Okay, so this one says ascorbic acid, which is also known as vitamin C, the molecular structure or the molecular formula is written there, is a diprodic acid and it gives us the K a, K a1 is this, K a2 is this, and it's found in citrus fruits. Calculate the equilibrium concentration of the ascorbic acid, the equilibrium concentration of the mono anion of ascorbic acid, the equilibrium concentration of the di-anion, which isn't written up here, but it would be C6H6O62 minus, and the pH of this solution if we've got an initial concentration of ascorbic acid at 0.050 molar. So first thing you're going to have to do is write this reaction equation down, okay? I just didn't do it because it takes a little bit of time and we were recording this, okay? We're kind of short on time in class. So second thing we want to do is write an high stable, okay? So remember with the high stable, cancel that out first. So initial concentration of ascorbic acid, like that, minus X, like that, 0, 0 plus X, plus X. So this is a weak organic acid, so we're going to be using that 5% rule eventually, okay? So even though we're going to write here, well what's the equilibrium concentration 0.050 minus X, X and X, okay? So it also, well I know, it doesn't give us a pH, but so 5% rule is that, right? So we just erased that X that. So let's write out what our equation for kA1 is. What would it be? You guys can write it out on your own and check up here and see if you get the same answer. Okay? So it's going to be, you guys got it? Okay, wonderful. So here, now what can we do, since we used the 5% rule we can put 0.050 down there, okay? And we know what kA is, okay? So we're looking for X effectively, okay? So let's rearrange this, well let's plug in all our stuff and then rearrange it and look for X, okay? So in other words, we know what kA is, so that's 1.0 times 10 to the negative 5 equals X times X on the top divided by 0.050, like that. Is everybody okay with me doing that? So the next thing I want to do is reduce this to X squared and then I'm going to solve for that. So solving for X squared, I'm going to get 0.050 times 1.0 times 10 to the negative 5. So that would be X squared, right? And then X would be the square root of that. Is everybody okay with that? So let's do that, 0.05. So I get 7.1 times 10 to the negative 4 and in this case it's going to be molar, okay? Why is it going to be molar? Because remember X is actually the concentration of H3O plus, like that, right? And since it's the concentration of H3O plus, it's also the concentration of that mono-anon, right? So X equals that, that, and that. So next thing I want to do is just kind of put, so can I erase some stuff up here? Everybody got everything right down. Okay, wonderful. So I'm going to erase this and put 7.1 times 10 to the negative 4. So now I'm going to erase all of this bottom stuff down. What were some of the things that asked us? Well, can we figure out the concentrations of this stuff? Well, what would be the concentration at equilibrium of H2C6H6O6? Well, we invoke the 5% rule, right? So at equilibrium it's going to be effectively the same concentration. So 0.050 whole, like that. Is everybody okay with that one? The next thing it wants us to do is figure out, well what's the concentration of this stuff? Here, okay. So the concentration of HC6O6C6H6O6 minus at equilibrium is going to be 7.1 times 10 to the negative 4 whole. So the one thing we haven't written yet is the second equation. So that's the next thing it's asking for. But let's figure out what is the pH of this solution. So remember I told you the pH comes from just this first equation here, okay? So let's figure out that first. So the pH of this solution is going to be the negative law just like any other solution of the hydrogen or hydronium ion concentration. In this case it's 7.1 times 10 to the negative 4. 5 is the pH of this solution. So now we're going to find out what is the concentration. The last thing it's looking for is the concentration at equilibrium of C6H6O6 2 minus. So it's looking for that one there. So how do we figure that out? Well, you guys don't need that up. I'm going to erase all of this stuff here, is that all right? Okay, one more. So in order to do that last portion you have to remember the second equation of this, right? Which is, well, we start with that intermediate, that monodicronated thing. So like that. So everybody will get this starting there. Adding H2O to that C6H6O6 minus 2 minus equals plus H3O plus. Okay, so from that we should be able to get the Ka2 equation. So everybody okay with that? So Ka2 equals the concentration of C6H6O6 2 minus times H3O plus. Concentration divided by this concentration up there. C6H6O6 like that. You okay there? So we're going to rearrange this equation to solve for this equilibrium. So when I do that I'm going to get H6O6 2 minus Ka2 times concentration of HC6H6O6 minus divided by the concentration of H3O plus. Like that. So everybody okay with that? Okay, wonderful. So when we plug in, so we're just now going to plug and show because we got all these numbers. Ka2 is 5 times 10 to the negative 12. This concentration is 7.1 times 10 to the negative 4. Divided by H3O plus concentration right is also 7.1 times 10 to the negative 4. Okay and you can plug that all in or you can just see that that's the same as that. Cancel each other out. So the concentration of this stuff is going to be 5 times 10 to the negative 12. So the only one you had to use that second equation for was the diam. Any questions on this one? Again just using the high tables that you already know how to use.