Rob Lederer: Sometimes you can't just regard X when you are trying to find the concentration of hydronium in these questions. So, when you can't do that, you need to be able to use the quadratic formula. Now, I will take you through the steps and show you the mathematics behind this. I'm going to tell you that the vast majority of questions that you would do in Alberta diploma exams would not require a quadratic, but it's a really good tool to have because in your classrooms, anywhere else, your teachers are going to make sure that they ask you questions like this. Here is a 0.15 mole per liter solution of oxalic acid, HOOCCOOH, which is the COOH in one direction and the other. It has two protons to donate in solution and that makes it diprotic, it's called diprotic acid. But for any polyprotic acid, any acid with more than one proton, the first dissociation of solution produces all the hydronium that you really need calculation0-wise in order to get the pH. You never worry about multiple steps or anything like that. Just take out one proton to calculate the pH, not for stoichiometry, later we'll talk about that, but just for calculating pH. Now, here is the oxalic acid loses a proton to the water to form hydronium and the hydrogen oxalate ion. Now, if you can find that hydronium ion concentration at equilibrium, you got the pH. So, 0.15 is what you initially have, nothing here and nothing here. This acid dissociates X amount to form those two Xs there; 0.15 minus X is how much this loses. We are looking for that X there. It's X here for the conjugate base ion in solution as well. I'll mention this again by the way later, but this is the Brønsted-Lowry acid, this is the base. But if you go back in this direction, that's the Brønsted-Lowry acid and that's the Brønsted-Lowry base. The Ka equals this times this divided by this, don't include the water. Then the Ka for oxalic acid is 5.6 times 10 to the negative two, that's 0.56' that's what that number actually equals, and that equals X square over, which is X square, these to multiply together times 0.15 minus X. So, in order to continue with this question, you need to realize that you can't disregard X here and you could try. If you took this K value here and divide it into that initial concentration of 0.15, you don't get a number that's greater than 1000. So, all better off, you got to include the X in the denominator and work out this math like this, right here now. So, we took that 0.15 minus X and multiply it through each side to get 0.056 times that, equals X-square. Now, we want to, after we foil this and multiply this times this to get this and this times negative X to get this, that equals X-square. We want to take this and then write it in standard form, which is having the square term, the X term, and then the number A, B, and C in that order. So, zero equals one X-square and A equals one, and B equals 0.56; remember we took this, then we added this to each side to get it on the X-square side, so it becomes positive. This number, which is C, goes over to this side as well. We have to subtract it from each side and it becomes negative. So, A is one, B is 0.56 and C is negative 8.4 times ten to the negative three. Those numbers A, B, and C then get plugged into the quadratic formula where X equals minus B or negative B, plus or minus the square root of B-square minus 4ac over 2a. You don't have to do the minus part, because that actually could give you a negative root and that's not what we are looking for here. We can't have a negative concentration. We only care about positives here, because this chemistry is real man. So, therefore when we just take this and what I do first if you - when I put this in the calculator, you can put brackets around everything, I always just start it here, punch in my B-square minus 4ac, take the square root of that, add it to negative B, then divided it by two; a is going to be the number one in front of the X. That will give the X. W