All right, continuing with acids, what about polyprotic acids? So the acid examples that we've done so far, it is just the one proton that that acid is putting into solution. But I mean, we know of polyprotic acids, right, especially sulfuric acid. We've already said that the first ionization is strong.
That means the second ionization is weak, and we might need to deal with that. So how do we deal with polyprotic acids? They ionize in successive steps.
So one proton comes off at a time. And each step will have its own Ka value. And those Ka values in the later steps are much smaller because the protons get harder to remove.
So as an example, I have sulfurous acid here. It's weak in both ionization steps. The first step...
Kicking out one proton to give HSO3-. That's Ka1, which is 1.6 times 10 to the minus 2. HSO3-, kicking out a proton to turn into SO3 That's the second ionization step. Ka2 is 6.4 times 10 to the minus 8th. So a lot smaller. It's harder to remove that second proton.
For sulfuric acid, Ka1, if you look at tables, especially for us dealing with all of these in water solutions, the Ka1 value might not be listed because we're just saying that it's strong, right? It 100% ionizes in water. You just split it up. It's strong. And then Ka2 is 1.2 times 10 to the minus second.
So Ka2... smaller than Ka1. It's harder to remove that second proton.
H3PO4, three protons. When you look it up, you will see that it has three Ka values. I have some good news.
Let's talk about calculating the pH. We can treat polyprotic acids using only Ka1 if a couple of conditions are met. One, If Ka1 is much greater than Ka2, so if it's about 300 times greater than Ka2, and then the second is if the concentration is greater than 0.05 molar. The Ka conditions are pretty much always met.
The concentration condition, right, you'll just have to read the problem. Most of the time, the concentration condition is also met. So in these cases, the H plus that forms in step one is much greater than the H plus formed in subsequent steps, which is one of the reasons why we can just deal with the first ionization. Also, if you have the first ionization and there is already H plus in the solution, Le Chatelier's principle...
will help us see that you won't get as much H plus from the second ionization. Now, we're talking about it in a slightly different order. But the way we looked at it with Le Chatelier's principle was like, if I just had the second ionization, it got to equilibrium, and then I perturbed it, what would happen?
Well, if you just have the second ionization, it gets to equilibrium. And then... We say, oh, oh wait, actually both ionizations are happening.
Let's put all the H plus from the first reaction in. That reaction we were looking at, that second ionization, would move to the left. Now we're saying we're going to start with some H plus in solution. And so what will happen is that because that H plus is already there, the second ionization will not go as far to the right. All right, we'll see this in some calculations.
Now the exception is if your solution is less than 0.05 molar. We're going to do an example of this. So I would like to calculate the pH of 0.010 molar H2SO4. We have to deal with all of the ionizations and deal with them appropriately. The very nice thing is that this first step is strong.
So in step one, the first ionization is strong. That means my... H plus concentration after the first ionization is 0.010 molar. First ionization, done.
Second ionization is when HSO4- acts like an acid and turns into H+, and sulfate. So it's weak. I'm going to set up a rice table. I have my reaction.
My initial conditions come from step one. So in step one, I took H2SO4. completely ionized it.
That gave me the HSO4 minus that is the reactant in my second ionization, but it also gave me some H+. So I have to fill in this initial concentration of 0.010 molar H+. But I don't have any sulfate because my HSO4 hasn't ionized yet.
I don't have one of my products. I still have to move to the right. I'm still going to lose one X reactant, make one X of each of the products, get my equilibrium row values.
And I have the equilibrium constant because I have the KA for this step. So products over reactants equals 1.2 times 10 to the minus two. Now you should practice doing out the algebra for this because you cannot assume that.
x is small. I mean 0.01 and 1.2 times 10 to the minus 2, those are basically the same number. They're not far apart enough at all for you to do x is small.
So algebra on a quadratic, out of this you will get negative 0.027, which we can reject, and 0.0045. So my h plus concentration in solution. is 0.010 plus x. So 0.010 plus 0.0045 is 0.0145. It is significant enough of a difference to change the pH. That's why we have to account for it in this case.
The negative log of 0.0145 is 1.84. So the pH of the solution is 1.84. All right. There are other cases. When we are very, very dilute, so concentrations closer to 1 x 10 to the minus 6th, 1 x 10 to the minus 7th, you would need to also account for the KW that comes from the autoionization of water.
I am not asking you to be responsible for those cases. So if you see practice problems with it, you don't need to do them. You don't need to worry about the reading. Just know for... low concentrations, right at this threshold of less than 0.05 molar, you do have to account for every ionization.
Now, here's the thing. It would take you longer to account for every ionization regardless of concentration, but you would not be wrong. You would just find that the subsequent steps that you calculate an H plus concentration contribution from that the contribution is so small, it doesn't change the overall result.
However, it would not be incorrect to do it for every ionization. I think you would just not want to spend that long doing it unless it was necessary.