Redox titrations | Chemical reactions | AP Chemistry | Khan Academy
Khan Academy Organic Chemistry
https://www.youtube.com/watch?v=EQJf8Gb8pg4

Transkript (automatisch erstellt)

0:01 - [Voiceover] We've already seen how to do an acid-base titration.
0:04 Now let's look at a redox titration. Let's say we have a solution
0:08 containing iron two plus cations. We don't know the concentration
0:13 of the iron two plus cations, but we can figure out the concentration
0:18 by doing a redox titration. Let's say we have 10 milliliters of our solution,
0:25 and let's say it's an acidic solution. You could have some sulfuric acid in there.
0:29 In solution, we have iron two plus cations and a source of protons from our acid.
0:36 To our iron two plus solution, we're going to add some potassium permanganate.
0:42 In here, we're going to have some potassium permanganate,
0:45 KMnO4. Let's say the concentration
0:49 of our potassium permanganate is .02 molar.
0:54 That's the concentration that we're starting with. Potassium permanganate is, of course,
0:59 the source of permanganate anions, because this would be K plus
1:03 and MnO4 minus. Down here, we have a source of permanganate anions.
1:10 We're going to drip in the potassium permanganate solution.
1:15 When we do that, we're going to get a redox reaction.
1:19 Here is the balanced redox reaction. If you're unsure
1:23 about how to balance a redox reaction, make sure to watch the video
1:27 on balancing redox reactions in acid. Let's look at some oxidation states really quickly
1:32 so we can see that this is a redox reaction. For oxygen, it would be negative two.
1:38 We have four oxygens, so negative two times four
1:41 is negative eight. Our total has to add up to equal negative one.
1:45 For manganese, we must have a plus seven, because plus seven and negative eight
1:50 give us negative one. Manganese has an oxidation state of plus seven.
1:57 Over here, for our products, we're going to make Mn two plus.
2:02 Manganese two plus cation in solution, so the oxidation state is plus two.
2:07 Manganese is going from an oxidation state of plus seven
2:10 to plus two. That's a decrease or a reduction
2:14 in the oxidation state. Therefore, manganese is being reduced
2:18 in our redox reaction. Let's look at iron two plus.
2:22 We have iron two plus as one of our reactants here.
2:25 That means the oxidation state is plus two. For our products,
2:30 we're making iron three plus, so an oxidation state of plus three.
2:35 Iron is going from plus two to plus three.
2:39 That's an increase in the oxidation state. Iron two plus is being oxidized
2:45 in our redox reaction. As we drip in our potassium permanganate,
2:52 we're forming our products over here. These ions are colorless in solution.
2:58 As the permanganate reacts, this purple color disappears
3:02 and we should have colorless, we should have a colorless solution.
3:07 Let's say we've added a lot of our permanganate. Everything is colorless.
3:12 But then we add one more drop, and a light purple color persists.
3:17 Everything was clear, but then we add one drop of permanganate
3:21 and then we get this light purple color. This indicates the endpoint of the titration.
3:28 The reason why this is the endpoint is because our products are colorless.
3:32 So if we get some purple color, that must mean we have some unreacted,
3:38 a tiny excess of unreacted permanganate ions in our solution.
3:43 That means we've completely reacted all the iron two plus
3:47 that we originally had present. So we stop our titration at this point.
3:52 We've reached the endpoint. We've used a certain volume
3:56 of our potassium permanganate solution. Let's say we finished down here.
4:01 If we started approximately there, we can see that we've used
4:05 a certain volume of our solution. Let's say it took 20 milliliters.
4:10 We used up 20 milliliters of our potassium permanganate solution
4:14 to completely titrate our iron two plus. Our goal was to find the concentration of iron two plus.
4:23 If we're going to find the concentration of iron two plus,
4:26 we could figure out how many moles of permanganate were necessary
4:30 to completely react with our iron two plus. We could figure out moles
4:35 from molarity and volume. Let's get some more room down here.
4:40 We know that molarity is equal to moles over liters.
4:45 The molarity of permanganate is .02. We have .02
4:50 for the concentration of permanganate ions. Moles is what we're solving for.
4:56 It took us 20 milliliters for our titration, which we move our decimal place one, two, three,
5:01 so we get .02 liters. So solve for moles.
5:06 .02 times .02 is equal to .0004.
5:13 So we have .0004. This is how many moles of permanganate
5:20 were needed to completely react with all of the iron two plus
5:25 that we originally had in our solution. It took .0004 moles of permanganate
5:31 to completely react with our iron. All right. Next, we need to figure out
5:36 how many moles of iron two plus that we originally started with.
5:40 To do that, we need to use our balance redox reaction.
5:44 We're going to look at the coefficients, because the coefficients tell us mole ratios.
5:48 The coefficient in front of permanganate is a one. The coefficient in front of iron two plus
5:54 is a five. If we're doing a mole ratio of permanganate
5:59 to iron two plus, permanganate would be a one
6:04 and iron would be a five. So we set up a proportion here.
6:08 One over five is equal to ... Well, we need to keep permanganate
6:13 in the numerator here. How many moles of permanganate were necessary
6:17 to react with the iron two plus? That was .0004.
6:22 So we have .0004 moles of permanganate. X would represent how many moles of iron two plus
6:31 we originally started with. We could cross-multiply here to solve for x.
6:35 Five times .0004 is equal to .002.
6:42 X is equal to .002. X represents the moles of iron two plus
6:51 that we originally had present. We're almost done,
6:55 because our goal was to find the concentration of iron two plus cations.
7:00 Now we have moles and we know the original volume,
7:03 which was 10 milliliters. To solve for the concentration
7:07 of iron two plus, we just take how many moles of iron two plus we have,
7:13 which is .002, so we have .002 moles of iron two plus.
7:18 We started with a total volume of 10 milliliters,
7:21 which is equal to .01 liters. So we have .01 liters here.
7:27 And .002 divided by .01 is equal to .2.
7:33 So this is equal to .2 molar. That was the original concentration
7:38 of iron two plus ions in solution. You could have used
7:43 the MV is equal to MV equation and modified it, because our ratio isn't one to one here.
7:49 That's another way to do it. But I prefer to actually sit down
7:52 and do these calculations and think about exactly what's happening.