 What we will do is have a quick look at the sort of questions that you're likely to get. So if we have a one-molar solution of ethanoic acid at 25 degrees, we know this is going to ionize to produce a solution with the acetate or ethanolate ion and H plus ions. And if we know that the solution has H plus ions of 4.18 times 10 to the minus 3, and we could do that simply by measuring the pH, then we can calculate both the Ka and the pKa. So we can calculate the Ka by looking at the concentration of products over reactants, leaving the water out. So this is going to be H plus multiplied by the concentration of CH3C00 minus over the concentration of CH3C00H. Now there's a couple of assumptions that I'm making here. The first is that the mole ratio is 1 to 1, and so therefore if the concentration of H plus is 4.18 times 10 to the minus 3, then I can assume the concentration of CH3C00 minus is going to be the same. So this is going to be 4.18 times 10 to the minus 3 multiplied by exactly the same value. Now the bottom value is the value of the concentration of the molecule itself. Now we know that it was a one molar solution, and we know that it's not going to be one molar, it's going to be one mole less the 4.18 times 10 to the minus 3 moles per liter that have reacted in order to form the ions, or at least have ionized in order to form the ions. But subtracting that away is going to give us a value which rounds very closely anyway. So there are two ways of doing this. You can either do it exactly, so 1 minus 4.18 times 10 to the minus 3, or you can round. And if you do it both ways, as you start, you'll actually see the difference in those numbers and see whether the rounding makes much difference. For today, to keep it simple, let's just ignore the fact that it's going to lose a little bit of that and divide it by 1, because it's the easiest thing. So therefore my Ka value is going to be 1.75 times 10 to the minus 1, 2, 3, 4, 5. Okay? We leave the units out like most of our equilibrium constants, we just ignore the units for now. In order to calculate the pKa, the pKa is found by going minus log base 10 of the kA value, which in this case is going to give me 4.76 as my pKa value, sorry. So my kA value is of the order of magnitude of 10 to the minus 5, and my pKa value is 4.76. So what you can see here is I actually have quite a weak acid. It's not fully ionized, only a very small amount of it has actually been produced. Why don't you do the full calculation, leaving the subtracted amount in and see what difference it makes to your calculation of kA and pKa, and of course we're going to have a look at a lot more examples in class. So thanks for watching.