Determine the pH of a 225 molar phosphoric acid solution. Here we're given 3K values since its triprotic. Now here step one is we're going to set up an ice chart for the weak Tri protic acid that has it reacting with water. Remember to use the Bronsted-lowry definition to predict the products formed. So here we have our weak triprotic acid reacting with a mole of water or molecule of water. Since it's an acid, it's going to donate an H plus the water. It becomes dihydrogen phosphate water gained in H plus, so it becomes H_{3}O^{+}.

We're dealing with an ice chart fence. It's a weak acid since it's K_{a} is less than one. And remember, an ice chart is initial change equilibrium. In a nice chart we ignore solids and liquids, so the water will be ignored. The initial concentration is 0.225 molar. Our products initially are 0, so remember we've filled out the initial row for step three. We lose reactants to make products. So using the change row, place A -, X for the reactants and A + X for the products, Minus X since we're losing this reactant and plus X since we're making them. We then bring down everything, so .225 -, X + X + X.

Our ice chart is completely filled in, so Step 4 says using the equilibrium row, set U the equilibrium constant expression with K_{a1} and solve for X K_{a1} is equal to products over reactants, so it equals dihydrogen phosphate ion times the hydronium ion divided by phosphoric acid. Here we place the numbers that we know for K_{a1} in the different concentrations here. So here K_{a1} is 7.5 * 10^{-3}. These would be X^{2} / .225 -, X. Now once we set up the expression, we can check to see if a shortcut can be utilized to avoid the quadratic formula. We do this by utilizing the 500 approximation rule, where we take the initial concentration of our weak triprotic acid, and in this case, divided by K_{a1}. If the ratio is greater than 500, we can ignore the minus X within our equilibrium expression.

So when we punch this number into our calculator, it gives us a number around 30, a number that's much lower than 500. So unfortunately, we have to keep the minus X and utilize our quadratic formula here. All right, so let's go back up and let's do that. We're going to cross multiply these, so we're going to say 7.5 * 10^{-3} distribute distribute. When I do that I'm going to get .0016875 -7.5 * 10^{-3} X = X^{2}. The X^{2} is the lead term, so everything has to move over to it. So you would add this to both sides and you will subtract this from both sides. Doing that would give us a new equation which is X^{2} and it's going to be plus 7.5 * 10^{-3} X minus 0.0016875.

This would be my A, my B and my C Setting up the quadratic formula, we'd have -7.5 * 10^{-3} ± 7.5 * 10^{-3}^{2} -, 4 * 1, negative sign, negative 0.0016875 divided by two. Solve for this portion here, and so we'd have -7.5 * 10^{-3} ± .0825 / 2. Because of the presence of plus or minus here, there's two possible answers for X1 where it's plus 0.0825 and one where it's negative or minus 0.0825. So X here could equal .0375 or X = -.045, the correct one. The correct X is the X where you can plug it anywhere in terms of the equilibrium row and your answer is always positive. That means the negative X. We cannot use because placing it here or here would give me a negative result.

Alright, so we just figured out the correct X to use, which is this one here that X that we just found. Well, that X is connected to H_{3}O^{+}, so we could take that X that we just found and find our pH. All right. So X = .0375, which is equal to the concentration of H_{3}O^{+} pH equals negative log of H_{3}O. Plug it in and when we do that we're going to get approximately 1.43. Add the pH for this week triprotic acid. So that would be our final answer.