 Okay guys, let's try the second portion of this titration problem, so we're still finding the pH during a weak acid strong base titration, but this is the second point in the curve. So we've added some NaOH, but it hasn't gotten to the equivalence point yet. Okay, so let's figure out how do we do this type of a problem. Okay, so the first thing we're going to want to do is write out a reaction equation of the things that are reactive. Okay, so the first thing we got here is the propionic acid, so CH3, CH2, CH8, aqueous, and that is reacting with NaOH at this point in time. Okay, so that's a strong base, so just react it with the OH minus anion. If you put the Na in there, it's going to come to you. So aqueous, remember this reaction is a straightforward reaction. Okay, so we're going to get CH3, CH2, COO minus aqueous plus H2O liquid. Okay, but notice, so unlike the ones we've done before, the volumes here are different. So we're going to have to calculate moles at this point, so this is where we use our volume. So this is a nice table, and we start with 0.100 molar and 40 mils. So let's figure out the number of moles of propionic acid. Okay, so 0.100 moles per one liter like that. And we've got 0.0400 liter. Cancel, cancel. So that's going to be 0.00400 moles. So this one, if you calculated it the same, the number of moles of NaOH, it's going to be 0.00300 moles. You're okay with doing that? Yeah, I don't have to show that calculation. So here we'll put 0.00300 moles. So we didn't start with any of this stuff, so zero there. Okay, so remember we said this is going to completely react with this, right? This is the limiting reagent in this particular reaction, so we're going to lose all of it. Does that make sense? It makes sense to you guys. 0.00300. That's going to give us zero there, right? So we're going to react it with this stuff, right? So we're going to lose the same amount of that stuff, right? So the acid, 0.00300. So what are we going to get down here? 1, 0, 0, 1, 0, 0, 0. Now remember, this is moles, not molarity. It usually is molarity in these high stages, and right now it's moles. Okay, so when we get that, right? Well, what happened over here? Well, we had to add 0.00300, okay? But are you okay with what we've done so far? When you guys are done, let me know. So the volume of our solution is going to be the equivalent, right? If we're looking for the volume of this or the volume of this, okay? So the volume will be equivalent because it's all in the same container, okay? So no longer are we comparing this volume to this volume, both volumes are 70 mils. Does everybody understand what I'm saying? So what I can do is use the Ka equation, right? Just using these numbers here, I don't have to change them to concentration values because it's the same ratio of the moles as it would be to the concentration because both volumes are the same, okay? So what we're going to do, I'm going to erase this stuff up here, okay? Do another ice table, but since we're doing the 5% rule we don't have to worry about worrying about what these concentrations would be, okay? The only thing we have to worry about is what X is down here, H3O plus, okay? So these are the concentrations and the water would be that, right? From the ice table. You want me to write up the ice table? You sure? Because I can. Okay? So these would be our concentrations from the, or the moles from the ice table. And remember Ka equals, well, the concentration of C, C6H5C00 minus times H3O plus, divided by C6H5CH, okay? Yeah. So like that. You got it wrong. C6H3CH2C00H, right? Oh. Oh, we're doing a different, sorry. Yeah, sorry guys. It is a different ass if you're absolutely wrong. So CH3, we're in the problem we did yesterday. And are those numbers from the ice, from the original ice table? No, from that ice table, this one here, that I didn't do. The water was... The don't worry. Okay. From this other ice table, okay? So I can start this recording over if you guys want. Okay? Okay? So this is the equation we're using here, okay? So I apologize for putting C6H5, I understand that. So we're going to rearrange this to solve for H3O plus, right? We're looking for pH. Equal kA times the concentration of CH3CH2C00H divided by H3CH2C00H, like that. Okay? But remember, these concentrations, it's going to be per 70 mils or 0.070 liters, right? Point divided by 0.070 liters. So both of those will cancel after this, okay? So we can use the moles here instead of just the molarity. So we don't have to calculate the molarity. So here, kA, 1.3 times 10 to the negative fifth times the moles of acid, 0.00100. Divided by the moles of acid, 0.00300. Like that. And then we're going to erase all of this. That's going to equal H3O plus. So 4.3 times 10 to the negative six. Like that. This is molar. Okay, so the pH equals the negative log of the H3O plus concentration. So the negative log of 4.3 times 10 to the negative six. And I'm just pushing Hanser in line so you might get a different pH. Like slightly different. So 5.36 to the negative solution. And you would expect it to have gone up because you added anti-aliased. Are there any questions about this? Questions, questions? Hold your peace, if no. Okay, wonderful.