So in this video we're going to learn how to calculate the concentration of the basic form of a weak diprotic acid. Now this is a special case, so pay close attention. We're going to say when given the initial concentration of our weak diprotic acid, the concentration of its basic form equals the K2 value. Now what is a question like this look like? Here in this example it says determine the concentration of the carbonate ion when given 0.115 molar of carbonic acid solution. So they're giving us the initial concentration of our weak diprotic acid. So it's acidic form and they're asking you to determine the concentration of its basic form.

Remember, for a weak diprotic acid we have the acidic form or it has both H+ ions. When it loses its first H+ ion, it becomes its intermediate form, and then when it loses that second H+, it becomes its basic form like we set up above. If they give you the initial concentration of the weak diprotic acid, then K2 equals the concentration of the basic form. So here we just simply say that the concentration of the carbonate ion equals K2, so it equals 5.6×10-11 molar. That will be our answer. Now if you don't want to continue on with this video, you can turn it off, because what we're about to do is we're about to show proof of why that is OK. But again, remember, if they give you the initial concentration of the weak diprotic acid, the concentration of the basic form is equal to K2.

For those of you who are continue on with this video, let's get started. So how do we prove this? Well, step one, we set up an ice chart to calculate the concentration of the intermediate form and the hydronium ion, and we're doing this by using K1. So the carbonic acid will react with water in its liquid phase. It donates an H+ to water to become the carbonate ion or bicarbonate ion. Sorry plus the hydronium ion. This is a nice chart. So it's initial change equilibrium. Remember with this we ignore solids and liquids. The initial concentration of my weak acid is 0.115 molar. These are zero. We lose reactants to make product. We bring down everything using the 500% approximation method. We would see that we could ignore the minus X.

If you're not familiar with this approach, make sure you go back and take a look at my video where I talk about the 500 approximation method. So here K1 equals products overreactants, so we plug in the numbers here. So K1 is 4.3×10-7 = X2 / 1 one five. When you cross multiply you see that X2 = 4.945×10-8, and when you take the square root of both sides you get X equals 2.224×10-4. Now that we've found X, we've found the concentrations of both the intermediate form, so your bicarbonate ion and the hydroxide ion.

With that information, we move on to Step 2, set U another ice chart and using the initial role, place the calculated amounts of the intermediate form and the hydronium ion concentration. So here we would place 2.224×10-4 molar. Here this is reacting with a second molecule of water creating the carbonate ion which is our basic form and we create more hydronium ion whose initial concentration would also be 2.224×10-4. Again this number is coming from the first ice chart where we calculated the equilibrium concentrations. Still a liquid, Place a zero for the concentration of the basic form, in this case carbonate ion set. Haul this up again. So this is initial change equilibrium. We're going to say here we lose reactants to make products, so usually change row. Again, we do minus X for reactants, plus X for products. Bring down everything in 2.224×10-4 + X.

So now here we're going to say, using the equilibrium row, set up the equilibrium constant expression. But now we're doing it with K2 and we're going to use it to solve for X. Now we're going to say we're going to check our shortcut, the 500 approximation method, to see if we can avoid the quadratic formula O. Here the initial concentration is not the very first initial concentration. If the initial concentration of our intermediate form, which is the bicarbonate and we're going to divide it by K2, when we do this, we're going to get 3964285.7. So by doing the initial concentration divided by our K2 in this in this case, we found a value greater than 500. But now we can ignore not only the minus X that we're used to seeing, but also plus X, because as we can see here, we had initial concentration of H3O plus, which gave us A + X here at the end.

So we'd ignore those two things. So we ignore this one, and we ignore this one. What else happens by ignoring the plus X and the minus X? What cancels out both of these two numbers because they're the same? And what do you have left at the end? You see that X equals the K2 value. Coming back up here in our eye chart, we can see X equals the concentration of the basic form of my diprotic acid. So again, it just simply equals K2. So this will be our answer. So all of its work was just proof to show that if they give us the initial concentration of the acidic form of the weak diprotic acid, then K2 is equal to the concentration of the basic form.

You don't need to do all this work. This is just to prove that OK, just remember, you see a question like this in the beginning, you can say, oh, this is a special case, The concentration of basic form is just simply equal to K2. Alright, so keep that in mind when dealing with this special circumstance, special case.