 Ok, so let's try this one. It says the pH of a 0.25 molar HF solution is 2.036. What are the values of Ka and pKa for hydrofluoric acid? So in order to do these Ka, pKa questions, it's asking you what is the kA? So in order to figure that out, you have to figure out, well what is the equation of HF or hydrofluoric acid reacting with water? So let's write that out first. So HF AgWis plus H2O liquid is an equilibrium with F minus Equial plus AgWis. So since we know that and we know how to figure out Ka, we should be able to put the two together. So what would be the expression for Ka? So recall it's the products, the reactants, right? So in this case the products are going to be a concentration of F minus times the concentration of H3O plus divided by the concentration of the reactants. Now remember with the reactants we only put the things that are either AgWis or gaseous in these equilibrium expressions so we're going to leave out water. So HF concentration of HF. So you might be tempted to take this concentration here and plug it into this value. But unfortunately you can't do that because this actually is the initial concentration of HF. So what you're going to have to do is set up what we call an ice table. So ice initial change equilibrium. And from this we can get the value of Ka. So the initial concentration of HF is presented to us in the problem at 0.250 molar. H2O we don't worry about. F minus the problem didn't us to put any in there. So it's 0 molar. H0 plus 0 molar. But once we added our HF it did react with water so something happened to it. It decreased by some amount. So that amount we're just going to say is X and the negative shows that it decreased. So since we have a one ratio of HF to F minus this F minus increased as much as this HF decreased. So we're going to put a positive X there. And of course the same thing happened to H3O plus so positive X there. So our equilibrium concentration which is what we're going to put into this expression for Ka is going to be for HF 0.250 molar minus X and F is going to be X and H3O plus is going to be X. So let's just go ahead and put these equilibrium values in for Ka. So F said it's X. H3O plus is also X. And HF is 0.250. There's no units in Ka so we'll just not put any minus X. So the problem also gives us this information though. PA H. So what do we know about the pH? pH is the negative log of the concentration of hydronium mine. So what are we looking for? Well one of the things we're looking for is the concentration of hydronium mine and it happens to equal X. So it'd be really nice to figure out well what is that concentration? So we rearrange this. H3O plus equals 10 to the negative pH which equals 10 to the negative 2.036. So we're going to have to plug that into our calculator. We're going to get 9.20 times 10 to the negative 3 molar. So notice that equals H3O plus and H3O plus equals X. So we can plug that value in for all of those things. So let's do that. So X times X is X squared. So I'm just going to write 9.20 times 10 to the negative 3rd squared divided by 0.250 minus 9.20 times 10 to the negative. So when we do that we get the KA and it's going to be 3.52 times 10 to the negative 4. No units. So that's the KA for hydrofluoric acid. And remember in chemistry P that's just another name for the function that's the negative log function. So if we're looking for something that's PKA that's going to be the negative log of the KA. So the negative log of that number that we just figured out, 2.52 times 10 to the negative 4, 3.454 when I do that. So remember your significant figure rules when you're going to and from logs. So we went from 4 so we went down to 3 and so we went from 3 here up to 4. So you've got to remember that rule. Go back to chapter 1 if you don't remember. So are there any questions on this one? Pretty long drawn out process but you've got to do it every time. So these ice tables are really the key for doing all of these types of problems. Let's kill them.