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## Titrations

Current time:0:00Total duration:7:58

# Redox titrations

AP.Chem:

SPQ‑4 (EU)

, SPQ‑4.B (LO)

, SPQ‑4.B.1 (EK)

## Video transcript

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