 Okay, so let's try this titration problem. So we're finding the pH during a weak acid, strong base titration, and this portion of the problem is before any base has been added. So hopefully you can see zero mills. So it says calculate the pH during the titration of 40 mills of propionic acid. Propionic acid, the k of it is 1.3 times 10 to the negative 5th after the following volumes of 0.10000 molar in AOH is added. And this first one is zero mills. So since there's no mills of hydroxide being added, this is just like doing a weak acid solution problem. So how do we do that? We have to figure out what the reaction equation is and then set up an ice table. There's actually, you know the quick and dirty way to do this, but we're going to do the whole shebang at this point. So first things first, right out of your acid, plus water, plus back and forth. Everybody OK with that? So remember this is we're able to remove x because k-h is very small, so 5% rule. So I'm going to erase the whole table. Is that OK? I'm just going to keep these values down there and then I'm going to write k-h out, OK? So can I do that if everybody got this done? OK. So what did we say? 5% allows us to do that. So k-h equals concentration of the conjugate base, H3O plus divided by an acid. So we're going to solve for, remember this is x and this is x, so we're going to solve for x squared here, OK? So when we do that and we solve for x squared and we take the square root of that, right? That is solving for H3O plus. Is everybody OK with that? Can I do that in one step or should I do both steps? Well, some people want to do both, so we'll do them in both, OK? So we're going to rearrange this and take this up to here, OK? And then these two both equal x, OK? So I'm going to erase these guys. Those are x's, so x and x. So we're going to plug in x squared for that, OK? So when we do that, right? That means this is x and this is x. So if we take the square root of that, then H3O plus equals the square root, right? Does that make sense? So H3O plus is going to equal the square root of k-h times the concentration of CH3CH2CO like that, k to the 5, concentration, 0.100, so 1.3 times 10 to the negative 6. So I get 1.1 times 10 to the negative 3 molar. Is everybody OK with that? So the pH, I'm going to erase this. pH, negative log of the H3O plus concentration, which is the negative log of 1.1 times 10 to the negative 3, 2. So that's the pH of the solution before any acid is added, or any base is added. Only the weak acid. So it's just like you would do a normal weak acid solution. OK? Are there any questions before I kill this one? That's the right one. Alex, that's 2.959. Oh, OK. Any other questions?