So here it says, calculate the equilibrium concentrations of phosphoric acid, dihydrogen phosphate, hydrogen phosphate, phosphate, and the hydronium ion for 0.35 molar of phosphoric acid. Alright. Here we're dealing with a triprotic or polyprotic acid and that's highlighted by the fact that we have 3 K_{a} values given to us. Here in this triprotic or polyprotic acid, we have all 3 acidic hydrogens. Therefore, we know that it's going to start off as an acid. It will react with water. Because it's in its acidic form, we know that it's going to donate an H^{+} to water. By doing this it becomes H_{2}PO_{4}^{-} and water becomes H_{3}O^{+}. Here we're going to say initial change equilibrium. We know that we're going to ignore solids and liquids. The initial concentration of this weak acid is 0.35 molar. These are initially 0. We're losing the reactant so that we can build up our products. Bring down everything. Since we're talking about removing the first acidic hydrogen or donating it to water, we know that we're dealing with
K_{1}
. Here that's equal to our products over our reactants. Times H_{3}O^{+} divided by our reactant. Now we plug in the values we know for each term. So
K_{1}
is
7.2×10-3
. Both of these are x so that's x squared on top divided by 0.35 minus x on the bottom. We use the 5% approximation method here so the initial concentration of 0.35 divided by the K_{a1} value we're using. When we do that, you should see that you don't get a value greater than 500. Because of that, you'll have to keep the minus x here and perform the quadratic formula. So we're just multiplying both sides by 0.35 minus x. So distribute, distribute. When we do that, we're gonna get 0.00252 minus 7.2 × 10^{-3} x equals x squared. Our lead term is the x squared since it has the highest power, So everything has to be moved over to its side. When we do that, we're gonna get x squared plus 7.2 × 10^{-3} x minus 0.00252. This represents my a, my b and my c. So using the quadratic formula, we're gonna have negative 7.2 × 10^{-3} plus or minus
7.2×10
-3
-4×1×-0.00252
divided by 2 times 1. Here, that's gonna give us negative 7.2 × 10^{-3} plus or minus 0.100657 divided by 2. Now remember, because it's plus or minus, that means we're going to get 2 different variables, answers for x. So x can equal 0.0467 molar. Or if you do the negative approach, x equals -0.0539 molar. Realize here that at equilibrium, you cannot have a negative value. It's not possible. That means that this negative x gets dropped out. That x represents our correct answer. Now we're gonna say at equilibrium, H_{3}PO_{4} equals 0.35 minus x. So plug in the value we just found for x. So that equals 0.3033 molar. At equilibrium, H_{2}PO_{4}^{-} equals H_{3}O^{+} because they both equal x. So their concentrations are 0.0467 molar. Alright. So, so far, we've found the concentration for our phosphoric acid, our dihydrogen phosphate, as well as our hydronium ion.
We still have to figure out what our hydrogen phosphate and our phosphate ions will be. Now, we need to realize at this point, I've created this first intermediate. Because it still possesses acidic hydrogens, it can continue to donate H^{+} to other water molecules, and that's what's going to happen. We're going to write a new equation. Now that dihydrogen phosphate will react with another water molecule. It'll donate an H^{+} to that new water molecule and therefore become HP_{4}^{2-} plus H_{3}O^{+} being created. We have again initial change equilibrium. Again, we ignore water because it's a liquid. Initially, HPO_{4}^{2-} this is the first time we're seeing this so initially it's 0. At equilibrium in the previous ICE chart we found out that both your dihydrogen phosphate and your hydronium ion equaled x which equaled this number. So they still equal that number and now we're creating a new ICE Chart, so their initial amounts are that number. Now, we're gonna say again we're still losing reactants here in order to make these products. Bring down everything. Now we're going to say here at this point, since we're dealing with the second acidic hydrogen being removed to create our second intermediate, we're dealing with
K_{2}
, which equals products over reactants. Coming back up here, let's look at what K_{a2} is. So K_{1} is
7.2×10-3
, Ka_{2} is
6.3×10-8
. Look at the differences in the powers, negative three to negative eight. Remember I've said before in the past that the first acidic hydrogen is the easiest and each hydrogen afterward becomes harder and harder to remove, which is why we see a big decrease in our Ka values. They're different by a magnitude of 5. What does that mean? Well, that means that H_{2}PO_{4}^{-}, although it acts as an acid and donates H^{+} because its Ka_{2} is so small, very, very little product is formed which means it's gonna lose a variable with x which means that this x is so small that it's not gonna really affect the final amount for H_{2}PO_{4}^{-}, and therefore it can be ignored. Same thing can be said for H_{3}O^{+}. This plus x, because x is such a small number, it's not going to really increase the concentration of H_{3}O^{+} so it can be ignored. This x we have to keep around because we're looking for the concentration of HPO_{4}^{2-}, and that's all we have for there. So going back again, Ka 2 is
6.3×10-8
.