 Okay, so let's try one of these weak acid problems where we're going to calculate the original molarity from the pH of the solution. So this one in particular says, what is the original molarity of a solution of formic acid HCOOH, it gives you, whose pH is 3.26 at equilibrium. So the other thing that you have to know for this problem is the Ka, which is 1.7 times 10 to the negative 4, and it actually gives you a table where you need to find formic acid and find its Ka. Okay, so I've already written that up there for you. So the first thing I like to do in these types of problems is to well figure out, well what does this mean, the pH is 3.26. So remember, the pH, that's a measurement of the hydronium ion concentration, okay? So let's just relate that. So the hydronium ion concentration H3O plus equals 10 to the negative pH, okay? So you guys got to remember how to do that, if you don't remember, go back to the pH list, video list, and figure it out. Okay, so then that's going to be 10 to the negative 3.26. So that's going to be our hydronium ion concentration. So let's do that together, 10 to the negative 3.26, and I get 5.5 times 10 to the negative 4 molarity, okay? So you got to do that first. So you can kind of think about, okay, where does this fit in, okay? So that means we're going, so we don't have that in the reaction yet, okay? So we got to find that in the reaction. So that means this acid has to react with the base, and the base it's going to react with is water. So that's why I like to do this first to remind myself, okay? That's what I was trying to tell you, put water in there, put water in there, okay? So we're going to have this equilibrium reaction, of course. Acid reacting with the base is going to give us our conjugate base, the formate ion ion, plus the conjugate acid, which is the business that we're looking for, okay, wonderful. So now we can do an ice table like this. We put in some amount of this, okay? Well, let's just, with liquids, we don't do anything. And we know what concentration of hydronium ion is at equilibrium, right? So let's just put that down here, alright. So did we have any of this to start out with? No. Right, it didn't say. It just said we put in formic acid. So did we put it in the conjugate base? No, because it told us we only put in formic acid, right? So we have zero and zero. So that means this must have been plus x, right? That must mean this is also plus x, right? So that means that if this is only x, and this is only x, then this is 5.5 times 10 to the negative 4th 2. So it's kind of like you got to do some detective work, you know what I'm saying? Kind of imagine yourself, you know, uncovering new clues, you know, every time and saying, okay, if this must be this, then this must be this also. Because this is a 1 to 1 ratio here, right? Okay, so watch what happens here. This we subtract x, right? So minus x there. And we're looking for that original concentration. So I'm just going to put, normally I'll put y there, okay? But I'm just going to put for this time, h c c o h original, like that, okay? So we won't get kind of confused, okay? So the equilibrium is going to be h c c h c o h original minus x, right? And we know what the value of x is already, right? So that is going to be minus 5.5 times 10 to the negative 4, like this, okay? So we can say this whole thing, if you want to, set this whole thing equal to y for a second, if you want to. So we can plug it into the other equation. And then we can set y back equal to that in a second, okay? So if you don't understand what I'm saying, don't worry about it, you'll see it in a second, okay? So now what else do we know about these equilibrium problems? That they have a k value at equilibrium, right? And then we haven't put that in there yet. So what is the k a of this reaction equation? Let's write the equation for k a. So remember that's the products raised to their coefficients, over the reactants raised to their coefficients. So what do we got? Concentration of the formate anion, or the conjugate base of formic acid there, times the concentration of H3O plus, like that, divided by the concentration of the formic acid. Okay? But this is at equilibrium, remember, these are all at equilibrium, okay? So we're going to stick y in for this bottom thing at equilibrium, okay? That's what I was talking about. And then eventually we'll figure out what the original concentration is, okay? So let's rearrange this whole thing, okay? Let's rearrange this whole thing to solve for this equilibrium concentration, okay? So let's do that. So the rearranged equation is going to be HcOOH at equilibrium equals concentration of HcOO minus at equilibrium times hydronium at equilibrium divided by k, okay? So remember that is y there, okay? So that's going to be y. Or if you prefer, just say at equilibrium, it doesn't matter to me either way, okay? So now we can kind of plug and check to figure out what the equilibrium concentration would be, right? So let's do that. 5.5 times 10 to the negative 4 times 5.5 times 10 to the negative 4 divided by 1.7 times 10 to the negative 4, like that, you got that part, very wonderful. Let's figure out what that is, so 5.5 times 10 to the negative 4, I'm just going to square that, okay? Is everybody okay with me just squaring that because it's the same thing and also wanting it, right? So divided by the ka, which is 1.7 times 10 to the negative 4, that's what I get. For the equilibrium concentration, right, this is not the original concentration, it's the equilibrium concentration, is 1 point, and I'm just going to take this to three significant figures for right now, okay, and we'll round it in a little bit, 7.8 times 10 to the negative 3 molar, that's the equilibrium concentration, but what do we know? The equilibrium concentration is the original minus x, right? So let's just rewrite because a lot of people like math equations, let's rewrite another math equation. So h c o o h at equilibrium, so that's going to be y there, okay, equals what? This thing, here, okay? So h c o o h o, or original, not minus 5.5 times 10 to the negative 4. Do we know the value for this at equilibrium? Yes, it's that, 1.78 times 10 to the negative 3, okay? So to figure out what the original concentration is, it's going to be that equals equilibrium value plus 5.5 times 10 to the negative 4, and this is molar, molar, okay? So this is going to be 1.78 times 10 to the negative 4 molar plus 5.5 times 10 to the negative 4 molar, and when we do that, we get 2, just to have these, I think it's 2.3, so plus 5.5, yeah, 2.3 times 10, oops, this should have been a third, sorry guys, times 10 to the negative 3 molar, like that, okay? So sorry, that third was from over there, I just wrote down that to the 4, okay? So hopefully that doesn't confuse you, so we don't have to re-record this, okay? Okay, are there any questions on that one? So it's kind of involved, you know, but again, it's more like if you know what this is, then you know what this is, and then from that, from the ice table, then you know what this is, so you must know what this whole thing is, okay? And all you need is this thing and this thing to figure it out. Are there any questions on that one? Questions? Okay, wonderful, good job guys.