 So let's try this one. It says a 5.00 gram quantity of a diprodic acid was dissolved in water and made up to exactly 500 milliliters. Calculate the molar mass of the acid if 25 milliliters of this solution required 11.1 mils of 1.00 molar potassium hydroxide for neutralization. Assume that both protons of the acid were titrated. So I've already taken the liberty of writing down all the information the problem gives us and what it's asking for, the molar mass of that diprodic acid. So we know that it's a diprodic acid. We know that we're reacting it with potassium hydroxide. And we know that it says that assume both of the protons were titrated. So from that, we should be able to write out a reaction equation. OK, so let's go ahead and do that first. So the reaction equation, and this is just a general diprodic acid. We don't know what the acid is. So I'm just going to put H2. So those are the two protons that are going to be removed. And just A are a general acid. Aqueous, we're going to react that with potassium hydroxide agents. So that's a strong base, so it's going to fully react, right? But it's got to react twice. It's only got one hydroxide group on it, going hydroxide, all the atomic anion. So we've got to put a 2 there so it can react with both of the protons. So when we do that, we can now do our equations, our equation. So it's going to be K2A increase plus. So that's the reaction that's going on in this solution. So the next thing I would do is figure out, well, how many moles did it take to fully titrate that 25-mil aliquot or portion of the solution that we removed, the portion of that 250 mils? So how do we do that? Well, so we're going to figure out the number of moles of potassium hydroxide. So we've got here the molarity, so 1.00 moles per one liter. And we've got the volume. So we're going to convert this to, we do that, cancel, cancel, like that. So now, if we do this, we can get 0.0111 moles of potassium hydroxide. So what do we know about the number of moles of potassium hydroxide relative to the number of moles of the dichrotic acid? Well, we have a mole ratio of 2 to 1. So for every two moles of this, we have one mole of the dichrotic acid. So let's figure out the number of moles of H2A we have. We get that from the reaction equation. So 2 moles of KOH, 1 mole H2A, cancel, cancel, like that. So that's going to be 5H2A. But that was in 25 milliliters of solution. So we could figure out, well, what's the concentration of that right now, the concentration of that solution. But we're going to eventually need to figure out what the concentration of the actual solution is. Why? Because 5 grams was put into 250 milliliters, not into 25 milliliters. So we don't know how many grams we've put into the 25 milliliters. So what we can do here is divide this by 25.0, like that, and multiply that by 1,000. And that's going to give us, here we'll say, the molarity, or since we have it in brackets, we'll do it this way, H2A in that portion that was removed for that valid one. So that's going to be divided by 0.222 molar H2A. So but that's the concentration that we found in the 25 milliliters. But we want to know, well, what's the concentration in 250 milliliters? So this would be in 1 liter. So is it a moles 1 liter? So let's multiply that by 0.25 to 0.25 liters, 250, because it's only the two-suit things there. 0, 5, 5, 5 moles of H2A in 250 milliliters. So this is now the total number of moles of H2A is, what do we say, 0.0555 H2A. But we want to figure out, well, what's the molar mass of H2A? Remember, the molar mass is the mass per moles, right? So mass divided by number of moles like that. So what do we have? Do we have the mass? Yes, there. Divided by my number of moles here, 0.2555 moles. Three significant figures there. The molar mass of my acid is 90.1 grams per mole. So if you look on our table of diprotic acids, well, you could figure out what acid this actually was. So if you look, the molar masses, you can see that the closest one to this would be lactic acid. So that's the identity of this. OK, any questions on that one? You got it? I think you could do it on the test. OK, good.