 So let's try this problem again with some different conditions. So it says calculate the pH of a buffer solution consisting of 0.50 molar, acetic acid in 0.50 molar, sodium acetate after adding 0.020 moles of solid sodium hydroxide to 1 liter of the buffer solution in part A. So all they're saying, all they want you to do with that number of moles and volume is to calculate the molarity of that sodium hydroxide. That's what they mean. So the concentration of sodium hydroxide is going to be, of course, 0.020 moles divided by 1.0 liters equals 0.020 moles. So let's draw the first reaction equation that we need to use. So notice, we're putting in a strong base here. That's going to react. Is that going to react with the sodium acetate here? Or is it going to react with the acetic acid? What do you think? The acetic acid, right? That's what it reacts with. So the first thing we've got to do is figure out what the extent of that reaction is. So we've got to figure out our new starting concentrations is what we're doing. So first, there's going to be two reaction equations we write for these buffers. So first one's going to be the reaction of that base with your weak acid. So whenever you have your weak acid and a strong base, so remember that sodium is just a spectator ion. So I'm not going to even put it in there. That goes all the way to product here. So when we do that, we're going to get CH3C00 minus plus H2O. Can everybody do that part of the equation? Part of that part of the problem. Got to do an ice table, that. So we started with 0.50 molar. Is everybody OK with that? 0.50 molar. What did we start here? 0.020 molar. What did we start here? Did we start here with something? 0.5. Yeah, 0.50 molar. So you see how this one's a little bit tougher because you've got to figure out something to put in all of them. So what happens when I react with this with this? This reacts completely. Everybody understand that? So I'm going to have to subtract 0.020 molar. When I do that, I get 0.48 molar. I'm also going to have to subtract 0.020 from that. Because once you react that with that, that's the correct answer. It's going to give me 0. Makes sense. What am I going to do here? That will be very good. So what am I going to get down here? 0.52. So can everybody do that portion? So that gives me new initial concentrations. So now I'm going to revamp this stuff up here. Is everybody OK with what I'm doing? OK, so this now is going to be 0.48 molar. I'm going to get that from there. And this one's going to be 0.52 molar, which I get from over there. Is everybody OK with what I've done there? Now it's just that same problem that we did last week. So can I erase this ice table? Did everybody get it? OK, cool. With water going back and forth plus H3O plus. And we're still looking for the pH truth. So what am I going to do? Draw an ice table. OK, what's here? Very good, plus X. So 0.522. X, 0.48 minus X. Can we use our 5% rule here? Yeah, so it's the same problem, right? So we can use it. So let's do it. So I'm going to erase this part down here to give me some room. So let's rearrange. Notice this is different. So it's not going to be the k8. So 1.8 into the negative 5. 0.4827 times to the negative fifth molar, H3O plus. So what you guys got? Very cool. And then I'm going to erase this and figure out the pH, OK? So I erase that. pH equals negative log of that number over there, 1.7 times 10 with the negative fifth and seven. Is that something you would have expected? It would be something like that, right? So up, right? Because it's studying my drop side, so that makes it more basic. But not very much, right? Because it's a buffered solution. So what did it go from 4.74 to 4.78? OK, is everybody OK with doing these buffers? So I want you guys to try the one by adding acid on your own, OK? Any questions? Questions over there. OK, cool.