"01:49.875","Text":"so that\u0027s 4 times minus 2 giving us minus 8,"},{"Start":"01:49.875 ","End":"01:55.160","Text":"but the total here must be minus 2 because it\u0027s an ion."},{"Start":"01:55.160 ","End":"02:01.070","Text":"Therefore, the oxidation state of chromium is plus 6."},{"Start":"02:01.070 ","End":"02:03.395","Text":"Here we have it written down."},{"Start":"02:03.395 ","End":"02:08.435","Text":"Chromium goes from plus 3 to plus 6 and it\u0027s oxidized."},{"Start":"02:08.435 ","End":"02:11.885","Text":"That means its oxidation state increases."},{"Start":"02:11.885 ","End":"02:15.200","Text":"It goes from chromium hydroxide to chromate."},{"Start":"02:15.200 ","End":"02:17.810","Text":"That\u0027s the beginning of our half-reaction,"},{"Start":"02:17.810 ","End":"02:20.030","Text":"which is not yet balanced."},{"Start":"02:20.030 ","End":"02:22.820","Text":"Now let\u0027s look at bromine."},{"Start":"02:22.820 ","End":"02:24.395","Text":"We have Br_2."},{"Start":"02:24.395 ","End":"02:28.430","Text":"Here the oxidation state of bromine is 0 because"},{"Start":"02:28.430 ","End":"02:32.840","Text":"it\u0027s an element and here it\u0027s minus 1 because it\u0027s an ion."},{"Start":"02:32.840 ","End":"02:35.465","Text":"So we write it down."},{"Start":"02:35.465 ","End":"02:41.020","Text":"Bromine goes from 0 to minus 1 and it\u0027s reduced."},{"Start":"02:41.020 ","End":"02:42.930","Text":"Here\u0027s the half reaction,"},{"Start":"02:42.930 ","End":"02:46.665","Text":"bromine 2 going to bromide Br^minus,"},{"Start":"02:46.665 ","End":"02:51.905","Text":"and it\u0027s reduced because the oxidation state of bromine has decreased."},{"Start":"02:51.905 ","End":"02:57.355","Text":"Now the next step is to balance all the atoms except H and O."},{"Start":"02:57.355 ","End":"03:00.750","Text":"If we look at the reaction with chromium,"},{"Start":"03:00.750 ","End":"03:04.790","Text":"we see there\u0027s 1 chromium on the left and 1 on the right."},{"Start":"03:04.790 ","End":"03:07.745","Text":"So that\u0027s balanced with respect to chromium."},{"Start":"03:07.745 ","End":"03:09.795","Text":"Look at the second 1."},{"Start":"03:09.795 ","End":"03:13.645","Text":"We\u0027ve Br_2 going to Br^minus."},{"Start":"03:13.645 ","End":"03:16.455","Text":"We have 2 bromines on the left,"},{"Start":"03:16.455 ","End":"03:17.715","Text":"only 1 on the right,"},{"Start":"03:17.715 ","End":"03:20.535","Text":"so we need to put in a 2 here."},{"Start":"03:20.535 ","End":"03:22.140","Text":"Here it\u0027s summarized."},{"Start":"03:22.140 ","End":"03:23.580","Text":"Chromium is balanced,"},{"Start":"03:23.580 ","End":"03:28.995","Text":"to balance Br we\u0027ve added bromine 2 to Br^minus."},{"Start":"03:28.995 ","End":"03:33.080","Text":"Step 3, balanced O and H in each half reaction"},{"Start":"03:33.080 ","End":"03:37.880","Text":"using water for the oxygen and H plus for the hydrogen."},{"Start":"03:37.880 ","End":"03:44.000","Text":"Just a reminder, we do this first by adding water and then we add H plus."},{"Start":"03:44.000 ","End":"03:47.150","Text":"So here\u0027s the chromium half-reaction."},{"Start":"03:47.150 ","End":"03:51.740","Text":"Chromium hydroxide going to chromate, that\u0027s what we had."},{"Start":"03:51.740 ","End":"03:56.675","Text":"Now we have 4 oxygens on the right and only 3 on the left."},{"Start":"03:56.675 ","End":"03:59.585","Text":"So we need to add another 1 to the left."},{"Start":"03:59.585 ","End":"04:01.910","Text":"We do this by adding water molecule."},{"Start":"04:01.910 ","End":"04:03.395","Text":"Here\u0027s the water molecule."},{"Start":"04:03.395 ","End":"04:05.740","Text":"Now once we\u0027ve done that,"},{"Start":"04:05.740 ","End":"04:08.400","Text":"let us look at the left-hand side now,"},{"Start":"04:08.400 ","End":"04:14.970","Text":"we have 2 hydrogens from the water and another 3 from the chromium hydroxide."},{"Start":"04:14.970 ","End":"04:16.800","Text":"That\u0027s 5 altogether."},{"Start":"04:16.800 ","End":"04:20.240","Text":"We need to add 5 to the right-hand side."},{"Start":"04:20.240 ","End":"04:26.840","Text":"Now this equation is balanced with respect to oxygen and hydrogen."},{"Start":"04:26.840 ","End":"04:31.175","Text":"Of course, the bromine equation is already balanced."},{"Start":"04:31.175 ","End":"04:35.810","Text":"Now we balance the charges in each half equation using electrons."},{"Start":"04:35.810 ","End":"04:37.820","Text":"Remember we always add electrons,"},{"Start":"04:37.820 ","End":"04:40.190","Text":"we never subtract electrons."},{"Start":"04:40.190 ","End":"04:46.220","Text":"Here\u0027s the equation as we had it ignoring the electrons at the moment."},{"Start":"04:46.220 ","End":"04:54.865","Text":"We have minus 2 for the chromate and plus 5 for the hydrogen giving us plus 3."},{"Start":"04:54.865 ","End":"04:56.675","Text":"On the left-hand side,"},{"Start":"04:56.675 ","End":"04:58.475","Text":"there are no charges at all."},{"Start":"04:58.475 ","End":"05:01.865","Text":"In order to get no charges at all on the right-hand side,"},{"Start":"05:01.865 ","End":"05:04.310","Text":"we have to add 3 electrons."},{"Start":"05:04.310 ","End":"05:08.075","Text":"That\u0027s like minus 3 that gives us 0."},{"Start":"05:08.075 ","End":"05:12.180","Text":"Now, on the second equation we have bromine 2 going to 2 Br^minus."},{"Start":"05:12.180 ","End":"05:18.140","Text":"We have 0 charge on the left and minus 2 on the right."},{"Start":"05:18.140 ","End":"05:20.405","Text":"In order to balance the equation,"},{"Start":"05:20.405 ","End":"05:23.089","Text":"we need to add 2 electrons to the left."},{"Start":"05:23.089 ","End":"05:25.790","Text":"That\u0027s like adding minus 2 to the left."},{"Start":"05:25.790 ","End":"05:33.730","Text":"Now, these equations are balanced with respect to both number of atoms and charges."},{"Start":"05:33.730 ","End":"05:35.250","Text":"Now step 5."},{"Start":"05:35.250 ","End":"05:37.610","Text":"If necessary, multiply each equation by"},{"Start":"05:37.610 ","End":"05:41.530","Text":"a factor so the same number of electrons appears in each 1."},{"Start":"05:41.530 ","End":"05:43.250","Text":"Now, if we look at these equations,"},{"Start":"05:43.250 ","End":"05:45.290","Text":"we have 3 electrons in the right-hand side of"},{"Start":"05:45.290 ","End":"05:50.660","Text":"the oxidation equation and 2 on the left-hand side of the reduction 1."},{"Start":"05:50.660 ","End":"05:52.775","Text":"In order to get the same number,"},{"Start":"05:52.775 ","End":"05:57.780","Text":"we multiply the top 1 by 2 and the bottom 1 by 3,"},{"Start":"05:57.780 ","End":"06:03.455","Text":"and then we\u0027ll have 6 electrons on the right here and 6 electrons on the left."},{"Start":"06:03.455 ","End":"06:04.949","Text":"Let\u0027s do that."},{"Start":"06:04.949 ","End":"06:07.465","Text":"If we multiply the top 1 by 2,"},{"Start":"06:07.465 ","End":"06:12.135","Text":"we get 2 water molecules plus 2 chromium hydroxide"},{"Start":"06:12.135 ","End":"06:18.885","Text":"giving 2 chromates plus 10 hydrogen ions plus 6 electrons."},{"Start":"06:18.885 ","End":"06:22.125","Text":"Multiplying the bromine equation by 3,"},{"Start":"06:22.125 ","End":"06:28.354","Text":"we get 6 electrons plus 3 bromine-2 giving 6 bromine minus."},{"Start":"06:28.354 ","End":"06:32.365","Text":"Now we can add the half equations to get the final equation."},{"Start":"06:32.365 ","End":"06:37.860","Text":"Here are the 2 half equations that we\u0027ve found out before."},{"Start":"06:37.860 ","End":"06:39.675","Text":"Here they are from here,"},{"Start":"06:39.675 ","End":"06:41.390","Text":"here they are down here,"},{"Start":"06:41.390 ","End":"06:43.040","Text":"and now we can add them."},{"Start":"06:43.040 ","End":"06:46.400","Text":"Before doing so, let\u0027s note there are 6 electrons in"},{"Start":"06:46.400 ","End":"06:50.435","Text":"the right-hand side here and 6 electrons in the left-hand side."},{"Start":"06:50.435 ","End":"06:52.345","Text":"So we can add them."},{"Start":"06:52.345 ","End":"06:58.710","Text":"We get 2 chromium hydroxide plus 3 bromine-2 plus 2 water molecules to"},{"Start":"06:58.710 ","End":"07:05.090","Text":"give 2 chromate plus 6 Br^minus plus 10 hydrogen ions."},{"Start":"07:05.090 ","End":"07:11.270","Text":"So that\u0027s equation balanced assuming that we have an acidic solution, that\u0027s H^plus."},{"Start":"07:11.270 ","End":"07:15.980","Text":"But now we want to correct for the fact that it\u0027s in fact a basic solution."},{"Start":"07:15.980 ","End":"07:22.595","Text":"We do this by adding OH^minus to both sides to neutralize H^plus."},{"Start":"07:22.595 ","End":"07:27.440","Text":"Here we\u0027ve added 10 hydroxides on each side."},{"Start":"07:27.440 ","End":"07:33.035","Text":"Now remember the H^plus plus OH^minus gives us water."},{"Start":"07:33.035 ","End":"07:36.740","Text":"We have here 10 water molecules,"},{"Start":"07:36.740 ","End":"07:39.325","Text":"but the left-hand side have 2."},{"Start":"07:39.325 ","End":"07:46.160","Text":"So we can take off the 2 on the left-hand side and write this as 8."},{"Start":"07:46.160 ","End":"07:49.445","Text":"We have 8 water molecules on the right-hand side."},{"Start":"07:49.445 ","End":"07:51.565","Text":"Here it\u0027s written out."},{"Start":"07:51.565 ","End":"07:55.435","Text":"See the 8 water molecules on the right-hand side."},{"Start":"07:55.435 ","End":"07:57.395","Text":"Just want to point out,"},{"Start":"07:57.395 ","End":"08:00.050","Text":"we\u0027ve could\u0027ve got the same answer if by adding"},{"Start":"08:00.050 ","End":"08:04.535","Text":"the OH^minus to the half reactions rather than waiting right to the end."},{"Start":"08:04.535 ","End":"08:06.590","Text":"Here\u0027s how it\u0027s done."},{"Start":"08:06.590 ","End":"08:12.025","Text":"Here\u0027s the original reaction and the second one."},{"Start":"08:12.025 ","End":"08:15.390","Text":"If we add 5 OH^minus here,"},{"Start":"08:15.390 ","End":"08:18.405","Text":"we\u0027ll get 5 water molecules,"},{"Start":"08:18.405 ","End":"08:20.415","Text":"but we\u0027ve got 1 on the right,"},{"Start":"08:20.415 ","End":"08:23.535","Text":"so that goes down to 4."},{"Start":"08:23.535 ","End":"08:30.440","Text":"Here it\u0027s written 5 OH^minus plus chromium hydroxide to give chromate plus"},{"Start":"08:30.440 ","End":"08:37.670","Text":"4 water molecules and still the same number of electrons plus 3 electrons."},{"Start":"08:37.670 ","End":"08:41.610","Text":"The second equation wasn\u0027t altered at all."},{"Start":"08:41.860 ","End":"08:47.510","Text":"The last stage, if we want to make sure we\u0027ve done the right thing,"},{"Start":"08:47.510 ","End":"08:50.060","Text":"is to check the final equation."},{"Start":"08:50.060 ","End":"08:51.770","Text":"You could do that yourselves."},{"Start":"08:51.770 ","End":"08:54.800","Text":"Just check that all the atoms are balanced and"},{"Start":"08:54.800 ","End":"08:58.685","Text":"charges are also balanced in the full equation."},{"Start":"08:58.685 ","End":"09:05.700","Text":"In this video, we learned how to balance redox equations in basic solutions."}],"ID":18117},{"Watched":false,"Name":"Exercise 1","Duration":"6m 23s","ChapterTopicVideoID":22900,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22900.jpeg","UploadDate":"2020-12-16T02:09:07.1930000","DurationForVideoObject":"PT6M23S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.750","Text":"Hi. We are we going to solve the following exercise."},{"Start":"00:03.750 ","End":"00:08.955","Text":"Indicate which of the following reactions are redox reactions and which are not."},{"Start":"00:08.955 ","End":"00:13.110","Text":"What we\u0027re going to do is we\u0027re going to calculate the oxidation states of"},{"Start":"00:13.110 ","End":"00:15.120","Text":"the elements in the reactions and then we\u0027re going to"},{"Start":"00:15.120 ","End":"00:18.130","Text":"see if these are redox reactions or not."},{"Start":"00:19.280 ","End":"00:23.940","Text":"Our first reaction in a is silver nitrate plus"},{"Start":"00:23.940 ","End":"00:29.250","Text":"sodium chloride gives us silver chloride plus sodium nitrate."},{"Start":"00:29.250 ","End":"00:31.905","Text":"Let\u0027s calculate the oxidation states."},{"Start":"00:31.905 ","End":"00:33.330","Text":"In silver nitrate,"},{"Start":"00:33.330 ","End":"00:38.655","Text":"nitrate has an oxidation state of minus 1 and silver has an oxidation state of plus 1."},{"Start":"00:38.655 ","End":"00:49.035","Text":"In sodium chloride, the chloride has a minus 1 oxidation state and the sodium is plus 1."},{"Start":"00:49.035 ","End":"00:55.865","Text":"In silver chloride we have minus 1 for the chlorine and the silver is plus 1."},{"Start":"00:55.865 ","End":"01:02.500","Text":"In sodium nitrate we have minus 1 for the nitrate and plus 1 for the sodium."},{"Start":"01:02.500 ","End":"01:05.240","Text":"Now, if we look at our whole reaction we can see"},{"Start":"01:05.240 ","End":"01:07.310","Text":"that there was no change in oxidation states,"},{"Start":"01:07.310 ","End":"01:10.790","Text":"meaning the sodium started at plus 1 and stayed at plus 1."},{"Start":"01:10.790 ","End":"01:14.600","Text":"The silver started at plus 1 and stayed at plus 1,"},{"Start":"01:14.600 ","End":"01:16.850","Text":"same as for chlorine and nitrate."},{"Start":"01:16.850 ","End":"01:21.455","Text":"Therefore since there was no change in oxidation states,"},{"Start":"01:21.455 ","End":"01:24.455","Text":"this is not a redox reaction."},{"Start":"01:24.455 ","End":"01:30.480","Text":"Answer for a is going to be not a redox reaction."},{"Start":"01:40.360 ","End":"01:44.240","Text":"That\u0027s our answer for a. Let\u0027s take a look at b."},{"Start":"01:44.240 ","End":"01:53.335","Text":"In b we have 2 aluminum plus 3 tin ion gives us 2 aluminum ion plus 3 tin."},{"Start":"01:53.335 ","End":"01:55.550","Text":"If we look at our oxidation states,"},{"Start":"01:55.550 ","End":"01:59.675","Text":"aluminum is going to have an oxidation state of 0."},{"Start":"01:59.675 ","End":"02:04.385","Text":"The tin ion has an oxidation state of plus 2."},{"Start":"02:04.385 ","End":"02:07.490","Text":"The aluminum ion has an oxidation state of plus 3,"},{"Start":"02:07.490 ","End":"02:12.630","Text":"and the tin solid has an oxidation state of 0."},{"Start":"02:12.640 ","End":"02:20.250","Text":"If we look at aluminum we can see that our oxidation state went from 0 to plus 3."},{"Start":"02:23.740 ","End":"02:32.230","Text":"If we look at tin we can see that our oxidation state went from plus 2 to 0."},{"Start":"02:32.230 ","End":"02:40.180","Text":"Meaning from aluminum to the aluminum ion we had an increase in oxidation state,"},{"Start":"02:40.180 ","End":"02:44.710","Text":"meaning the aluminum was oxidized."},{"Start":"02:47.360 ","End":"02:50.180","Text":"If we look at the tin we can see that we"},{"Start":"02:50.180 ","End":"02:54.980","Text":"have a decrease in oxidation state from plus 2 to 0,"},{"Start":"02:54.980 ","End":"02:57.780","Text":"meaning that tin was reduced."},{"Start":"03:00.950 ","End":"03:05.240","Text":"As we can see here we have a reduction and an oxidation."},{"Start":"03:05.240 ","End":"03:08.885","Text":"Therefore this is a redox reaction."},{"Start":"03:08.885 ","End":"03:11.540","Text":"As we can see we have an oxidation and reduction,"},{"Start":"03:11.540 ","End":"03:14.880","Text":"meaning b is a redox reaction."},{"Start":"03:28.880 ","End":"03:34.420","Text":"Again, b is a redox reaction and now we\u0027re going to go on to c. In c we have"},{"Start":"03:34.420 ","End":"03:40.430","Text":"hydrogen chloride plus potassium hydroxide gives us potassium chloride plus water."},{"Start":"03:44.990 ","End":"03:48.245","Text":"Let\u0027s take a look at the oxidation states."},{"Start":"03:48.245 ","End":"03:49.935","Text":"In hydrogen chloride,"},{"Start":"03:49.935 ","End":"03:53.810","Text":"the chlorine is minus 1 and the hydrogen is plus 1."},{"Start":"03:53.810 ","End":"03:57.895","Text":"In potassium hydroxide we have a plus 1 for the potassium,"},{"Start":"03:57.895 ","End":"03:59.410","Text":"minus 2 for the oxygen,"},{"Start":"03:59.410 ","End":"04:01.610","Text":"and then plus 1 for the hydrogen."},{"Start":"04:01.610 ","End":"04:04.420","Text":"In potassium chloride we have a plus 1 for the potassium,"},{"Start":"04:04.420 ","End":"04:06.040","Text":"minus 1 for the chlorine."},{"Start":"04:06.040 ","End":"04:11.135","Text":"In water we have a plus 1 for the hydrogen and minus 2 for the oxygen."},{"Start":"04:11.135 ","End":"04:14.915","Text":"If we look at our oxidation states we can see that they were all kept."},{"Start":"04:14.915 ","End":"04:21.975","Text":"The hydrogen which began with plus 1 is also plus 1 in the product side."},{"Start":"04:21.975 ","End":"04:25.215","Text":"Our chlorine is minus 1 throughout the reaction."},{"Start":"04:25.215 ","End":"04:28.310","Text":"Potassium is plus 1 throughout the reaction."},{"Start":"04:28.310 ","End":"04:30.710","Text":"Therefore there was no change in oxidation states,"},{"Start":"04:30.710 ","End":"04:35.390","Text":"meaning this is not a redox reaction."},{"Start":"04:40.820 ","End":"04:44.730","Text":"C is not a redox reaction."},{"Start":"04:44.730 ","End":"04:50.795","Text":"Let\u0027s take a look at d. In d we have carbon plus oxygen,"},{"Start":"04:50.795 ","End":"04:52.370","Text":"gives us carbon dioxide."},{"Start":"04:52.370 ","End":"04:55.790","Text":"If we look at our oxidation states we can see that carbon has"},{"Start":"04:55.790 ","End":"05:01.730","Text":"an oxidation state of 0 and oxygen has an oxidation state of 0."},{"Start":"05:01.730 ","End":"05:05.510","Text":"In carbon dioxide, the carbon has an oxidation state of"},{"Start":"05:05.510 ","End":"05:10.160","Text":"plus 4 and the oxygen has an oxidation state of minus 2."},{"Start":"05:10.310 ","End":"05:17.990","Text":"If we look at our carbon we can see that our oxidation state increase from 0 to plus 4."},{"Start":"05:17.990 ","End":"05:22.340","Text":"If we look at our oxidation states of oxygen we can"},{"Start":"05:22.340 ","End":"05:27.590","Text":"see that the oxidation state decrease from 0 to minus 2."},{"Start":"05:27.590 ","End":"05:31.100","Text":"Again, in carbon there is an increase in oxidation states,"},{"Start":"05:31.100 ","End":"05:34.170","Text":"meaning the carbon was oxidized."},{"Start":"05:39.800 ","End":"05:44.990","Text":"In oxygen we have a decrease in oxidation states,"},{"Start":"05:46.980 ","End":"05:50.660","Text":"meaning the oxygen was reduced."},{"Start":"05:53.910 ","End":"05:58.030","Text":"We can see the carbon was oxidized and oxygen was reduced."},{"Start":"05:58.030 ","End":"05:59.650","Text":"We have an oxidation and a reduction,"},{"Start":"05:59.650 ","End":"06:03.255","Text":"meaning this is a redox reaction."},{"Start":"06:03.255 ","End":"06:08.650","Text":"D is a redox reaction."},{"Start":"06:18.480 ","End":"06:20.590","Text":"That is our final answer."},{"Start":"06:20.590 ","End":"06:23.450","Text":"Thank you very much for watching."}],"ID":23721},{"Watched":false,"Name":"Exercise 2 - Part a","Duration":"4m 4s","ChapterTopicVideoID":22901,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22901.jpeg","UploadDate":"2020-12-16T02:09:17.3830000","DurationForVideoObject":"PT4M4S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.150 ","End":"00:06.525","Text":"Complete and balance the following half equations."},{"Start":"00:06.525 ","End":"00:08.340","Text":"Now before we begin with a,"},{"Start":"00:08.340 ","End":"00:11.310","Text":"I just want to mention that there are a number of methods that can be"},{"Start":"00:11.310 ","End":"00:15.270","Text":"used to balance half equations in redox reactions."},{"Start":"00:15.270 ","End":"00:18.180","Text":"I\u0027m going to use 1 way which I think is very"},{"Start":"00:18.180 ","End":"00:22.680","Text":"good but just for you to know there are other methods."},{"Start":"00:22.680 ","End":"00:27.600","Text":"Let\u0027s begin with a."},{"Start":"00:27.600 ","End":"00:33.580","Text":"In a we can see that we have the manganate ion and we have the manganese ion."},{"Start":"00:34.520 ","End":"00:37.680","Text":"Now the first step in balancing"},{"Start":"00:37.680 ","End":"00:43.520","Text":"half equations is to balance the elements other than oxygen and hydrogen."},{"Start":"00:43.520 ","End":"00:48.215","Text":"We can see that we have 1 manganese on the left and 1 manganese on the right,"},{"Start":"00:48.215 ","End":"00:50.989","Text":"meaning that the manganese is balanced."},{"Start":"00:50.989 ","End":"00:54.440","Text":"Now the next step is to balance the oxygens."},{"Start":"00:54.440 ","End":"00:59.370","Text":"Now on the left we have 4 oxygens and on the right we have no oxygen,"},{"Start":"00:59.370 ","End":"01:01.815","Text":"so we have to add 4 oxygen to the right."},{"Start":"01:01.815 ","End":"01:05.010","Text":"In order to add oxygens we actually add water molecules we\u0027re going"},{"Start":"01:05.010 ","End":"01:10.105","Text":"to add 4 water molecules."},{"Start":"01:10.105 ","End":"01:14.105","Text":"Now the next step is to balance the hydrogen."},{"Start":"01:14.105 ","End":"01:17.660","Text":"On the left we could see that we have no hydrogen,"},{"Start":"01:17.660 ","End":"01:22.100","Text":"and on the right we have 2 hydrogen times 4,"},{"Start":"01:22.100 ","End":"01:24.870","Text":"meaning we have 8 hydrogens in all."},{"Start":"01:25.190 ","End":"01:29.145","Text":"We need to add 8 hydrogens to the left,"},{"Start":"01:29.145 ","End":"01:31.620","Text":"and we do this by adding protons."},{"Start":"01:31.620 ","End":"01:34.619","Text":"We\u0027re going to add 8H plus,"},{"Start":"01:34.619 ","End":"01:36.085","Text":"which is protons,"},{"Start":"01:36.085 ","End":"01:38.495","Text":"in order to balance our hydrogens."},{"Start":"01:38.495 ","End":"01:41.555","Text":"Now we can see that the manganese is balanced."},{"Start":"01:41.555 ","End":"01:43.200","Text":"The oxygens are balanced,"},{"Start":"01:43.200 ","End":"01:45.510","Text":"we have 4 on the left side and 4 on"},{"Start":"01:45.510 ","End":"01:48.330","Text":"the right side and we can see that the hydrogens are also balanced,"},{"Start":"01:48.330 ","End":"01:51.720","Text":"we have 8 on the left side and 8 on the right side."},{"Start":"01:52.250 ","End":"01:54.910","Text":"Now that our atoms are balanced,"},{"Start":"01:54.910 ","End":"01:58.260","Text":"we have to balance for electric charge."},{"Start":"01:58.400 ","End":"02:02.030","Text":"Here if we look at our hydrogens in our oxygens,"},{"Start":"02:02.030 ","End":"02:07.100","Text":"the oxidation state of the hydrogen is going to be plus 1 in both cases."},{"Start":"02:07.100 ","End":"02:12.390","Text":"In oxidation state of the oxygen is going to be minus 2 in both cases."},{"Start":"02:12.880 ","End":"02:16.287","Text":"An oxidation state of manganese is what changes."},{"Start":"02:16.287 ","End":"02:20.390","Text":"So we can see it on the right that we have a plus 2 oxidation state and now we\u0027re going"},{"Start":"02:20.390 ","End":"02:26.520","Text":"to calculate the oxidation state of manganese and the manganese ion."},{"Start":"02:27.170 ","End":"02:33.855","Text":"We just going to say manganese plus 4,"},{"Start":"02:33.855 ","End":"02:39.080","Text":"because we have 4 oxygens times minus 2 because the oxidation state of oxygen is"},{"Start":"02:39.080 ","End":"02:45.440","Text":"minus 2 is going to give us minus 1 because minus 1 is the overall charge."},{"Start":"02:45.590 ","End":"02:49.258","Text":"The manganese oxidation state is going to come out to,"},{"Start":"02:49.258 ","End":"02:56.300","Text":"this is manganese minus 8 equals minus 1 so this is going to be manganese equals plus 7."},{"Start":"02:56.300 ","End":"03:00.035","Text":"The oxidation state is going to be plus 7."},{"Start":"03:00.035 ","End":"03:02.510","Text":"We can see that we have the manganese and"},{"Start":"03:02.510 ","End":"03:07.655","Text":"a plus 7 oxidation state and then we have it in a plus 2 oxidation state,"},{"Start":"03:07.655 ","End":"03:10.340","Text":"meaning it received 5 electrons."},{"Start":"03:10.340 ","End":"03:16.530","Text":"So we\u0027re going to add 5 electrons to the left side of our equation."},{"Start":"03:17.320 ","End":"03:20.990","Text":"At this point our half equation is balanced."},{"Start":"03:20.990 ","End":"03:23.585","Text":"We have to look at 1 more thing."},{"Start":"03:23.585 ","End":"03:26.855","Text":"Now at this point our half equation is balanced."},{"Start":"03:26.855 ","End":"03:28.340","Text":"We just have to look at 1 more thing,"},{"Start":"03:28.340 ","End":"03:30.770","Text":"we have to look at the medium in which"},{"Start":"03:30.770 ","End":"03:35.080","Text":"the reaction is in so here we have it in acidic solution."},{"Start":"03:35.080 ","End":"03:37.170","Text":"Now if it\u0027s in acidic solution,"},{"Start":"03:37.170 ","End":"03:39.035","Text":"we don\u0027t have to take any other steps."},{"Start":"03:39.035 ","End":"03:41.540","Text":"This is going to be our balanced half equation,"},{"Start":"03:41.540 ","End":"03:45.200","Text":"and if it\u0027s in basic solution soon we\u0027re going to see what we need to do."},{"Start":"03:45.200 ","End":"03:48.930","Text":"We have another step in balancing our equation."},{"Start":"03:48.980 ","End":"03:55.560","Text":"For a the balanced half equation is the manganate ion plus 8 protons plus"},{"Start":"03:55.560 ","End":"03:58.640","Text":"5 electrons is going to give us manganese ion"},{"Start":"03:58.640 ","End":"04:02.315","Text":"plus 4 water molecules and that\u0027s our answer for a."},{"Start":"04:02.315 ","End":"04:04.530","Text":"Now we\u0027re going to go on to b."}],"ID":23722},{"Watched":false,"Name":"Exercise 2 - Part b","Duration":"4m 58s","ChapterTopicVideoID":22902,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22902.jpeg","UploadDate":"2020-12-16T02:09:29.7670000","DurationForVideoObject":"PT4M58S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.470 ","End":"00:02.805","Text":"Now we\u0027re going to take a look at b."},{"Start":"00:02.805 ","End":"00:08.475","Text":"B gives us nitrogen in basic solution."},{"Start":"00:08.475 ","End":"00:11.700","Text":"This time it\u0027s in basic solution remember."},{"Start":"00:11.700 ","End":"00:15.345","Text":"We\u0027re going to start off the same way we started a."},{"Start":"00:15.345 ","End":"00:19.950","Text":"First of all, we\u0027re going to balance the elements which are not hydrogen or oxygen."},{"Start":"00:19.950 ","End":"00:22.065","Text":"We\u0027re going to take a look at our nitrogens."},{"Start":"00:22.065 ","End":"00:23.970","Text":"Now you can see that we have 1 nitrogen on"},{"Start":"00:23.970 ","End":"00:27.105","Text":"the left side and 2 nitrogens on the right side."},{"Start":"00:27.105 ","End":"00:32.580","Text":"We\u0027re going to multiply the hydroxyl by 2."},{"Start":"00:32.580 ","End":"00:36.875","Text":"Now we have 2 nitrogens on the left side and 2 nitrogens on the right side."},{"Start":"00:36.875 ","End":"00:40.790","Text":"Now the second step is to balance the oxygens."},{"Start":"00:40.790 ","End":"00:43.355","Text":"If we look on the left side, we can see that we have"},{"Start":"00:43.355 ","End":"00:47.405","Text":"2 oxygens and we have no oxygens on the right side."},{"Start":"00:47.405 ","End":"00:52.385","Text":"We\u0027re going to add 2 water molecules in order to balance the oxygens."},{"Start":"00:52.385 ","End":"00:55.820","Text":"Now we have 2 nitrogens on each side and 2 oxygens on each side,"},{"Start":"00:55.820 ","End":"00:57.905","Text":"and we have to look at our hydrogens."},{"Start":"00:57.905 ","End":"01:01.820","Text":"If we look on the left, we can see that we have 2 hydrogens times 2,"},{"Start":"01:01.820 ","End":"01:04.190","Text":"which is 4 plus another 2."},{"Start":"01:04.190 ","End":"01:08.360","Text":"We have 6 hydrogens in all on our left side."},{"Start":"01:08.360 ","End":"01:11.350","Text":"Let\u0027s just write that here for a second."},{"Start":"01:11.350 ","End":"01:13.130","Text":"We have 6 hydrogens on the left."},{"Start":"01:13.130 ","End":"01:15.710","Text":"If we look on the right, we have 2 hydrogens times 2,"},{"Start":"01:15.710 ","End":"01:17.944","Text":"we have 4 hydrogens."},{"Start":"01:17.944 ","End":"01:25.430","Text":"Therefore, we need to add 2 protons to the right side in order to have the same amount,"},{"Start":"01:25.430 ","End":"01:28.025","Text":"6 hydrogens and 6 hydrogens."},{"Start":"01:28.025 ","End":"01:31.795","Text":"Now the next step is to balance for electric charge."},{"Start":"01:31.795 ","End":"01:34.070","Text":"We can see on the right we have nitrogen,"},{"Start":"01:34.070 ","End":"01:37.500","Text":"meaning the oxidation state is 0."},{"Start":"01:37.790 ","End":"01:42.740","Text":"On the left, the oxidation state of nitrogen is, let\u0027s calculate it."},{"Start":"01:42.740 ","End":"01:45.335","Text":"We have nitrogen and oxidation state of nitrogen."},{"Start":"01:45.335 ","End":"01:50.220","Text":"Plus we have 3 hydrogens,"},{"Start":"01:50.220 ","End":"01:52.125","Text":"so it\u0027s going to be plus 3,"},{"Start":"01:52.125 ","End":"01:56.250","Text":"and we have 1 oxygen which has a minus 2 oxidation state."},{"Start":"01:56.250 ","End":"02:01.925","Text":"It\u0027s going to be minus 2 equals 0 since we\u0027re talking about a neutral molecule."},{"Start":"02:01.925 ","End":"02:09.055","Text":"The oxidation state of the nitrogen in this molecule comes out 2 minus 1."},{"Start":"02:09.055 ","End":"02:15.560","Text":"The oxidation state of nitrogen on the left is minus 1 and on the right is a 0."},{"Start":"02:15.560 ","End":"02:18.570","Text":"But remember we have 2 nitrogens."},{"Start":"02:18.680 ","End":"02:20.810","Text":"If we had 1 nitrogen,"},{"Start":"02:20.810 ","End":"02:23.780","Text":"we would have to add 1 electron to our right side because"},{"Start":"02:23.780 ","End":"02:27.380","Text":"0 electrons plus 1 is going to give us our minus 1 on the left side,"},{"Start":"02:27.380 ","End":"02:29.735","Text":"however, we have 2 nitrogens."},{"Start":"02:29.735 ","End":"02:35.770","Text":"Therefore, we\u0027re going to add 2 electrons to the right side of our equation."},{"Start":"02:35.770 ","End":"02:37.925","Text":"Now at this point,"},{"Start":"02:37.925 ","End":"02:42.360","Text":"if we had our equation in acidic medium,"},{"Start":"02:43.390 ","End":"02:45.680","Text":"that would be our final step."},{"Start":"02:45.680 ","End":"02:47.900","Text":"However, we have it in basic solution,"},{"Start":"02:47.900 ","End":"02:50.460","Text":"meaning this is basic medium."},{"Start":"02:51.340 ","End":"02:54.960","Text":"We\u0027re going to write the equation again."},{"Start":"03:12.050 ","End":"03:16.070","Text":"Now let\u0027s change the equation from acidic medium to basic medium,"},{"Start":"03:16.070 ","End":"03:17.600","Text":"which is what we need to do here."},{"Start":"03:17.600 ","End":"03:20.150","Text":"What we need to do is add"},{"Start":"03:20.150 ","End":"03:25.775","Text":"hydroxides with the same number of protons as we have in our equation meaning,"},{"Start":"03:25.775 ","End":"03:29.695","Text":"in this equation we have 2 protons, 2H plus."},{"Start":"03:29.695 ","End":"03:36.465","Text":"We need to add 2 hydroxides to the right side of our equation."},{"Start":"03:36.465 ","End":"03:38.420","Text":"Obviously, we\u0027re going to have to add them to"},{"Start":"03:38.420 ","End":"03:41.630","Text":"the left side because they have to be equal."},{"Start":"03:41.630 ","End":"03:44.190","Text":"We\u0027re not going to change our equation."},{"Start":"03:45.350 ","End":"03:48.620","Text":"When we add these 2 hydroxides to the protons,"},{"Start":"03:48.620 ","End":"03:50.840","Text":"what we get is a water molecule."},{"Start":"03:50.840 ","End":"03:53.495","Text":"We have 2 protons plus 2 hydroxides."},{"Start":"03:53.495 ","End":"03:58.050","Text":"That\u0027s going to give us 2 water molecules."},{"Start":"03:58.940 ","End":"04:01.185","Text":"Now we need to just simplify."},{"Start":"04:01.185 ","End":"04:08.250","Text":"On the left side we have 2 of them,"},{"Start":"04:08.250 ","End":"04:12.925","Text":"plus 2OH minus, which is 2 hydroxides."},{"Start":"04:12.925 ","End":"04:16.195","Text":"This is going to give us nitrogen."},{"Start":"04:16.195 ","End":"04:20.460","Text":"Plus we can see that we have 2 water molecules plus 2 more water molecules."},{"Start":"04:20.460 ","End":"04:27.640","Text":"It\u0027s going to be a plus 4 water molecules plus 2 electrons."},{"Start":"04:27.640 ","End":"04:31.980","Text":"That\u0027s our final balanced equation."},{"Start":"04:33.040 ","End":"04:37.070","Text":"Again, when you need to balance the equation in basic medium,"},{"Start":"04:37.070 ","End":"04:42.590","Text":"what we do is we balance the equation in acidic medium, the regular way."},{"Start":"04:42.590 ","End":"04:46.040","Text":"The last step is to add the number of hydroxides,"},{"Start":"04:46.040 ","End":"04:48.200","Text":"which equals the number of protons which we have in"},{"Start":"04:48.200 ","End":"04:53.600","Text":"our equation to both sides of the equation, and to simplify."},{"Start":"04:53.600 ","End":"04:55.820","Text":"That\u0027s our answer for b,"},{"Start":"04:55.820 ","End":"04:58.229","Text":"and now we\u0027re going to go on to c."}],"ID":23723},{"Watched":false,"Name":"Exercise 2 - Part c","Duration":"3m 35s","ChapterTopicVideoID":22903,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22903.jpeg","UploadDate":"2020-12-16T02:09:38.8030000","DurationForVideoObject":"PT3M35S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.680 ","End":"00:05.370","Text":"Now we\u0027re going to go on to c. C we"},{"Start":"00:05.370 ","End":"00:10.470","Text":"have iron on our left side and on the right side we have iron hydroxide."},{"Start":"00:10.470 ","End":"00:14.490","Text":"Again, we\u0027re going to balance the elements other than the oxygens and the hydrogens,"},{"Start":"00:14.490 ","End":"00:17.730","Text":"meaning the iron and then in this case we have 1 on our left side,"},{"Start":"00:17.730 ","End":"00:19.665","Text":"1 on the right side, so they\u0027re balanced."},{"Start":"00:19.665 ","End":"00:21.690","Text":"Now we\u0027re going to go to the oxygens to balance"},{"Start":"00:21.690 ","End":"00:23.490","Text":"them and when we have a 2 on the right side,"},{"Start":"00:23.490 ","End":"00:24.990","Text":"that\u0027s 1 times 2."},{"Start":"00:24.990 ","End":"00:27.300","Text":"On our left side, we have none."},{"Start":"00:27.300 ","End":"00:36.485","Text":"We\u0027re going to add 2 water molecules in order to add 2 oxygen to our left side."},{"Start":"00:36.485 ","End":"00:38.765","Text":"The next step is to balance the hydrogen."},{"Start":"00:38.765 ","End":"00:40.220","Text":"On the left side, now we have 4,"},{"Start":"00:40.220 ","End":"00:44.090","Text":"since we have 2 hydrogens times 2 that gives us 4 hydrogens."},{"Start":"00:44.090 ","End":"00:47.455","Text":"On the right side, we have 2 hydrogens because that\u0027s 1 times 2."},{"Start":"00:47.455 ","End":"00:51.670","Text":"We\u0027re going to add 2 protons to our right side."},{"Start":"00:52.030 ","End":"00:55.760","Text":"At this point we have the iron, the oxygen,"},{"Start":"00:55.760 ","End":"00:57.530","Text":"and the hydrogen is balanced on both sides"},{"Start":"00:57.530 ","End":"00:59.660","Text":"and we\u0027re going to balance for electric charge."},{"Start":"00:59.660 ","End":"01:01.550","Text":"On the left side of the equation again,"},{"Start":"01:01.550 ","End":"01:03.230","Text":"if we look our hydrogens and oxygens,"},{"Start":"01:03.230 ","End":"01:06.590","Text":"we only have a plus 1 oxidation state for our hydrogens"},{"Start":"01:06.590 ","End":"01:10.980","Text":"and a minus 2 oxidation state for our oxygens."},{"Start":"01:11.260 ","End":"01:14.310","Text":"That doesn\u0027t change."},{"Start":"01:14.590 ","End":"01:20.820","Text":"What is going to change is our oxidation state for the iron."},{"Start":"01:20.820 ","End":"01:25.610","Text":"On the left, we have an oxidation state of 0 and on the right,"},{"Start":"01:25.610 ","End":"01:33.140","Text":"let\u0027s see the oxidation state of iron and now we have 2 hydroxides."},{"Start":"01:33.140 ","End":"01:37.265","Text":"As you can see, hydroxide is minus 2 plus 1,"},{"Start":"01:37.265 ","End":"01:39.860","Text":"because we have 1 oxygen and 1 hydrogen."},{"Start":"01:39.860 ","End":"01:43.490","Text":"That\u0027s going to give us some minus 1 times 2,"},{"Start":"01:43.490 ","End":"01:49.250","Text":"going to be plus 2 times minus 1 and the overall molecule is neutral,"},{"Start":"01:49.250 ","End":"01:50.945","Text":"so it\u0027s going to give us a 0."},{"Start":"01:50.945 ","End":"01:56.135","Text":"The oxidation state of iron in this case is going to equal plus 2."},{"Start":"01:56.135 ","End":"02:00.820","Text":"We can see that on the left side of the equation we have an oxidation state of 0 for"},{"Start":"02:00.820 ","End":"02:04.720","Text":"iron and on the right side we have an oxidation state of plus 2,"},{"Start":"02:04.720 ","End":"02:07.420","Text":"meaning we need to add 2 electrons to the right side of"},{"Start":"02:07.420 ","End":"02:12.865","Text":"our equation in order to balance electric charge."},{"Start":"02:12.865 ","End":"02:14.680","Text":"Now again, if we look at c,"},{"Start":"02:14.680 ","End":"02:17.620","Text":"we can see that this equation is in basic solution,"},{"Start":"02:17.620 ","End":"02:22.270","Text":"basic medium, meaning that we need to add now hydroxides."},{"Start":"02:22.270 ","End":"02:25.000","Text":"As you can see, we have 2 protons on our right side,"},{"Start":"02:25.000 ","End":"02:26.890","Text":"so we\u0027re going to add 2 hydroxides to"},{"Start":"02:26.890 ","End":"02:31.210","Text":"this side and we\u0027re going to add 2 hydroxides to the other side because you"},{"Start":"02:31.210 ","End":"02:37.860","Text":"can\u0027t just add hydroxides to one sides of the equation,"},{"Start":"02:37.860 ","End":"02:39.285","Text":"it\u0027s going to change your equation."},{"Start":"02:39.285 ","End":"02:42.695","Text":"We\u0027re going to add this to both sides this way it doesn\u0027t change our equation."},{"Start":"02:42.695 ","End":"02:46.460","Text":"Now the protons plus the hydroxide is going to give us water."},{"Start":"02:46.460 ","End":"02:49.770","Text":"Instead of 2 protons and 2 hydroxides,"},{"Start":"02:49.770 ","End":"02:52.540","Text":"we\u0027re going to get 2 water molecules."},{"Start":"02:53.800 ","End":"02:57.155","Text":"Now our next step is to simplify."},{"Start":"02:57.155 ","End":"03:00.200","Text":"If you look, you can see that on the right side of the equation we have"},{"Start":"03:00.200 ","End":"03:03.470","Text":"2 water molecules and on the left side we also have 2 water molecules,"},{"Start":"03:03.470 ","End":"03:08.025","Text":"meaning I\u0027m going to take these off of our equation"},{"Start":"03:08.025 ","End":"03:15.070","Text":"and we\u0027re going to be left with iron plus 2 hydroxides,"},{"Start":"03:16.220 ","End":"03:20.620","Text":"is going to give us iron hydroxide."},{"Start":"03:21.320 ","End":"03:28.305","Text":"What we have left here is plus 2 electrons."},{"Start":"03:28.305 ","End":"03:35.440","Text":"That\u0027s our answer for c. Now we\u0027re going to go on to d."}],"ID":23724},{"Watched":false,"Name":"Exercise 2 - Part d","Duration":"2m 40s","ChapterTopicVideoID":22895,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22895.jpeg","UploadDate":"2020-12-16T02:07:11.6400000","DurationForVideoObject":"PT2M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:06.990","Text":"Now we\u0027re going to take a look at d. On the left side of the equation,"},{"Start":"00:06.990 ","End":"00:10.950","Text":"we have our sulfate ion and on the right side we have sulfur dioxide."},{"Start":"00:10.950 ","End":"00:13.890","Text":"Again, we\u0027re going to take a look at the elements other than oxygen and hydrogen."},{"Start":"00:13.890 ","End":"00:15.660","Text":"We have 1 sulfur atom on the left,"},{"Start":"00:15.660 ","End":"00:17.910","Text":"we have 1 sulfur on the right."},{"Start":"00:17.910 ","End":"00:20.430","Text":"The sulfur is balanced."},{"Start":"00:20.430 ","End":"00:22.200","Text":"Now we\u0027re going to take a look at our oxygens."},{"Start":"00:22.200 ","End":"00:24.750","Text":"As you can see, we have 4 on our left and 2 on the right,"},{"Start":"00:24.750 ","End":"00:27.780","Text":"so we\u0027re going to add 2 water molecules"},{"Start":"00:27.780 ","End":"00:31.200","Text":"to the right side of the equation in order to balance the number of oxygens."},{"Start":"00:31.200 ","End":"00:33.690","Text":"Now we have 4 on our left and 4 on the right."},{"Start":"00:33.690 ","End":"00:37.125","Text":"The next step is to balance the number of hydrogens."},{"Start":"00:37.125 ","End":"00:39.175","Text":"We have 4 on the right side,"},{"Start":"00:39.175 ","End":"00:42.800","Text":"2 times 2, and we have none left side,"},{"Start":"00:42.800 ","End":"00:45.960","Text":"so we\u0027re just going to add 4 protons."},{"Start":"00:46.450 ","End":"00:51.230","Text":"Now that our sulfur or oxygen and our hydrogen is balanced,"},{"Start":"00:51.230 ","End":"00:54.350","Text":"the next step is to balance for electric charge."},{"Start":"00:54.350 ","End":"00:57.125","Text":"Let\u0027s calculate the oxidation states."},{"Start":"00:57.125 ","End":"01:00.080","Text":"If we take the oxidation state of sulfur."},{"Start":"01:00.080 ","End":"01:01.550","Text":"Here we have 4 oxygens,"},{"Start":"01:01.550 ","End":"01:04.640","Text":"meaning we add 4 times the oxidation state of oxygen,"},{"Start":"01:04.640 ","End":"01:06.185","Text":"which is minus 2."},{"Start":"01:06.185 ","End":"01:12.315","Text":"This is going to give us a minus 2 charge since that\u0027s what we have on our ion."},{"Start":"01:12.315 ","End":"01:16.220","Text":"The sulfur minus 8 is going to give us a minus 2,"},{"Start":"01:16.220 ","End":"01:20.560","Text":"meaning the sulfur oxidation state is going to be a plus 6."},{"Start":"01:20.560 ","End":"01:23.855","Text":"That\u0027s for the left side of the equation."},{"Start":"01:23.855 ","End":"01:29.900","Text":"If we look at the right side of the equation at our sulfur dioxide,"},{"Start":"01:29.900 ","End":"01:34.805","Text":"we have the oxidation state of sulfur plus 2 oxygens times minus 2,"},{"Start":"01:34.805 ","End":"01:36.515","Text":"which is the oxidation state of oxygen,"},{"Start":"01:36.515 ","End":"01:39.480","Text":"is going to give us a 0 since we have a neutral molecule."},{"Start":"01:39.480 ","End":"01:44.920","Text":"The oxidation state of sulfur in this case it\u0027s going to be plus 4."},{"Start":"01:44.920 ","End":"01:48.110","Text":"Since we have sulfur minus 4 giving us 0."},{"Start":"01:48.110 ","End":"01:50.820","Text":"That\u0027s going to be a plus 4."},{"Start":"01:51.730 ","End":"01:56.600","Text":"The oxidation state of sulfur on the left side of the equations is plus 6,"},{"Start":"01:56.600 ","End":"01:58.655","Text":"and on the right side of the equation is plus 4."},{"Start":"01:58.655 ","End":"02:01.040","Text":"We have 1 sulfur atom,"},{"Start":"02:01.040 ","End":"02:04.300","Text":"meaning we need to add 2 electrons to"},{"Start":"02:04.300 ","End":"02:09.540","Text":"the left side of our equation in order to balance the electric charge."},{"Start":"02:09.820 ","End":"02:14.300","Text":"Now remember in d we need it in acidic solution, in acidic medium,"},{"Start":"02:14.300 ","End":"02:16.700","Text":"meaning that this is the last step of"},{"Start":"02:16.700 ","End":"02:20.120","Text":"balancing our equation and our equation is balanced."},{"Start":"02:20.120 ","End":"02:22.610","Text":"If we take a look at our balanced equation,"},{"Start":"02:22.610 ","End":"02:27.305","Text":"we have 2 electrons plus our sulfate ion plus"},{"Start":"02:27.305 ","End":"02:34.510","Text":"4 protons is going to give us sulfur dioxide plus 2 water molecules."},{"Start":"02:34.510 ","End":"02:41.070","Text":"That\u0027s our balanced equation for d. Thank you very much for watching."}],"ID":23716},{"Watched":false,"Name":"Exercise 3 - Part a","Duration":"10m 16s","ChapterTopicVideoID":22896,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22896.jpeg","UploadDate":"2020-12-16T02:07:37.7500000","DurationForVideoObject":"PT10M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:03.885","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.885 ","End":"00:10.120","Text":"Balance the following equations for redox reactions occurring in acidic solution."},{"Start":"00:10.610 ","End":"00:13.080","Text":"We\u0027re going to start with a."},{"Start":"00:13.080 ","End":"00:16.140","Text":"Before we begin, I just want to remind you that there are a number of"},{"Start":"00:16.140 ","End":"00:19.019","Text":"different ways to balance redox reactions,"},{"Start":"00:19.019 ","End":"00:21.425","Text":"and we\u0027re just going to use one,"},{"Start":"00:21.425 ","End":"00:24.345","Text":"but if you see other ways just keep that in mind."},{"Start":"00:24.345 ","End":"00:27.440","Text":"In a we have hydrogen peroxide plus"},{"Start":"00:27.440 ","End":"00:33.200","Text":"the dichromate ion gives us oxygen plus the chromium ion."},{"Start":"00:33.200 ","End":"00:36.440","Text":"The first step in balancing"},{"Start":"00:36.440 ","End":"00:40.615","Text":"this redox reaction is going to be dividing it into 2 half equations."},{"Start":"00:40.615 ","End":"00:44.840","Text":"We can call them skeletal half equations. Let\u0027s see."},{"Start":"00:44.990 ","End":"00:48.200","Text":"We have a. One of the equations is going to"},{"Start":"00:48.200 ","End":"00:52.230","Text":"be hydrogen peroxide is going to give us oxygen."},{"Start":"00:56.210 ","End":"01:00.200","Text":"We will write the second one over here,"},{"Start":"01:00.200 ","End":"01:05.005","Text":"and that\u0027s the dichromate ion,"},{"Start":"01:05.005 ","End":"01:10.790","Text":"it\u0027s going to give us the chromium ion."},{"Start":"01:14.910 ","End":"01:19.375","Text":"We\u0027re going to balance these just as we did in the last exercise."},{"Start":"01:19.375 ","End":"01:24.670","Text":"Remember the first step is to balance the elements which are not oxygen and hydrogen."},{"Start":"01:24.670 ","End":"01:27.430","Text":"In one we don\u0027t have that step."},{"Start":"01:27.430 ","End":"01:31.965","Text":"The second step is to balance the oxygens by adding water."},{"Start":"01:31.965 ","End":"01:34.060","Text":"But here if you look on the left side we have"},{"Start":"01:34.060 ","End":"01:37.180","Text":"2 oxygens and on the other side we also have 2 oxygens."},{"Start":"01:37.180 ","End":"01:42.490","Text":"We\u0027re going to go to the third step which is balancing the hydrogens by adding protons."},{"Start":"01:42.490 ","End":"01:44.530","Text":"On the left we have 2 hydrogens,"},{"Start":"01:44.530 ","End":"01:49.100","Text":"and we\u0027re just going to add 2 protons to the right side of our equation."},{"Start":"01:51.170 ","End":"01:54.680","Text":"Now we\u0027re going to balance for electric charge."},{"Start":"01:54.680 ","End":"01:58.250","Text":"If we look at our oxidation states, let\u0027s start with the right."},{"Start":"01:58.250 ","End":"02:04.160","Text":"We can see that the hydrogen has a plus 1 and our oxygen is going to have a 0."},{"Start":"02:04.160 ","End":"02:05.960","Text":"If we look at the left,"},{"Start":"02:05.960 ","End":"02:10.645","Text":"we see that the hydrogen here is also plus 1."},{"Start":"02:10.645 ","End":"02:16.865","Text":"Therefore, the oxygen which is usually minus 2 gets a minus 1 oxidation state."},{"Start":"02:16.865 ","End":"02:19.760","Text":"The hydrogen\u0027s oxidation state doesn\u0027t change,"},{"Start":"02:19.760 ","End":"02:21.140","Text":"but the oxygen does change."},{"Start":"02:21.140 ","End":"02:25.610","Text":"We have minus 1 on the left side and 0 on the right side,"},{"Start":"02:25.610 ","End":"02:27.965","Text":"meaning we need to add electrons to the right side."},{"Start":"02:27.965 ","End":"02:31.055","Text":"We need to add 1 electron per oxygen atom,"},{"Start":"02:31.055 ","End":"02:35.550","Text":"and we have 2, therefore we\u0027re going to add 2 electrons."},{"Start":"02:36.530 ","End":"02:41.530","Text":"Now the first step is balanced and we\u0027re going to go on to the second step."},{"Start":"02:41.530 ","End":"02:44.085","Text":"We\u0027re going to balance it in the exact same way."},{"Start":"02:44.085 ","End":"02:47.595","Text":"We\u0027re going to start with the chromium atom."},{"Start":"02:47.595 ","End":"02:50.580","Text":"You see we have 2 on the left and only 1 on the right,"},{"Start":"02:50.580 ","End":"02:53.415","Text":"so we\u0027re going to multiply the right side by 2."},{"Start":"02:53.415 ","End":"02:55.820","Text":"That is the first step. Now we\u0027re going to look at our oxygens."},{"Start":"02:55.820 ","End":"02:59.780","Text":"We see that we have 7 oxygens on our left side and none on our right,"},{"Start":"02:59.780 ","End":"03:06.935","Text":"so we\u0027re going to add 7 water molecules to the right side of the equation."},{"Start":"03:06.935 ","End":"03:09.185","Text":"Now we\u0027re going to balance for the hydrogens."},{"Start":"03:09.185 ","End":"03:14.015","Text":"Now we see that we have 14 hydrogens on the right side,"},{"Start":"03:14.015 ","End":"03:15.500","Text":"none on the left side."},{"Start":"03:15.500 ","End":"03:22.545","Text":"So we\u0027re going to add 14 protons to our left side."},{"Start":"03:22.545 ","End":"03:26.275","Text":"What we have left is to balance for electric charge."},{"Start":"03:26.275 ","End":"03:31.390","Text":"If we look here, our oxygen is going to be minus 2,"},{"Start":"03:31.390 ","End":"03:33.270","Text":"and our chromium, let\u0027s see,"},{"Start":"03:33.270 ","End":"03:36.260","Text":"we have 2 times oxidation state of chromium plus"},{"Start":"03:36.260 ","End":"03:40.385","Text":"7 times oxidation state of oxygen which is minus 2,"},{"Start":"03:40.385 ","End":"03:42.410","Text":"is going to give us a minus 2 because"},{"Start":"03:42.410 ","End":"03:46.075","Text":"the overall charge of the dichromate ion is minus 2."},{"Start":"03:46.075 ","End":"03:49.880","Text":"2 times oxidation state of chromium is going give us 12,"},{"Start":"03:49.880 ","End":"03:54.065","Text":"an oxidation state of chromium is going to be plus 6."},{"Start":"03:54.065 ","End":"03:56.350","Text":"Oxidation state of plus 6,"},{"Start":"03:56.350 ","End":"03:58.655","Text":"oxygen is minus 2 as usual,"},{"Start":"03:58.655 ","End":"04:00.260","Text":"and hydrogen is plus 1."},{"Start":"04:00.260 ","End":"04:02.840","Text":"On the other side we also have hydrogen plus 1,"},{"Start":"04:02.840 ","End":"04:05.430","Text":"oxygen minus 2 as usual."},{"Start":"04:05.680 ","End":"04:09.679","Text":"The only atom which changes its oxidation state is our chromium."},{"Start":"04:09.679 ","End":"04:12.710","Text":"Again, we can see that on the left side we have plus 6,"},{"Start":"04:12.710 ","End":"04:14.935","Text":"on the right side we have plus 3,"},{"Start":"04:14.935 ","End":"04:20.315","Text":"meaning we need to add 3 electrons to each chromium atom on the left side."},{"Start":"04:20.315 ","End":"04:22.760","Text":"However, remember we have 2 chromium atoms,"},{"Start":"04:22.760 ","End":"04:25.770","Text":"therefore, we\u0027re going to add 6 electrons."},{"Start":"04:28.100 ","End":"04:30.650","Text":"We have our 2 half equations balanced."},{"Start":"04:30.650 ","End":"04:34.040","Text":"The next step we have to do is to have"},{"Start":"04:34.040 ","End":"04:38.985","Text":"the same number of electrons for both half equations."},{"Start":"04:38.985 ","End":"04:43.355","Text":"For example, we can see that in equation 2 we have"},{"Start":"04:43.355 ","End":"04:47.915","Text":"6 electrons on our left side and in equation 1 we have 2 electrons on the right side."},{"Start":"04:47.915 ","End":"04:50.940","Text":"We want these electrons to cancel out at the end."},{"Start":"04:50.940 ","End":"04:52.395","Text":"What we\u0027re going to do is we\u0027re going to take"},{"Start":"04:52.395 ","End":"04:58.015","Text":"our first half equation and we\u0027re just going to multiply the whole equation by 3."},{"Start":"04:58.015 ","End":"05:01.731","Text":"That way we will have 6 electrons on the right side,"},{"Start":"05:01.731 ","End":"05:05.900","Text":"and in the second equation we have 6 on the left,"},{"Start":"05:05.900 ","End":"05:07.235","Text":"so they\u0027re going to cancel out."},{"Start":"05:07.235 ","End":"05:11.130","Text":"Let\u0027s just do this. I\u0027m going to multiply"},{"Start":"05:11.130 ","End":"05:16.025","Text":"the first half equation by 3 and we\u0027re going to add it to the second half equation."},{"Start":"05:16.025 ","End":"05:19.340","Text":"This is going to be 3 times hydrogen peroxide."},{"Start":"05:19.340 ","End":"05:21.063","Text":"You know what? Let\u0027s just write it down here,"},{"Start":"05:21.063 ","End":"05:22.550","Text":"it\u0027s going to be more comfortable assists."},{"Start":"05:22.550 ","End":"05:26.660","Text":"3 times hydrogen peroxide is going to give us"},{"Start":"05:26.660 ","End":"05:36.425","Text":"3 oxygen plus 6 protons plus 6 electrons."},{"Start":"05:36.425 ","End":"05:42.700","Text":"Now at this point we can add the first half equation to the second one."},{"Start":"05:42.700 ","End":"05:46.990","Text":"This is going to give us 3 hydrogen peroxide,"},{"Start":"05:47.780 ","End":"05:51.610","Text":"plus the dichromate ion,"},{"Start":"05:53.630 ","End":"06:02.260","Text":"plus 14 protons, plus 6 electrons."},{"Start":"06:04.130 ","End":"06:08.680","Text":"It\u0027s going to give us 3 oxygen"},{"Start":"06:09.230 ","End":"06:16.175","Text":"plus 6 protons plus 6 electrons."},{"Start":"06:16.175 ","End":"06:27.370","Text":"I\u0027m just going to continue here, plus 2 chromium ions plus 7 water molecules."},{"Start":"06:27.540 ","End":"06:30.310","Text":"First of all, again, we can see that we have"},{"Start":"06:30.310 ","End":"06:32.425","Text":"the 6 electrons that are left side and on the right."},{"Start":"06:32.425 ","End":"06:34.705","Text":"That\u0027s just going to cancel out."},{"Start":"06:34.705 ","End":"06:37.000","Text":"Also if you look you could see that we have"},{"Start":"06:37.000 ","End":"06:39.589","Text":"6 protons on our right side and 14 on our left side,"},{"Start":"06:39.589 ","End":"06:42.400","Text":"so we\u0027re just going to subtract 6 protons from each side,"},{"Start":"06:42.400 ","End":"06:51.510","Text":"and this is going to give us a total of 8 protons on the left side."},{"Start":"07:01.220 ","End":"07:08.110","Text":"What we\u0027re left with is 3 hydrogen peroxide plus the dichromate ion,"},{"Start":"07:08.210 ","End":"07:12.130","Text":"plus 8 protons,"},{"Start":"07:12.530 ","End":"07:17.760","Text":"it\u0027s going to give us 3 oxygen,"},{"Start":"07:17.760 ","End":"07:23.790","Text":"plus 2 chromium ions"},{"Start":"07:23.790 ","End":"07:29.524","Text":"plus 7 water molecules."},{"Start":"07:29.524 ","End":"07:33.815","Text":"Just remember also this is an acidic medium."},{"Start":"07:33.815 ","End":"07:36.020","Text":"We don\u0027t have to add hydroxide at this point,"},{"Start":"07:36.020 ","End":"07:38.185","Text":"we can leave it the way it is."},{"Start":"07:38.185 ","End":"07:39.915","Text":"Our answer for a is,"},{"Start":"07:39.915 ","End":"07:45.000","Text":"3 hydrogen peroxide plus the dichromate ion plus 8 protons is going to"},{"Start":"07:45.000 ","End":"07:50.405","Text":"give us 3 oxygen plus 2 chromium ion plus 7 water molecules."},{"Start":"07:50.405 ","End":"07:52.180","Text":"That is our final answer for a,"},{"Start":"07:52.180 ","End":"07:54.515","Text":"now we\u0027re going to take a look at b."},{"Start":"07:54.515 ","End":"07:57.560","Text":"However, before we take a look at b,"},{"Start":"07:57.560 ","End":"08:00.485","Text":"there is something that we should do."},{"Start":"08:00.485 ","End":"08:03.920","Text":"At this point we can check for charge and"},{"Start":"08:03.920 ","End":"08:07.790","Text":"atoms to see that our reaction really is balanced,"},{"Start":"08:07.790 ","End":"08:10.580","Text":"that we\u0027ve balanced it like it was supposed to be."},{"Start":"08:10.580 ","End":"08:11.960","Text":"If we look at the atoms,"},{"Start":"08:11.960 ","End":"08:13.670","Text":"we can see that on both sides we have,"},{"Start":"08:13.670 ","End":"08:19.785","Text":"if we look at our chromium we have 2 chromium atoms, 2 and 2."},{"Start":"08:19.785 ","End":"08:22.730","Text":"Now if we look at the number of oxygens,"},{"Start":"08:22.730 ","End":"08:27.650","Text":"we can see that on the left side we have 3 times 2 which is 6 plus 7."},{"Start":"08:27.650 ","End":"08:30.300","Text":"Meaning we have 13 oxygens."},{"Start":"08:30.920 ","End":"08:37.610","Text":"On the right side we have 3 times 2 which is 6 plus 7 times 1,"},{"Start":"08:37.610 ","End":"08:40.475","Text":"which again gives us 13 oxygen."},{"Start":"08:40.475 ","End":"08:45.330","Text":"Again, on both sides we have 2 chromium, 13 oxygen."},{"Start":"08:46.510 ","End":"08:48.800","Text":"Let\u0027s look at our hydrogens,"},{"Start":"08:48.800 ","End":"08:52.610","Text":"we have 3 times 2 which is 6 hydrogens plus 8."},{"Start":"08:52.610 ","End":"08:58.724","Text":"That\u0027s giving us a total of 14 hydrogen atoms on the left side of the equation."},{"Start":"08:58.724 ","End":"09:01.180","Text":"On the right side we have again,"},{"Start":"09:01.180 ","End":"09:03.190","Text":"2 times 7 hydrogens,"},{"Start":"09:03.190 ","End":"09:05.255","Text":"which is also 14."},{"Start":"09:05.255 ","End":"09:09.290","Text":"For atoms we can see that this equation is definitely balanced."},{"Start":"09:09.290 ","End":"09:12.200","Text":"Now we\u0027re going to take a look at our charge to see if we have"},{"Start":"09:12.200 ","End":"09:16.025","Text":"the same charge on the left and the right side of our equation."},{"Start":"09:16.025 ","End":"09:19.640","Text":"If you take a look at the left side of it reaction,"},{"Start":"09:19.640 ","End":"09:29.340","Text":"we can see that we have a minus 2 for the dichromate ion plus 8 times plus 1."},{"Start":"09:29.810 ","End":"09:35.925","Text":"Since we have 8 atoms of H plus of the protons."},{"Start":"09:35.925 ","End":"09:38.485","Text":"That\u0027s plus 8 times plus 1."},{"Start":"09:38.485 ","End":"09:43.115","Text":"This is going to give us a total of 6 because we have minus 2 plus 8, so that\u0027s 6."},{"Start":"09:43.115 ","End":"09:47.810","Text":"We want to see if this is equal to the charge on the right side."},{"Start":"09:47.810 ","End":"09:52.130","Text":"On the right side we can see we only have charge from our chromium ion."},{"Start":"09:52.130 ","End":"09:57.120","Text":"Here we have 2 times plus 3 charge."},{"Start":"09:58.040 ","End":"10:01.245","Text":"That\u0027s also giving us a 6."},{"Start":"10:01.245 ","End":"10:03.475","Text":"We can see that 6 equals 6."},{"Start":"10:03.475 ","End":"10:07.570","Text":"We know that the reaction is balanced also for atoms and also for charge."},{"Start":"10:07.570 ","End":"10:12.715","Text":"At this point, we can say that we have a balanced redox reaction."},{"Start":"10:12.715 ","End":"10:16.420","Text":"That\u0027s all for a, now we\u0027re going to go on to b."}],"ID":23717},{"Watched":false,"Name":"Exercise 3 - Part b","Duration":"9m 21s","ChapterTopicVideoID":22897,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22897.jpeg","UploadDate":"2020-12-16T02:08:01.7070000","DurationForVideoObject":"PT9M21S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.265","Text":"Now we\u0027re going to take a look at b."},{"Start":"00:02.265 ","End":"00:09.615","Text":"In b we have lead oxide plus iodine gives us the lead ion plus iodate ion."},{"Start":"00:09.615 ","End":"00:14.129","Text":"Let\u0027s begin by dividing this into 2 half equations."},{"Start":"00:14.129 ","End":"00:17.790","Text":"First of all, we\u0027re going to have the lead oxide,"},{"Start":"00:17.790 ","End":"00:21.220","Text":"it\u0027s going to give us the lead ion."},{"Start":"00:22.340 ","End":"00:27.520","Text":"Second equation is going to be iodine is going to give us the iodide ion."},{"Start":"00:33.400 ","End":"00:37.410","Text":"We\u0027re going to balance the first half equation."},{"Start":"00:37.510 ","End":"00:39.890","Text":"First of all, if we look at our lead,"},{"Start":"00:39.890 ","End":"00:41.960","Text":"we have 1 on the right side and 1 on the left side,"},{"Start":"00:41.960 ","End":"00:43.325","Text":"so we don\u0027t need to balance the led."},{"Start":"00:43.325 ","End":"00:44.570","Text":"We\u0027re going to go to the oxygen,"},{"Start":"00:44.570 ","End":"00:47.280","Text":"so we have 2 on our left side and none on our right,"},{"Start":"00:47.280 ","End":"00:51.455","Text":"so we\u0027re going to add 2 water molecules in order to balance the oxygens."},{"Start":"00:51.455 ","End":"00:53.450","Text":"Next step is to balance the hydrogens."},{"Start":"00:53.450 ","End":"00:56.270","Text":"We have 4 on our right side which is 2 times 2 and we have"},{"Start":"00:56.270 ","End":"01:00.450","Text":"none on our left side and we\u0027re going to add 4 protons to our left side."},{"Start":"01:00.790 ","End":"01:05.360","Text":"Last but not least, we\u0027re going to balance the electric charge."},{"Start":"01:05.360 ","End":"01:08.070","Text":"If we take a look, we can see that hydrogen is plus 1"},{"Start":"01:08.070 ","End":"01:11.105","Text":"on our left side and oxygen is minus 2."},{"Start":"01:11.105 ","End":"01:13.420","Text":"Unless I can look at the oxidation state of the lead."},{"Start":"01:13.420 ","End":"01:16.280","Text":"We have the oxidation state of the lead plus 2,"},{"Start":"01:16.280 ","End":"01:18.780","Text":"which is 2 oxygens times minus 2,"},{"Start":"01:18.780 ","End":"01:20.710","Text":"which is the oxidation state of oxygen,"},{"Start":"01:20.710 ","End":"01:26.780","Text":"is going to give us 0 because our lead oxide is neutral."},{"Start":"01:28.140 ","End":"01:36.370","Text":"The oxidation state of lead is going to equal plus 4."},{"Start":"01:37.470 ","End":"01:41.590","Text":"Oxidation state of lead is plus 4 on"},{"Start":"01:41.590 ","End":"01:45.025","Text":"our left side of our equation and on the rest of it we have a plus 2,"},{"Start":"01:45.025 ","End":"01:46.570","Text":"the hydrogen is still plus 1,"},{"Start":"01:46.570 ","End":"01:49.130","Text":"and oxygen is still minus 2."},{"Start":"01:50.780 ","End":"01:54.040","Text":"In order to balance the electric charge,"},{"Start":"01:54.040 ","End":"01:56.020","Text":"we\u0027re just going to take a look at our lead."},{"Start":"01:56.020 ","End":"01:58.705","Text":"We have plus 4 on the left side,"},{"Start":"01:58.705 ","End":"01:59.960","Text":"plus 2 on the right side,"},{"Start":"01:59.960 ","End":"02:06.880","Text":"meaning that for each lead and we have to add 2 electrons to the left side."},{"Start":"02:07.430 ","End":"02:10.035","Text":"Since we have only 1 lead atom,"},{"Start":"02:10.035 ","End":"02:11.985","Text":"we\u0027re going to add 2 electrons."},{"Start":"02:11.985 ","End":"02:15.080","Text":"At this point the first half equation is balanced."},{"Start":"02:15.080 ","End":"02:17.150","Text":"We are going to look at the second one."},{"Start":"02:17.150 ","End":"02:18.680","Text":"If we look at our left side of our equation,"},{"Start":"02:18.680 ","End":"02:21.740","Text":"we have 2 iodines and on the right side we only have 1."},{"Start":"02:21.740 ","End":"02:25.430","Text":"Therefore, we\u0027re going to multiply the iodate ion by 2."},{"Start":"02:25.430 ","End":"02:28.910","Text":"At this point, we\u0027re going to balance the oxygens here on"},{"Start":"02:28.910 ","End":"02:32.770","Text":"the right side we have 6 oxygens and switches 3 times 2 is 6."},{"Start":"02:32.770 ","End":"02:34.060","Text":"On the left side we have none,"},{"Start":"02:34.060 ","End":"02:37.280","Text":"so we\u0027re going to add 6 water molecules."},{"Start":"02:39.110 ","End":"02:41.450","Text":"Now we\u0027re going to balance the hydrogen."},{"Start":"02:41.450 ","End":"02:43.910","Text":"On the left side we have 2 times 6,"},{"Start":"02:43.910 ","End":"02:45.485","Text":"which is 12 hydrogens."},{"Start":"02:45.485 ","End":"02:47.660","Text":"We\u0027re going to add 12 protons to the right side."},{"Start":"02:47.660 ","End":"02:50.549","Text":"We had to have no hydrogens."},{"Start":"02:51.610 ","End":"02:54.800","Text":"Now we\u0027re going to balance for electric charge."},{"Start":"02:54.800 ","End":"02:58.250","Text":"First of all, on the left side of the equation, in the water,"},{"Start":"02:58.250 ","End":"03:01.280","Text":"the hydrogen is as usual plus 1 and oxygen is minus"},{"Start":"03:01.280 ","End":"03:05.135","Text":"2 and this doesn\u0027t change throughout the equation."},{"Start":"03:05.135 ","End":"03:07.219","Text":"Now we\u0027re going to take a look at that iodine."},{"Start":"03:07.219 ","End":"03:11.690","Text":"On the left side we have an oxidation state of 0 for each iodine atom."},{"Start":"03:11.690 ","End":"03:15.650","Text":"On the right side, let\u0027s take a look."},{"Start":"03:15.650 ","End":"03:22.970","Text":"We have the oxidation state of iodine plus 3 times the oxidation state of oxygen,"},{"Start":"03:22.970 ","End":"03:24.695","Text":"which is minus 2,"},{"Start":"03:24.695 ","End":"03:27.440","Text":"needs to give us minus 1."},{"Start":"03:27.440 ","End":"03:31.440","Text":"Since the charge on the iodate ion is minus 1."},{"Start":"03:33.170 ","End":"03:36.270","Text":"Oxidation state of iodine in this case,"},{"Start":"03:36.270 ","End":"03:40.840","Text":"it\u0027s going to be minus 1 plus 6 is going to be plus 5."},{"Start":"03:44.060 ","End":"03:48.710","Text":"For each iodine and we need to add 5 electrons on"},{"Start":"03:48.710 ","End":"03:53.180","Text":"the right side of the equation since we have plus 5 on the right side and 0 on the left."},{"Start":"03:53.180 ","End":"03:55.625","Text":"Remember we have 2 iodine atoms."},{"Start":"03:55.625 ","End":"04:00.240","Text":"Therefore, we need to add 10 electrons to the right side of the equation."},{"Start":"04:00.640 ","End":"04:04.340","Text":"Now at this point our equations are balanced and remember,"},{"Start":"04:04.340 ","End":"04:06.905","Text":"before we can add them 1 to the other,"},{"Start":"04:06.905 ","End":"04:10.430","Text":"we need to balance that number of electrons on both sides."},{"Start":"04:10.430 ","End":"04:13.865","Text":"Now you can see we have 10 electrons on the right side."},{"Start":"04:13.865 ","End":"04:15.575","Text":"On the left side,"},{"Start":"04:15.575 ","End":"04:16.820","Text":"we have 2 electrons."},{"Start":"04:16.820 ","End":"04:19.160","Text":"So we\u0027re just going to multiply the first equation by"},{"Start":"04:19.160 ","End":"04:23.280","Text":"5 and that\u0027s going to give us 10 electrons on the left side."},{"Start":"04:23.280 ","End":"04:33.510","Text":"It\u0027s going to give us 5 lead oxide plus 20 protons so multiplying everything by 5"},{"Start":"04:33.510 ","End":"04:38.550","Text":"is plus 10 electrons is going to give us"},{"Start":"04:38.550 ","End":"04:48.920","Text":"5 lead ions plus 10 water molecules because that\u0027s 2 times 5."},{"Start":"04:50.660 ","End":"04:53.675","Text":"Now that we have our 2 half equations,"},{"Start":"04:53.675 ","End":"04:55.280","Text":"the number of electrons is equal."},{"Start":"04:55.280 ","End":"04:57.425","Text":"We can just add them 1 to the other."},{"Start":"04:57.425 ","End":"04:59.570","Text":"This is going to give us on"},{"Start":"04:59.570 ","End":"05:03.090","Text":"the left side 5 lead oxide"},{"Start":"05:04.570 ","End":"05:13.620","Text":"plus 20 protons plus 10 electrons,"},{"Start":"05:16.660 ","End":"05:21.480","Text":"plus the iodine molecule,"},{"Start":"05:22.070 ","End":"05:25.990","Text":"plus 6 water molecules."},{"Start":"05:27.380 ","End":"05:31.220","Text":"This gives us, on the right side we have"},{"Start":"05:31.220 ","End":"05:40.540","Text":"5 lead ions plus 10 water molecules."},{"Start":"05:42.250 ","End":"05:45.740","Text":"We\u0027re going to continue to write down here,"},{"Start":"05:45.740 ","End":"05:48.570","Text":"plus 2 iodate ion,"},{"Start":"05:52.060 ","End":"05:59.105","Text":"plus 12 protons plus 10 electrons."},{"Start":"05:59.105 ","End":"06:06.480","Text":"First of all, we know that the electrons do cancel out."},{"Start":"06:06.530 ","End":"06:09.350","Text":"Next, we can see first of all,"},{"Start":"06:09.350 ","End":"06:13.159","Text":"that we have water molecules on both sides of the reaction."},{"Start":"06:13.159 ","End":"06:16.370","Text":"We have 6 on the left side and 10 on the right side."},{"Start":"06:16.370 ","End":"06:19.280","Text":"We\u0027re going to subtract 6 water molecules from each side."},{"Start":"06:19.280 ","End":"06:22.440","Text":"That\u0027s going to give us a 0 on our left side."},{"Start":"06:23.750 ","End":"06:26.250","Text":"Instead of 10,"},{"Start":"06:26.250 ","End":"06:30.810","Text":"we\u0027re going to have 4 water molecules on our right side."},{"Start":"06:31.060 ","End":"06:34.340","Text":"We also have protons on both sides of the reaction."},{"Start":"06:34.340 ","End":"06:37.880","Text":"We have 12 on the right side and 20 on the left side."},{"Start":"06:37.880 ","End":"06:40.160","Text":"We\u0027re going to subtract 12 from both sides."},{"Start":"06:40.160 ","End":"06:42.230","Text":"This is going to give us a 0 on the right side."},{"Start":"06:42.230 ","End":"06:46.285","Text":"On the left side, we\u0027re going to have 20 protons minus 12 protons."},{"Start":"06:46.285 ","End":"06:49.480","Text":"That\u0027s going to give us 8 protons."},{"Start":"06:50.840 ","End":"06:54.875","Text":"Now we\u0027re going to see what we have leftover from our reaction."},{"Start":"06:54.875 ","End":"07:02.435","Text":"We have a 5 lead oxide plus we have"},{"Start":"07:02.435 ","End":"07:12.330","Text":"8 protons plus we have an iodine molecule."},{"Start":"07:12.550 ","End":"07:21.480","Text":"This is going to give us 5 lead ions plus"},{"Start":"07:21.480 ","End":"07:31.390","Text":"4 water molecules plus 2 iodate ion."},{"Start":"07:33.140 ","End":"07:36.080","Text":"This point, let\u0027s check if our reaction is balanced."},{"Start":"07:36.080 ","End":"07:37.835","Text":"Let\u0027s check for other elements."},{"Start":"07:37.835 ","End":"07:41.705","Text":"We have 5 lead on their left side and 5 on the right side."},{"Start":"07:41.705 ","End":"07:43.010","Text":"Let\u0027s see how many oxygens we have."},{"Start":"07:43.010 ","End":"07:46.330","Text":"We have 2 times 5,"},{"Start":"07:46.330 ","End":"07:54.000","Text":"that\u0027s giving us 10 oxygens on the left side and on the right side we have"},{"Start":"07:54.000 ","End":"08:01.350","Text":"4 times oxygen plus 3 times 2 is 6 oxygen meaning 4 plus 6."},{"Start":"08:01.350 ","End":"08:04.900","Text":"That\u0027s going to give us 10 oxygens also on the right side."},{"Start":"08:04.900 ","End":"08:06.860","Text":"If we look at our iodine,"},{"Start":"08:06.860 ","End":"08:08.930","Text":"we have 2 on our left side and 2 on our right side and"},{"Start":"08:08.930 ","End":"08:11.420","Text":"now what we have left is to look at it for the hydrogens."},{"Start":"08:11.420 ","End":"08:15.145","Text":"On the left side we have 8 hydrogens atoms."},{"Start":"08:15.145 ","End":"08:18.110","Text":"On the right side, we also have 2 times 4,"},{"Start":"08:18.110 ","End":"08:20.180","Text":"which is 8 hydrogen atoms."},{"Start":"08:20.180 ","End":"08:26.915","Text":"We see that our reaction is balanced for the atoms."},{"Start":"08:26.915 ","End":"08:29.705","Text":"Now let\u0027s look at the charge."},{"Start":"08:29.705 ","End":"08:32.375","Text":"On the left side, the only charged we have is from the protons."},{"Start":"08:32.375 ","End":"08:34.040","Text":"That\u0027s going to be 8 times plus 1."},{"Start":"08:34.040 ","End":"08:36.755","Text":"That\u0027s going to give us a plus 8 charge."},{"Start":"08:36.755 ","End":"08:39.930","Text":"On the right side we have,"},{"Start":"08:40.210 ","End":"08:42.710","Text":"first of all, from our lead ion,"},{"Start":"08:42.710 ","End":"08:45.330","Text":"we have 5 times plus 2,"},{"Start":"08:46.630 ","End":"08:48.800","Text":"from the iodide anion,"},{"Start":"08:48.800 ","End":"08:51.450","Text":"we have 2 times minus 1."},{"Start":"08:51.800 ","End":"08:54.480","Text":"That\u0027s 10 minus 2."},{"Start":"08:54.480 ","End":"08:56.220","Text":"That\u0027s also a plus 8."},{"Start":"08:56.220 ","End":"08:58.880","Text":"Our reaction is balanced for charge."},{"Start":"08:58.880 ","End":"09:02.750","Text":"Now let\u0027s just take a look at what we got for our reaction."},{"Start":"09:02.750 ","End":"09:07.610","Text":"We have 5 lead oxide plus 8 protons plus"},{"Start":"09:07.610 ","End":"09:14.075","Text":"the iodine molecule is going to give us 5 lead ions plus 4 water plus 2 iodine ion."},{"Start":"09:14.075 ","End":"09:18.380","Text":"That is our balanced reaction for b,"},{"Start":"09:18.380 ","End":"09:19.550","Text":"and that is our final answer."},{"Start":"09:19.550 ","End":"09:21.990","Text":"Thank you very much for watching."}],"ID":23718},{"Watched":false,"Name":"Exercise 4 - Part a","Duration":"11m 26s","ChapterTopicVideoID":22898,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22898.jpeg","UploadDate":"2020-12-16T02:08:31.3770000","DurationForVideoObject":"PT11M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.519","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.519 ","End":"00:04.500","Text":"Balance the following equations for"},{"Start":"00:04.500 ","End":"00:08.940","Text":"redox reactions occurring in basic solutions. We\u0027re going to start with a."},{"Start":"00:08.940 ","End":"00:12.090","Text":"In a, we have the manganate ion plus"},{"Start":"00:12.090 ","End":"00:17.745","Text":"the oxalate ion gives us manganese dioxide plus carbon dioxide."},{"Start":"00:17.745 ","End":"00:22.260","Text":"First of all, as we did in the redox reactions occurring in acidic solution,"},{"Start":"00:22.260 ","End":"00:26.589","Text":"we\u0027re going to divide a redox reaction into 2 equations."},{"Start":"00:26.960 ","End":"00:30.300","Text":"The first equation is going to be the manganate ion"},{"Start":"00:30.300 ","End":"00:40.780","Text":"and it\u0027s going to give us manganese dioxide,"},{"Start":"00:40.910 ","End":"00:44.700","Text":"and our second reaction is going to be the oxalate ion"},{"Start":"00:44.700 ","End":"00:52.019","Text":"and this is going to give us carbon dioxide."},{"Start":"00:54.560 ","End":"00:56.705","Text":"Now that we have our 2 equations,"},{"Start":"00:56.705 ","End":"00:59.630","Text":"we\u0027re going to balance each separately."},{"Start":"00:59.630 ","End":"01:01.970","Text":"Let\u0027s start with our first equation."},{"Start":"01:01.970 ","End":"01:05.585","Text":"Remember, first of all, we balance elements which are not oxygen and not hydrogen."},{"Start":"01:05.585 ","End":"01:08.990","Text":"We have 1 manganese on the left side of the reaction and"},{"Start":"01:08.990 ","End":"01:12.725","Text":"1 manganese on the right side of the equation and therefore, they\u0027re equal."},{"Start":"01:12.725 ","End":"01:14.263","Text":"We\u0027re going on to the oxygens."},{"Start":"01:14.263 ","End":"01:16.070","Text":"To balance the oxygens, we have 4 on"},{"Start":"01:16.070 ","End":"01:20.210","Text":"our left side and 2 on the right side of the equation."},{"Start":"01:20.210 ","End":"01:25.430","Text":"Therefore, we\u0027re going to add 2 water molecules in order to balance our oxygens."},{"Start":"01:25.430 ","End":"01:29.165","Text":"That way we\u0027re adding 2 oxygens to our equation."},{"Start":"01:29.165 ","End":"01:32.600","Text":"Now we have to balance our hydrogen so we can see on the right that we"},{"Start":"01:32.600 ","End":"01:35.645","Text":"have 4 hydrogens now, 2 times 2."},{"Start":"01:35.645 ","End":"01:39.655","Text":"On the left we have none, so we\u0027re going to add 4 protons."},{"Start":"01:39.655 ","End":"01:43.010","Text":"We\u0027re going to add 4 protons in order to balance our hydrogens."},{"Start":"01:43.010 ","End":"01:44.780","Text":"At this point,"},{"Start":"01:44.780 ","End":"01:46.880","Text":"we have all our elements balanced,"},{"Start":"01:46.880 ","End":"01:49.145","Text":"and now we have to balance the electric charge."},{"Start":"01:49.145 ","End":"01:51.680","Text":"Let\u0027s look at our oxidation states."},{"Start":"01:51.680 ","End":"01:56.540","Text":"First of all, the hydrogen is plus 1 on our left and plus 1 in water,"},{"Start":"01:56.540 ","End":"02:00.890","Text":"and oxygen is minus 2 on the left and minus"},{"Start":"02:00.890 ","End":"02:06.080","Text":"2 in water and minus 2 also in the manganese dioxide."},{"Start":"02:06.080 ","End":"02:08.885","Text":"Let\u0027s start with our manganese dioxide."},{"Start":"02:08.885 ","End":"02:16.190","Text":"The oxidation state of manganese plus 2 times minus 2,"},{"Start":"02:16.190 ","End":"02:17.900","Text":"because it\u0027s the oxidation state of oxygen,"},{"Start":"02:17.900 ","End":"02:20.090","Text":"we have 2 oxygens,"},{"Start":"02:20.090 ","End":"02:25.205","Text":"equals 0 since the manganese dioxide is a neutral molecule."},{"Start":"02:25.205 ","End":"02:29.720","Text":"The manganese oxidation state is going to equal plus 4."},{"Start":"02:29.720 ","End":"02:32.390","Text":"Since we have manganese minus 4, it\u0027s going to equal 0."},{"Start":"02:32.390 ","End":"02:37.280","Text":"We have a plus 4 for oxidation state for manganese on the right side of the equation."},{"Start":"02:37.280 ","End":"02:38.420","Text":"On the left side of the equation,"},{"Start":"02:38.420 ","End":"02:45.065","Text":"let\u0027s see the oxidation state of manganese plus 4 since we have 4 oxygens times minus 2,"},{"Start":"02:45.065 ","End":"02:47.360","Text":"it\u0027s going to give us a minus 1 since"},{"Start":"02:47.360 ","End":"02:50.405","Text":"the overall charge of the whole molecules is minus 1."},{"Start":"02:50.405 ","End":"02:53.915","Text":"We have manganese minus 8 equals minus 1."},{"Start":"02:53.915 ","End":"02:56.930","Text":"The manganese oxidation state is going to come out to plus 7."},{"Start":"02:56.930 ","End":"02:59.990","Text":"We\u0027re just going to write that over our manganese here."},{"Start":"02:59.990 ","End":"03:02.645","Text":"As you can see, oxidation state for manganese on"},{"Start":"03:02.645 ","End":"03:05.750","Text":"the left side is plus 7 and on our right side is plus 4,"},{"Start":"03:05.750 ","End":"03:12.665","Text":"meaning we need to add 3 electrons to the left side in order to balance electric charge."},{"Start":"03:12.665 ","End":"03:15.515","Text":"Now we\u0027re going to go on to the second equation."},{"Start":"03:15.515 ","End":"03:20.270","Text":"We have the oxalate ion and it gives us carbon dioxide."},{"Start":"03:20.270 ","End":"03:23.470","Text":"First, we\u0027re going to look at the elements other than oxygen and hydrogen."},{"Start":"03:23.470 ","End":"03:24.769","Text":"We have our carbon."},{"Start":"03:24.769 ","End":"03:27.305","Text":"On the left side, we have 2 carbons and then the right side we have 1."},{"Start":"03:27.305 ","End":"03:30.265","Text":"We\u0027re going to multiply our carbon dioxide by 2."},{"Start":"03:30.265 ","End":"03:32.520","Text":"That\u0027s going to give us 2 carbons on the left side"},{"Start":"03:32.520 ","End":"03:34.395","Text":"and 2 carbons on the right side of the equation."},{"Start":"03:34.395 ","End":"03:35.630","Text":"If we look at our oxygens,"},{"Start":"03:35.630 ","End":"03:39.695","Text":"we have 4 on left side and on the right side we have 2 times 2,"},{"Start":"03:39.695 ","End":"03:42.455","Text":"which is 4, giving us also 4 oxygens."},{"Start":"03:42.455 ","End":"03:44.525","Text":"We don\u0027t have to balance for hydrogens."},{"Start":"03:44.525 ","End":"03:47.285","Text":"We\u0027re just going to balance for electric charge now."},{"Start":"03:47.285 ","End":"03:50.375","Text":"If we look at our carbon dioxide,"},{"Start":"03:50.375 ","End":"03:52.805","Text":"the oxidation state of carbon plus 2,"},{"Start":"03:52.805 ","End":"03:55.085","Text":"since we have 2 oxygens times minus 2,"},{"Start":"03:55.085 ","End":"03:57.885","Text":"which is the oxidation state of oxygen,"},{"Start":"03:57.885 ","End":"04:01.280","Text":"equals 0 since this is a neutral molecule."},{"Start":"04:01.280 ","End":"04:06.020","Text":"The oxidation state of carbon comes out to plus 4,"},{"Start":"04:06.020 ","End":"04:08.705","Text":"since we have carbon minus 4 equaling 0."},{"Start":"04:08.705 ","End":"04:11.480","Text":"It\u0027s plus 4 on our right side."},{"Start":"04:11.480 ","End":"04:15.950","Text":"On the left side, let\u0027s look at our oxidation states."},{"Start":"04:15.950 ","End":"04:17.870","Text":"We have 2 carbons,"},{"Start":"04:17.870 ","End":"04:23.480","Text":"meaning 2 times the oxidation state of carbon, plus 4."},{"Start":"04:23.480 ","End":"04:25.460","Text":"We have 4 oxygens times minus 2,"},{"Start":"04:25.460 ","End":"04:27.455","Text":"which is the oxidation state of oxygen,"},{"Start":"04:27.455 ","End":"04:32.580","Text":"equals minus 2 because it\u0027s overall charge of our whole molecule."},{"Start":"04:32.580 ","End":"04:34.700","Text":"2 carbons is going to give"},{"Start":"04:34.700 ","End":"04:39.785","Text":"us 2 times oxidation state of carbon minus 8 is going to give us minus 2."},{"Start":"04:39.785 ","End":"04:42.890","Text":"2 times oxidation state of carbon is going to give us 6."},{"Start":"04:42.890 ","End":"04:47.165","Text":"Oxidation state of carbon is going to come out to plus 3."},{"Start":"04:47.165 ","End":"04:49.160","Text":"We have plus 3 on the left side of"},{"Start":"04:49.160 ","End":"04:52.555","Text":"the equation and plus 4 on the right side of the equation."},{"Start":"04:52.555 ","End":"04:56.960","Text":"If we have plus 4 on the right side of the equation and plus 3 on the left,"},{"Start":"04:56.960 ","End":"05:03.590","Text":"we have to add 1 electron to each carbon on the right side of the equation."},{"Start":"05:03.590 ","End":"05:04.730","Text":"But since we have 2 carbons,"},{"Start":"05:04.730 ","End":"05:08.795","Text":"we\u0027re going to add 2 electrons to the right side of the equation."},{"Start":"05:08.795 ","End":"05:12.920","Text":"Again, 1 electron for every carbon to go from plus 4 to plus 3,"},{"Start":"05:12.920 ","End":"05:16.655","Text":"but this is times 2 since we have 2 carbons."},{"Start":"05:16.655 ","End":"05:21.650","Text":"Now remember, what we\u0027re going to do here is we\u0027re going to multiply the equations in"},{"Start":"05:21.650 ","End":"05:26.570","Text":"order so that electrons on each side of the equations are going to be equal."},{"Start":"05:26.570 ","End":"05:30.230","Text":"On the left side we see we have 3 electrons and then the right side we have 2 electrons,"},{"Start":"05:30.230 ","End":"05:34.710","Text":"meaning we have to multiply the top equation by 2 so we get"},{"Start":"05:34.710 ","End":"05:40.450","Text":"6 electrons in all and the bottom equation by 3 so you also get 6 electrons at all."},{"Start":"05:40.450 ","End":"05:43.460","Text":"Let\u0027s multiply the top equation by 2."},{"Start":"05:43.460 ","End":"05:53.850","Text":"We have 2 times the manganate ion plus 8 protons,"},{"Start":"05:53.850 ","End":"05:55.914","Text":"since that\u0027s 4 times 2,"},{"Start":"05:55.914 ","End":"05:59.070","Text":"plus 6 electrons is going to give us"},{"Start":"05:59.070 ","End":"06:07.429","Text":"2 times manganese dioxide plus 4 water molecules."},{"Start":"06:07.429 ","End":"06:09.830","Text":"That\u0027s going to be a tough equation."},{"Start":"06:09.830 ","End":"06:11.420","Text":"Now the bottom equation,"},{"Start":"06:11.420 ","End":"06:12.615","Text":"we\u0027re going to multiply it by 3."},{"Start":"06:12.615 ","End":"06:17.790","Text":"We\u0027re going to get 3 oxalate ion is going to give"},{"Start":"06:17.790 ","End":"06:24.850","Text":"us 3 times 2 is 6 carbon dioxide plus 6 electrons."},{"Start":"06:25.010 ","End":"06:28.280","Text":"Our body equation is 3 times oxalate ion is"},{"Start":"06:28.280 ","End":"06:31.055","Text":"going to give us 6 carbon dioxide plus 6 electrons."},{"Start":"06:31.055 ","End":"06:35.500","Text":"Now we have to add 1 equation to the other. Let\u0027s add them."},{"Start":"06:35.500 ","End":"06:45.070","Text":"We have 2 manganate ion plus 8 protons plus"},{"Start":"06:45.070 ","End":"06:54.420","Text":"6 electrons plus 3 times oxalate ion is going to give us"},{"Start":"06:54.420 ","End":"07:00.480","Text":"2 manganese dioxide plus"},{"Start":"07:00.480 ","End":"07:05.880","Text":"4 water molecules plus"},{"Start":"07:05.880 ","End":"07:10.980","Text":"6 carbon dioxide plus 6 electrons."},{"Start":"07:10.980 ","End":"07:14.230","Text":"First of all, we can take off the 6 electrons."},{"Start":"07:14.720 ","End":"07:19.960","Text":"Now if we look, we don\u0027t see any other molecule in common on both sides."},{"Start":"07:19.960 ","End":"07:23.725","Text":"But remember, we have to balance this reaction in basic medium."},{"Start":"07:23.725 ","End":"07:26.790","Text":"Right now, we have it balanced in acidic medium."},{"Start":"07:26.790 ","End":"07:30.880","Text":"What we need to do in order to balance in basic medium is like we did before."},{"Start":"07:30.880 ","End":"07:35.230","Text":"We need to add the same number of hydroxides as we have protons."},{"Start":"07:35.230 ","End":"07:36.970","Text":"On the left side of the reaction,"},{"Start":"07:36.970 ","End":"07:38.680","Text":"we have 8 protons."},{"Start":"07:38.680 ","End":"07:41.420","Text":"We\u0027re going to add 8 hydroxides."},{"Start":"07:41.970 ","End":"07:44.530","Text":"We\u0027re going to add that to the right side of"},{"Start":"07:44.530 ","End":"07:47.545","Text":"the equation too because we don\u0027t want to change anything here."},{"Start":"07:47.545 ","End":"07:50.160","Text":"8 hydroxide to the left and to the right."},{"Start":"07:50.160 ","End":"07:53.200","Text":"Now our protons plus our hydroxide is going to give us water,"},{"Start":"07:53.200 ","End":"07:57.555","Text":"so that\u0027s going to be plus 8 water molecules."},{"Start":"07:57.555 ","End":"07:59.350","Text":"Now if we look,"},{"Start":"07:59.350 ","End":"08:01.270","Text":"we can see that on the left side of the equation we have"},{"Start":"08:01.270 ","End":"08:03.640","Text":"8 water molecules and on the right side we have 4."},{"Start":"08:03.640 ","End":"08:07.340","Text":"We\u0027re going to subtract 4 water molecules from each side."},{"Start":"08:09.770 ","End":"08:14.110","Text":"On the left side, we\u0027re going to get 8 minus 4 water molecules."},{"Start":"08:14.110 ","End":"08:17.330","Text":"That\u0027s going to give us 4 water molecules."},{"Start":"08:18.900 ","End":"08:30.165","Text":"The reaction comes out to 2 manganate ions plus"},{"Start":"08:30.165 ","End":"08:37.950","Text":"3 times oxalate ion gives us"},{"Start":"08:37.950 ","End":"08:44.790","Text":"2 manganese dioxide plus"},{"Start":"08:44.790 ","End":"08:51.370","Text":"8 hydroxides plus 6 carbon dioxide."},{"Start":"08:51.740 ","End":"08:58.465","Text":"At this point, our reaction should be balanced in basic medium."},{"Start":"08:58.465 ","End":"09:01.585","Text":"Now we just want to check that it really is balanced."},{"Start":"09:01.585 ","End":"09:02.920","Text":"Let\u0027s look first at the elements."},{"Start":"09:02.920 ","End":"09:05.110","Text":"We can see that we have 2 manganese atoms"},{"Start":"09:05.110 ","End":"09:07.720","Text":"on the left side and 2 manganese atoms on the right side."},{"Start":"09:07.720 ","End":"09:10.210","Text":"Now let\u0027s take a look at the carbons we have."},{"Start":"09:10.210 ","End":"09:13.000","Text":"On the left side, we have 2 times 3 giving us 6."},{"Start":"09:13.000 ","End":"09:16.120","Text":"On the right side we also have 6. That\u0027s good."},{"Start":"09:16.120 ","End":"09:24.080","Text":"Now our oxygens, we have 4 times 2 giving us 8 plus 1 times 4 giving us 4,"},{"Start":"09:24.080 ","End":"09:26.450","Text":"meaning 8 plus 4 equals 12."},{"Start":"09:26.450 ","End":"09:28.500","Text":"Let\u0027s just write that in here. Oxygens,"},{"Start":"09:28.500 ","End":"09:31.490","Text":"we have 8 plus 4 plus."},{"Start":"09:31.490 ","End":"09:35.475","Text":"Let\u0027s take a look. Here we have 4 times 3 giving us 12."},{"Start":"09:35.475 ","End":"09:39.740","Text":"This in all is 24 oxygens."},{"Start":"09:40.170 ","End":"09:45.850","Text":"If we take a look at the right side of the equations, for the oxygens,"},{"Start":"09:45.850 ","End":"09:50.380","Text":"we have 4 because we have 2 times 2 from the manganese dioxide plus 8 for"},{"Start":"09:50.380 ","End":"09:56.505","Text":"my hydroxide plus 2 times 6 from our carbon dioxide, giving us 12."},{"Start":"09:56.505 ","End":"09:58.570","Text":"That\u0027s 4 plus 8 plus 12,"},{"Start":"09:58.570 ","End":"10:01.495","Text":"so that equals 24 oxygens also."},{"Start":"10:01.495 ","End":"10:03.640","Text":"On the left and right of the equation,"},{"Start":"10:03.640 ","End":"10:05.710","Text":"we have 24 oxygens."},{"Start":"10:05.710 ","End":"10:08.570","Text":"Now let\u0027s check our hydrogens."},{"Start":"10:08.660 ","End":"10:11.850","Text":"On the left side of the equation we have 2 times 4,"},{"Start":"10:11.850 ","End":"10:14.180","Text":"so that\u0027s 8 hydrogens."},{"Start":"10:14.180 ","End":"10:15.940","Text":"On the right side of the equation,"},{"Start":"10:15.940 ","End":"10:17.785","Text":"we also have 1 times 8,"},{"Start":"10:17.785 ","End":"10:19.950","Text":"giving us 8 hydrogens."},{"Start":"10:19.950 ","End":"10:24.670","Text":"The elements we can see are equal because remember we had"},{"Start":"10:24.670 ","End":"10:30.560","Text":"2 manganese in the beginning and also 6 carbons."},{"Start":"10:33.810 ","End":"10:36.910","Text":"The elements are equal. Now we have to take a look at"},{"Start":"10:36.910 ","End":"10:41.230","Text":"our ions and see if they equal on both sides."},{"Start":"10:41.230 ","End":"10:46.655","Text":"On the left side of the equation we have 2 times minus 1, which is minus 2,"},{"Start":"10:46.655 ","End":"10:50.160","Text":"plus we have 3 times minus 2,"},{"Start":"10:50.160 ","End":"10:55.540","Text":"so plus minus 6 giving us a total of minus 8."},{"Start":"10:55.540 ","End":"10:58.025","Text":"On the right side of the equation,"},{"Start":"10:58.025 ","End":"10:59.330","Text":"we have the hydroxides,"},{"Start":"10:59.330 ","End":"11:03.770","Text":"we have 8 times minus 1 and that also gives us minus 8."},{"Start":"11:03.770 ","End":"11:06.665","Text":"We can see that the left side and the right side are equal."},{"Start":"11:06.665 ","End":"11:10.645","Text":"Therefore, we know that the reaction we have is balanced."},{"Start":"11:10.645 ","End":"11:15.780","Text":"Again, we have 2 manganate ion plus 4 water molecules plus 3 oxalate ion"},{"Start":"11:15.780 ","End":"11:21.045","Text":"gives us 2 manganese dioxide plus 8 hydroxides plus 6 carbon dioxides."},{"Start":"11:21.045 ","End":"11:23.150","Text":"That is our final answer for a."},{"Start":"11:23.150 ","End":"11:25.680","Text":"Now we\u0027re going to go onto b."}],"ID":23719},{"Watched":false,"Name":"Exercise 4 - Part b","Duration":"7m 3s","ChapterTopicVideoID":22899,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/22899.jpeg","UploadDate":"2020-12-16T02:08:49.5670000","DurationForVideoObject":"PT7M3S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:02.445","Text":"Now we\u0027re going to continue with b."},{"Start":"00:02.445 ","End":"00:07.750","Text":"In b we have hydrazine plus copper hydroxide gives us nitrogen plus copper."},{"Start":"00:08.000 ","End":"00:12.645","Text":"Again, first of all, we\u0027re going to write the 2 half equations."},{"Start":"00:12.645 ","End":"00:15.465","Text":"The first one is going to be hydrazine,"},{"Start":"00:15.465 ","End":"00:18.340","Text":"is going to give us nitrogen."},{"Start":"00:20.720 ","End":"00:26.130","Text":"The second one is going to be copper hydroxide is going to give us copper."},{"Start":"00:34.210 ","End":"00:36.440","Text":"We\u0027re going to look first of all at"},{"Start":"00:36.440 ","End":"00:40.920","Text":"the first half equation and we\u0027re going to balance it."},{"Start":"00:42.230 ","End":"00:45.270","Text":"First of all, if we look at our nitrogen we can see that"},{"Start":"00:45.270 ","End":"00:47.330","Text":"we have 2 on our left side and 2 on the right side."},{"Start":"00:47.330 ","End":"00:52.235","Text":"Remember you start with the elements that are not oxygen and hydrogen."},{"Start":"00:52.235 ","End":"00:54.260","Text":"Next, if we look at our oxygens,"},{"Start":"00:54.260 ","End":"00:57.950","Text":"we have none on the right side and none on the left side."},{"Start":"00:57.950 ","End":"01:02.675","Text":"Meaning that we don\u0027t need to balance oxygens because there are none."},{"Start":"01:02.675 ","End":"01:04.280","Text":"Now we look at our hydrogens."},{"Start":"01:04.280 ","End":"01:06.950","Text":"The left side we have 4 hydrogens on the right side we have none."},{"Start":"01:06.950 ","End":"01:10.410","Text":"Therefore, we\u0027re going to add 4 protons to the right side."},{"Start":"01:11.800 ","End":"01:14.555","Text":"Now our hydrogens are balanced."},{"Start":"01:14.555 ","End":"01:17.780","Text":"The next step is to balance the electric charge."},{"Start":"01:17.780 ","End":"01:20.750","Text":"If we look at our nitrogen on the right side,"},{"Start":"01:20.750 ","End":"01:24.145","Text":"we can see that the nitrogen has an oxidation state of"},{"Start":"01:24.145 ","End":"01:28.460","Text":"0 and our hydrogen has an oxidation state of plus 1."},{"Start":"01:28.460 ","End":"01:30.725","Text":"Now we\u0027re going to look at hydrazine."},{"Start":"01:30.725 ","End":"01:33.700","Text":"We know that the hydrogen has an oxidation state of plus 1"},{"Start":"01:33.700 ","End":"01:36.410","Text":"and now we\u0027re going to calculate the oxidation state of the nitrogen."},{"Start":"01:36.410 ","End":"01:41.390","Text":"We have 2 times the oxidation state of nitrogen plus 4 since we have"},{"Start":"01:41.390 ","End":"01:47.305","Text":"4 hydrogens times plus 1=0."},{"Start":"01:47.305 ","End":"01:50.750","Text":"Because this is a neutral molecule."},{"Start":"01:50.750 ","End":"01:56.045","Text":"2 times the oxidation state of nitrogen equals minus 4."},{"Start":"01:56.045 ","End":"02:01.850","Text":"Therefore, the oxidation state of nitrogen equals minus 2 in hydrogen."},{"Start":"02:01.850 ","End":"02:06.809","Text":"Therefore, the oxidation state of nitrogen equals minus 2 in hydrazine."},{"Start":"02:06.820 ","End":"02:10.790","Text":"If we look at our oxidation states of the nitrogen,"},{"Start":"02:10.790 ","End":"02:14.150","Text":"consider that on the left side we have minus 2 and on the right side,"},{"Start":"02:14.150 ","End":"02:15.200","Text":"we have 0,"},{"Start":"02:15.200 ","End":"02:21.945","Text":"meaning we need to add 2 electrons to the right side for every nitrogen."},{"Start":"02:21.945 ","End":"02:26.140","Text":"Since we have 2 nitrogens we are going to add 4 electrons."},{"Start":"02:26.780 ","End":"02:29.035","Text":"Now the first equation,"},{"Start":"02:29.035 ","End":"02:30.540","Text":"our first half equation is balanced."},{"Start":"02:30.540 ","End":"02:32.230","Text":"Let\u0027s go to the second one."},{"Start":"02:32.230 ","End":"02:35.825","Text":"Again, we\u0027re going to look at the elements other than oxygen and hydrogen."},{"Start":"02:35.825 ","End":"02:38.950","Text":"We can see that the copper there\u0027s one on our left side, one on the right side."},{"Start":"02:38.950 ","End":"02:41.560","Text":"That\u0027s fine. Now let\u0027s take a look at our oxygen."},{"Start":"02:41.560 ","End":"02:45.070","Text":"We have 2 on the left side because it\u0027s 2 times 1."},{"Start":"02:45.070 ","End":"02:47.385","Text":"On the right side, we have none."},{"Start":"02:47.385 ","End":"02:53.215","Text":"We\u0027re going to add 2 water molecules in order to balance our oxygens."},{"Start":"02:53.215 ","End":"02:54.970","Text":"Now the oxygens are balanced."},{"Start":"02:54.970 ","End":"02:56.230","Text":"We\u0027re going to balance our hydrogens."},{"Start":"02:56.230 ","End":"03:00.010","Text":"We can see that on the right side we have 2 times 2, meaning 4 hydrogens."},{"Start":"03:00.010 ","End":"03:04.390","Text":"On the left side, we have 2 times 1,"},{"Start":"03:04.390 ","End":"03:05.430","Text":"which is 2 hydrogens."},{"Start":"03:05.430 ","End":"03:12.565","Text":"We\u0027re going to add 2 protons to the left side and now our hydrogens are balanced."},{"Start":"03:12.565 ","End":"03:14.995","Text":"Now let\u0027s take a look at our oxidation states."},{"Start":"03:14.995 ","End":"03:18.385","Text":"The oxidation states of the oxygen and the hydrogen are,"},{"Start":"03:18.385 ","End":"03:21.670","Text":"on oxygen they\u0027re minus 2 on both sides,"},{"Start":"03:21.670 ","End":"03:23.710","Text":"and on the hydrogen they\u0027re plus 1."},{"Start":"03:23.710 ","End":"03:26.500","Text":"Let\u0027s calculate the oxidation states of the copper."},{"Start":"03:26.500 ","End":"03:28.945","Text":"On the right side, it\u0027s 0."},{"Start":"03:28.945 ","End":"03:31.330","Text":"The oxidation state of copper is 0."},{"Start":"03:31.330 ","End":"03:33.850","Text":"On the left side, we have the oxidation state of"},{"Start":"03:33.850 ","End":"03:39.100","Text":"copper plus 2 times the oxidation state of the hydroxide."},{"Start":"03:39.100 ","End":"03:44.830","Text":"But the hydroxide is minus 2 plus 1 because we have an oxygen and a hydrogen."},{"Start":"03:44.830 ","End":"03:46.660","Text":"Minus 2 plus 1 is a minus 1."},{"Start":"03:46.660 ","End":"03:49.480","Text":"The oxidation state of the hydroxide is minus"},{"Start":"03:49.480 ","End":"03:55.640","Text":"1 and this equals 0 because it\u0027s a neutral molecule."},{"Start":"03:55.640 ","End":"03:59.360","Text":"The oxidation state of the copper minus 2 is going to give us 0"},{"Start":"03:59.360 ","End":"04:04.290","Text":"so the oxidation state of copper equals plus 2."},{"Start":"04:07.850 ","End":"04:10.530","Text":"Again we have one copper."},{"Start":"04:10.530 ","End":"04:13.700","Text":"On our left side, we have an oxidation state of plus 2 and on the right side,"},{"Start":"04:13.700 ","End":"04:15.560","Text":"we have an oxidation state of 0."},{"Start":"04:15.560 ","End":"04:18.710","Text":"Meaning we need to add 2 electrons to"},{"Start":"04:18.710 ","End":"04:22.580","Text":"the left side in order to"},{"Start":"04:22.580 ","End":"04:27.875","Text":"balance the electric charge and our second equation is balanced."},{"Start":"04:27.875 ","End":"04:30.710","Text":"Now we need to look at the number of electrons."},{"Start":"04:30.710 ","End":"04:34.595","Text":"On the right side, we have 4 and on the left side, we have 2."},{"Start":"04:34.595 ","End":"04:37.970","Text":"Meaning we need to multiply the second half equation by 2,"},{"Start":"04:37.970 ","End":"04:41.730","Text":"therefore will have 4 electrons on both sides."},{"Start":"04:42.550 ","End":"04:46.070","Text":"Let\u0027s multiply the second half equation by 2."},{"Start":"04:46.070 ","End":"04:49.410","Text":"It\u0027s going to give us 2 copper hydroxide"},{"Start":"04:49.960 ","End":"04:56.915","Text":"plus 4 protons plus 4 electrons,"},{"Start":"04:56.915 ","End":"05:03.095","Text":"is going to give us 2 copper plus 4 water molecules."},{"Start":"05:03.095 ","End":"05:07.940","Text":"Now we\u0027re going to add the first half equation to the second one."},{"Start":"05:07.940 ","End":"05:17.650","Text":"You\u0027re going to get hydrazine plus 2 copper hydroxide"},{"Start":"05:20.140 ","End":"05:30.970","Text":"plus 4 protons plus 4 electrons is going to give us nitrogen"},{"Start":"05:30.970 ","End":"05:38.790","Text":"plus 4 protons plus"},{"Start":"05:38.790 ","End":"05:45.400","Text":"2 copper plus 4 water molecules."},{"Start":"05:47.220 ","End":"05:49.810","Text":"Now if we look at both sides of the equation,"},{"Start":"05:49.810 ","End":"05:52.540","Text":"we can see that we have 4 protons on our left and 4 on our right,"},{"Start":"05:52.540 ","End":"05:55.610","Text":"so we can cancel them out."},{"Start":"05:56.100 ","End":"06:01.270","Text":"We need to add another 4 electrons to the right side of the equation."},{"Start":"06:01.270 ","End":"06:03.890","Text":"Let\u0027s just add them in here."},{"Start":"06:05.000 ","End":"06:08.770","Text":"We also have on our left side and on the right side 4 electrons,"},{"Start":"06:08.770 ","End":"06:11.180","Text":"so we are going to cancel those out."},{"Start":"06:12.000 ","End":"06:14.245","Text":"Now let\u0027s see what we\u0027re left with."},{"Start":"06:14.245 ","End":"06:20.520","Text":"We have our hydrazine plus"},{"Start":"06:20.520 ","End":"06:28.540","Text":"2 copper hydroxide is going to"},{"Start":"06:28.540 ","End":"06:37.820","Text":"give us nitrogen plus 2 copper plus 4 water molecules."},{"Start":"06:37.820 ","End":"06:43.300","Text":"Here, remember that we need to balance for our basic medium,"},{"Start":"06:43.300 ","End":"06:45.115","Text":"but since we have no protons,"},{"Start":"06:45.115 ","End":"06:48.080","Text":"we don\u0027t need to add any hydroxides."},{"Start":"06:48.530 ","End":"06:53.895","Text":"What we\u0027re left with is hydrazine plus 2 copper hydroxide"},{"Start":"06:53.895 ","End":"06:59.185","Text":"give us nitrogen plus 2 copper plus 4 water molecules."},{"Start":"06:59.185 ","End":"07:01.805","Text":"That\u0027s our balanced reaction."},{"Start":"07:01.805 ","End":"07:04.920","Text":"Thank you very much for watching."}],"ID":23720},{"Watched":false,"Name":"Disproportionation Reactions","Duration":"5m 40s","ChapterTopicVideoID":17371,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/17371.jpeg","UploadDate":"2019-03-03T01:09:31.2170000","DurationForVideoObject":"PT5M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"In this video, we will discuss disproportionation reactions."},{"Start":"00:05.340 ","End":"00:08.835","Text":"Here it is disproportionation reactions."},{"Start":"00:08.835 ","End":"00:14.490","Text":"Disproportionation reactions are oxidation-reduction reactions in which,"},{"Start":"00:14.490 ","End":"00:19.110","Text":"at least one reactant undergoes both oxidation and reduction."},{"Start":"00:19.110 ","End":"00:23.910","Text":"The same species undergoes both oxidation and reduction."},{"Start":"00:23.910 ","End":"00:25.845","Text":"Let\u0027s take an example."},{"Start":"00:25.845 ","End":"00:29.415","Text":"Balance the equation CL_2 plus water,"},{"Start":"00:29.415 ","End":"00:31.980","Text":"giving HCl plus HClO."},{"Start":"00:31.980 ","End":"00:35.655","Text":"The first step is to write the half equations."},{"Start":"00:35.655 ","End":"00:39.015","Text":"To do this, we need to know the oxidation states."},{"Start":"00:39.015 ","End":"00:40.515","Text":"Here\u0027s our equation."},{"Start":"00:40.515 ","End":"00:45.620","Text":"The oxidation state of chlorine is 0 because it\u0027s an element."},{"Start":"00:45.620 ","End":"00:50.360","Text":"In HCl, the oxidation state of hydrogen is plus 1."},{"Start":"00:50.360 ","End":"00:53.855","Text":"The oxidation state of chlorine must be minus 1."},{"Start":"00:53.855 ","End":"01:02.305","Text":"In HClO, the oxidation state of hydrogen is plus 1 and that of oxygen is minus 2."},{"Start":"01:02.305 ","End":"01:04.940","Text":"That gives us a total of minus 1."},{"Start":"01:04.940 ","End":"01:09.280","Text":"That means the oxidation state of chlorine must be plus 1."},{"Start":"01:09.280 ","End":"01:11.165","Text":"Let\u0027s write that out."},{"Start":"01:11.165 ","End":"01:18.740","Text":"The oxidation state of chlorine is 0 and CL_2 minus 1 in HCl plus 1 in HCLO."},{"Start":"01:18.740 ","End":"01:21.730","Text":"We can write the oxidation reduction,"},{"Start":"01:21.730 ","End":"01:25.700","Text":"here\u0027s the oxidation reaction CL_2 going to HClO,"},{"Start":"01:25.700 ","End":"01:28.010","Text":"that\u0027s oxidation, an increase in"},{"Start":"01:28.010 ","End":"01:33.110","Text":"the oxidation state and the reduction of CL_2 going to HCL."},{"Start":"01:33.110 ","End":"01:38.480","Text":"That\u0027s our lower oxidation state in the products than in the reactants,"},{"Start":"01:38.480 ","End":"01:42.140","Text":"has gone from 0 to minus 1."},{"Start":"01:42.140 ","End":"01:46.340","Text":"The next step is to balance all atoms except H and O."},{"Start":"01:46.340 ","End":"01:47.660","Text":"Let\u0027s look at the equations."},{"Start":"01:47.660 ","End":"01:51.190","Text":"We have two CLs on the left only one on the right,"},{"Start":"01:51.190 ","End":"01:53.720","Text":"two CLs in the left only one of the right."},{"Start":"01:53.720 ","End":"01:56.989","Text":"We have to multiply it by 2 in both cases."},{"Start":"01:56.989 ","End":"01:58.940","Text":"Here\u0027s the result."},{"Start":"01:58.940 ","End":"02:01.790","Text":"CL_2 going to 2HClO,"},{"Start":"02:01.790 ","End":"02:04.495","Text":"CL_2 going to 2HCL."},{"Start":"02:04.495 ","End":"02:08.900","Text":"Step 3 is to balance O and H in each half equation"},{"Start":"02:08.900 ","End":"02:14.000","Text":"using water for the oxygens and H plus for the hydrogens."},{"Start":"02:14.000 ","End":"02:19.615","Text":"Remember, we work with H2O first and then with H plus."},{"Start":"02:19.615 ","End":"02:21.500","Text":"Let\u0027s look at the equations here."},{"Start":"02:21.500 ","End":"02:25.220","Text":"We have 2 oxygens on the right and none on the left."},{"Start":"02:25.220 ","End":"02:29.500","Text":"We have to add 2H2O to the left."},{"Start":"02:29.500 ","End":"02:38.175","Text":"We have 2H2O plus CL_2 to 2HClO."},{"Start":"02:38.175 ","End":"02:43.370","Text":"Then we have 4 hydrogens left and two on the right."},{"Start":"02:43.370 ","End":"02:47.674","Text":"We have to add another two hydrogen atoms."},{"Start":"02:47.674 ","End":"02:52.985","Text":"Two hydrogen ions on the right."},{"Start":"02:52.985 ","End":"02:54.870","Text":"Here, it\u0027s written out,"},{"Start":"02:54.870 ","End":"03:00.855","Text":"2H2O plus CL_2 given to HClO plus 2H plus,"},{"Start":"03:00.855 ","End":"03:03.315","Text":"what about the second equation?"},{"Start":"03:03.315 ","End":"03:06.835","Text":"We have CL_2 go into 2HCl."},{"Start":"03:06.835 ","End":"03:10.805","Text":"We need to add two hydrogens to the left-hand side."},{"Start":"03:10.805 ","End":"03:12.845","Text":"There are no oxygens here involved."},{"Start":"03:12.845 ","End":"03:15.005","Text":"We have 2H plus,"},{"Start":"03:15.005 ","End":"03:20.110","Text":"plus the CL_2, giving 2HCl."},{"Start":"03:20.110 ","End":"03:25.350","Text":"Here it\u0027s written out 2H plus plus CL_2 giving 2H."},{"Start":"03:25.350 ","End":"03:31.340","Text":"Step 4 is to balance the charges in each half equation by adding electrons."},{"Start":"03:31.340 ","End":"03:33.649","Text":"If we look at the top equation,"},{"Start":"03:33.649 ","End":"03:35.570","Text":"you see there are no charges at all on"},{"Start":"03:35.570 ","End":"03:40.070","Text":"the left-hand side and just two positive charges on the right-hand side."},{"Start":"03:40.070 ","End":"03:45.535","Text":"We need to add two electrons to neutralize those two positive charges."},{"Start":"03:45.535 ","End":"03:47.630","Text":"Look at the second equation."},{"Start":"03:47.630 ","End":"03:50.525","Text":"There are no charges at all on the right-hand side"},{"Start":"03:50.525 ","End":"03:54.200","Text":"and two positives on the left-hand side."},{"Start":"03:54.200 ","End":"03:59.090","Text":"We have to add two electrons to neutralize the two positive charges."},{"Start":"03:59.090 ","End":"04:03.290","Text":"We have to add two electrons on the left-hand side."},{"Start":"04:03.290 ","End":"04:07.084","Text":"Here it\u0027s written out here\u0027s the oxidation equation,"},{"Start":"04:07.084 ","End":"04:10.280","Text":"which the oxidation equation,"},{"Start":"04:10.280 ","End":"04:12.890","Text":"the electrons appear on the right-hand side and"},{"Start":"04:12.890 ","End":"04:16.115","Text":"the reduction equation the electrons appear on the left-hand side."},{"Start":"04:16.115 ","End":"04:20.885","Text":"Step 5 is to multiply each equation by a factor,"},{"Start":"04:20.885 ","End":"04:23.875","Text":"so the same number of electrons appears in each equation."},{"Start":"04:23.875 ","End":"04:27.650","Text":"Here there\u0027s no necessity because of two electrons at the right-hand side,"},{"Start":"04:27.650 ","End":"04:29.510","Text":"two electrodes, the left-hand side,"},{"Start":"04:29.510 ","End":"04:33.410","Text":"they will cancel when we add up so we don\u0027t need to do anything."},{"Start":"04:33.410 ","End":"04:39.695","Text":"Now we can add the half equations to get the final equation here so 2.5 equations."},{"Start":"04:39.695 ","End":"04:41.285","Text":"Now we can add them."},{"Start":"04:41.285 ","End":"04:43.220","Text":"Two electrons cancel."},{"Start":"04:43.220 ","End":"04:47.060","Text":"Anything else? It cancels the 2H plus on the right,"},{"Start":"04:47.060 ","End":"04:48.845","Text":"2H plus on the left."},{"Start":"04:48.845 ","End":"04:50.600","Text":"Now we can add up."},{"Start":"04:50.600 ","End":"04:54.295","Text":"We get 2H2O plus 2CL_2,"},{"Start":"04:54.295 ","End":"04:57.465","Text":"CL_2 from here and CL_2 from here."},{"Start":"04:57.465 ","End":"05:03.165","Text":"Giving 2HClO,2 HClO plus 2HCl plus 2HCl."},{"Start":"05:03.165 ","End":"05:06.410","Text":"Now we notice 2 appears in everywhere,"},{"Start":"05:06.410 ","End":"05:10.110","Text":"so we can divide by 2 to make it more compact."},{"Start":"05:10.110 ","End":"05:11.985","Text":"Here\u0027s our final equation,"},{"Start":"05:11.985 ","End":"05:13.675","Text":"H_2_O plus CL_2,"},{"Start":"05:13.675 ","End":"05:16.085","Text":"giving HClO plus HCl,"},{"Start":"05:16.085 ","End":"05:22.910","Text":"which is indeed exactly equal to the equation we were asked to balance."},{"Start":"05:22.910 ","End":"05:26.480","Text":"In fact, the original equation was balanced."},{"Start":"05:26.480 ","End":"05:28.595","Text":"But by doing it in this way,"},{"Start":"05:28.595 ","End":"05:32.225","Text":"we\u0027ve seen that it\u0027s a disproportionation reaction,"},{"Start":"05:32.225 ","End":"05:35.465","Text":"and that oxidation-reduction reaction."},{"Start":"05:35.465 ","End":"05:40.950","Text":"In this video, we learned about disproportionation reactions."}],"ID":18118},{"Watched":false,"Name":"Oxidizing and Reducing Agents","Duration":"8m 15s","ChapterTopicVideoID":17372,"CourseChapterTopicPlaylistID":86822,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.com/Images/Videos_Thumbnails/17372.jpeg","UploadDate":"2019-03-03T01:10:52.6770000","DurationForVideoObject":"PT8M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:03.300","Text":"In previous videos, we discussed"},{"Start":"00:03.300 ","End":"00:08.655","Text":"oxidation-reduction reactions and learned how to balance your equations."},{"Start":"00:08.655 ","End":"00:14.955","Text":"In this video we\u0027ll discuss the related concepts of oxidizing and reducing agents."},{"Start":"00:14.955 ","End":"00:20.910","Text":"First, let\u0027s summarize what we\u0027ve already learned about oxidation-reduction equations."},{"Start":"00:20.910 ","End":"00:24.150","Text":"The first part is that oxidation reduction are"},{"Start":"00:24.150 ","End":"00:28.260","Text":"complimentary processes that occur in the same reaction,"},{"Start":"00:28.260 ","End":"00:30.450","Text":"you can\u0027t have 1 without the other,"},{"Start":"00:30.450 ","End":"00:32.325","Text":"they always go together."},{"Start":"00:32.325 ","End":"00:37.820","Text":"The equations for redox processes can be divided into 2 half equations,"},{"Start":"00:37.820 ","End":"00:41.275","Text":"1 for oxidation and 1 for reduction."},{"Start":"00:41.275 ","End":"00:44.180","Text":"In oxidation there is an increase in"},{"Start":"00:44.180 ","End":"00:49.040","Text":"the oxidation state of 1 of the reactants and electrons are lost,"},{"Start":"00:49.040 ","End":"00:52.175","Text":"they appear on the right-hand side of the equation."},{"Start":"00:52.175 ","End":"00:54.800","Text":"In reduction there\u0027s a decrease in"},{"Start":"00:54.800 ","End":"00:58.969","Text":"the oxidation state of 1 of the reactants and electrons are gained,"},{"Start":"00:58.969 ","End":"01:02.555","Text":"that is the appear on the left-hand side of the equation."},{"Start":"01:02.555 ","End":"01:06.154","Text":"Now we\u0027re going to talk about oxidizing and reducing"},{"Start":"01:06.154 ","End":"01:11.000","Text":"agents which are related to oxidation-reduction reactions."},{"Start":"01:11.000 ","End":"01:12.440","Text":"Let\u0027s define them."},{"Start":"01:12.440 ","End":"01:16.490","Text":"An oxidizing agent causes another substance to be"},{"Start":"01:16.490 ","End":"01:21.080","Text":"oxidized while it itself gains electrons and is"},{"Start":"01:21.080 ","End":"01:26.150","Text":"reduced and the reducing agent causes another substance"},{"Start":"01:26.150 ","End":"01:31.715","Text":"to be reduced while itself loses electrons and is oxidized."},{"Start":"01:31.715 ","End":"01:35.060","Text":"So an oxidizing agent can also be called an oxidant"},{"Start":"01:35.060 ","End":"01:38.090","Text":"and an oxidizer and the reducing agent can"},{"Start":"01:38.090 ","End":"01:41.090","Text":"also be called a reductant or reducer but we\u0027ll"},{"Start":"01:41.090 ","End":"01:44.990","Text":"stick to oxidizing agent and reducing agent."},{"Start":"01:44.990 ","End":"01:48.050","Text":"In order to understand this better,"},{"Start":"01:48.050 ","End":"01:53.055","Text":"we\u0027re going to reconsider the examples we took before,"},{"Start":"01:53.055 ","End":"01:59.540","Text":"remember that we balanced equations in an acidic medium and in a basic medium."},{"Start":"01:59.540 ","End":"02:05.530","Text":"So the first example is the 1 we considered before in the acidic medium, here it is,"},{"Start":"02:05.530 ","End":"02:13.715","Text":"formic acid reacted with permanganate in the presence of acid to give us carbon dioxide,"},{"Start":"02:13.715 ","End":"02:17.735","Text":"manganese, cation and water."},{"Start":"02:17.735 ","End":"02:21.860","Text":"We divided this equation into 2 half reactions and"},{"Start":"02:21.860 ","End":"02:26.075","Text":"now we\u0027re going to consider the equations for each of the half equations."},{"Start":"02:26.075 ","End":"02:27.720","Text":"Here\u0027s the first one,"},{"Start":"02:27.720 ","End":"02:34.480","Text":"here permanganate is reduced to manganese cation."},{"Start":"02:34.480 ","End":"02:37.880","Text":"It\u0027s reduced because its oxidation state went from"},{"Start":"02:37.880 ","End":"02:43.340","Text":"7-2 and we see also there\u0027s electrons are gained,"},{"Start":"02:43.340 ","End":"02:45.365","Text":"there on the left-hand side."},{"Start":"02:45.365 ","End":"02:54.085","Text":"Here permanganate is reduced and gains electrons so we call it an oxidizing agent."},{"Start":"02:54.085 ","End":"02:56.415","Text":"Let\u0027s see the second reaction,"},{"Start":"02:56.415 ","End":"03:01.819","Text":"here formic acid is oxidized to carbon dioxide,"},{"Start":"03:01.819 ","End":"03:07.475","Text":"is oxidation state increases and electrons are lost."},{"Start":"03:07.475 ","End":"03:12.935","Text":"The formic acid loses electrons so they\u0027re on the right-hand side of the equation."},{"Start":"03:12.935 ","End":"03:20.815","Text":"So formic acid is oxidized and loses electrons and we call it a reducing agent."},{"Start":"03:20.815 ","End":"03:25.365","Text":"So we can summarize this by saying"},{"Start":"03:25.365 ","End":"03:34.170","Text":"permanganate oxidizes the formic acid and formic acids reduces the permanganate,"},{"Start":"03:34.170 ","End":"03:36.444","Text":"they always go in pairs."},{"Start":"03:36.444 ","End":"03:38.810","Text":"Let\u0027s take the second example,"},{"Start":"03:38.810 ","End":"03:44.570","Text":"which is the one we considered when we balanced equation in a basic medium."},{"Start":"03:44.570 ","End":"03:50.750","Text":"So chromium hydroxide reacts with bromine in the presence of"},{"Start":"03:50.750 ","End":"03:58.110","Text":"a base that\u0027s OH minus to give chromate and bromide and water,"},{"Start":"03:58.110 ","End":"04:02.495","Text":"we wrote half reactions for this reaction."},{"Start":"04:02.495 ","End":"04:04.520","Text":"In the reduction step,"},{"Start":"04:04.520 ","End":"04:08.120","Text":"bromine was reduced to bromide."},{"Start":"04:08.120 ","End":"04:12.680","Text":"It\u0027s reduction because it went from the oxidation state of 0 to"},{"Start":"04:12.680 ","End":"04:17.615","Text":"minus 1 and electrons were gained there on the left-hand side,"},{"Start":"04:17.615 ","End":"04:23.955","Text":"bromine is reduced and gains electrons and so it is an oxidizing agent."},{"Start":"04:23.955 ","End":"04:29.739","Text":"Let\u0027s look at the second part of the equation."},{"Start":"04:29.739 ","End":"04:35.300","Text":"Chromium hydroxide is oxidized to"},{"Start":"04:35.300 ","End":"04:42.390","Text":"chromate in the presence of the base and electrons are lost,"},{"Start":"04:42.390 ","End":"04:45.575","Text":"they appear on the right-hand side of the equation."},{"Start":"04:45.575 ","End":"04:53.390","Text":"So chromium hydroxide is oxidized and loses electrons so we say it\u0027s a reducing agent."},{"Start":"04:53.390 ","End":"04:58.819","Text":"We can summarize that by saying that in this reaction,"},{"Start":"04:58.819 ","End":"05:03.095","Text":"bromine oxidizes chromium hydroxide and"},{"Start":"05:03.095 ","End":"05:08.765","Text":"chromium hydroxide reduces bromine they go together as a pair."},{"Start":"05:08.765 ","End":"05:12.950","Text":"Now sometimes we\u0027re asked to decide whether something is"},{"Start":"05:12.950 ","End":"05:16.670","Text":"more likely to become an oxidizing agent or"},{"Start":"05:16.670 ","End":"05:25.040","Text":"a reducing agent and we can find out which is more likely using the oxidation states."},{"Start":"05:25.040 ","End":"05:29.315","Text":"If the oxidation state is high for a particular element,"},{"Start":"05:29.315 ","End":"05:33.380","Text":"then the substance is probably an oxidizing agent."},{"Start":"05:33.380 ","End":"05:40.385","Text":"Why? Because if the oxidation state is high,"},{"Start":"05:40.385 ","End":"05:45.320","Text":"then it\u0027s more likely to go down than up."},{"Start":"05:45.320 ","End":"05:48.830","Text":"So we say that it is more likely to be"},{"Start":"05:48.830 ","End":"05:53.315","Text":"an oxidizing agent where the oxidation state goes down."},{"Start":"05:53.315 ","End":"05:57.290","Text":"If the oxidation state is low for a particular element,"},{"Start":"05:57.290 ","End":"06:00.170","Text":"then the substance is probably a reducing agent."},{"Start":"06:00.170 ","End":"06:02.780","Text":"The oxidation state increases,"},{"Start":"06:02.780 ","End":"06:05.900","Text":"rather decreases so it\u0027s low down,"},{"Start":"06:05.900 ","End":"06:08.495","Text":"it\u0027s more likely to go up."},{"Start":"06:08.495 ","End":"06:11.300","Text":"Now what about intermediate cases?"},{"Start":"06:11.300 ","End":"06:15.410","Text":"If the oxidation state is intermediate for a particular element,"},{"Start":"06:15.410 ","End":"06:16.910","Text":"then the substance can be"},{"Start":"06:16.910 ","End":"06:21.020","Text":"either an oxidizing agent or reducing agent depending on the reaction,"},{"Start":"06:21.020 ","End":"06:23.770","Text":"it depends what it\u0027s reacting with."},{"Start":"06:23.770 ","End":"06:26.010","Text":"Here\u0027s an example,"},{"Start":"06:26.010 ","End":"06:30.985","Text":"which of the following are strong oxidizing or reducing agents?"},{"Start":"06:30.985 ","End":"06:34.085","Text":"In order to know which they are,"},{"Start":"06:34.085 ","End":"06:37.190","Text":"whether they\u0027re strong oxidizing or reducing agents,"},{"Start":"06:37.190 ","End":"06:40.385","Text":"we need to work out the oxidation states."},{"Start":"06:40.385 ","End":"06:46.400","Text":"So here are the substances nitrate NO_3 minus"},{"Start":"06:46.400 ","End":"06:53.070","Text":"nitrogen monoxide NO this is called hydrazine,"},{"Start":"06:53.070 ","End":"06:57.870","Text":"N_2H_4 and ammonia which is NH_3."},{"Start":"06:57.870 ","End":"07:02.285","Text":"First thing to do is to work out their oxidation states."},{"Start":"07:02.285 ","End":"07:03.500","Text":"So what are they?"},{"Start":"07:03.500 ","End":"07:07.220","Text":"Where I have 3 oxygens here in NO_3 minus,"},{"Start":"07:07.220 ","End":"07:10.500","Text":"that\u0027s minus 6 we want to be left with minus"},{"Start":"07:10.500 ","End":"07:14.780","Text":"1 so N has to have the oxidation state of plus 5,"},{"Start":"07:14.780 ","End":"07:22.255","Text":"in NO the oxidation state of O is minus 2 so that N has to be plus 2."},{"Start":"07:22.255 ","End":"07:26.915","Text":"In N_2H_4 the oxidation state of hydrogen is 1."},{"Start":"07:26.915 ","End":"07:31.910","Text":"So 4 hydrogens it\u0027s 4 and each nitrogen must be"},{"Start":"07:31.910 ","End":"07:38.130","Text":"minus 2 and in NH_3 the 3 hydrogens each plus 1,"},{"Start":"07:38.130 ","End":"07:42.030","Text":"that\u0027s plus 3 so nitrogen must be minus 3,"},{"Start":"07:42.030 ","End":"07:44.185","Text":"here we have it written out."},{"Start":"07:44.185 ","End":"07:48.320","Text":"Let\u0027s look at the highest number of plus 5 and we"},{"Start":"07:48.320 ","End":"07:52.085","Text":"said if it\u0027s a high number, high oxidation state,"},{"Start":"07:52.085 ","End":"07:58.429","Text":"it has to be an oxidizing agent and If it\u0027s a low number,"},{"Start":"07:58.429 ","End":"08:00.530","Text":"then it has to be a reducing agent."},{"Start":"08:00.530 ","End":"08:03.140","Text":"So we expect the NO_3 minus will be"},{"Start":"08:03.140 ","End":"08:08.360","Text":"a strong oxidizing agent and NH_3 a strong reducing agent."},{"Start":"08:08.360 ","End":"08:14.760","Text":"In this video we\u0027ve learned about oxidizing and reducing agents."}],"ID":18119}],"Thumbnail":null,"ID":86822}]

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