 So, let's do one of these equilibria of acid based buffer system. So, this one says calculate the pH of an acetic acid buffer solution consisting of 0.50 molar acetic acid and 0.50 molar sodium acid. So when we do this, we've got to remember the reaction equation. So, on these, since we're doing Ka, put the acid first, okay? So, CH3COOH, and then it reacts with water in this CH3COO-minor. Is everybody okay with doing that one? Okay? I stable. But look here at what we've done now. Original concentration or initial concentration of both of them is 0.5, okay? So instead of putting a 0 over here like we normally have done, we're going to put a 0.5, okay? So, 0.50, 0.50, like that. This one is still 0. This is minus x, plus x, plus x, like that, minus 0, minus 5, 0, plus x. Is everybody okay with that? Okay? Can we use the 5% rule on this one? Give me an answer, tell me. Yeah, we can, right? Why? Because we have a small number here, okay? So, since we can use the 5% rule, what can we do? Get rid of that x there, and get rid of that x, okay? So it really helps out. So now we have to remember what is Ka, the equation for it? Let's figure that out, okay? So Ka equals, arrange it to figure out the H3O plus concentration. Just plug in, right, and this would be x. Ka, we've got, and we've got this and this. So remember, what did I say, Lynn? This and this are the same concentration, right? Our buffer capacity is the best, okay? So let's just plug these values in. So 1.8 times 10, the negative 5 times 0.50 divided by 0.50. Well, this cancels with that, right? Everybody sees that? So when you've got these two concentrations equivalent, right? Your hydronium concentration equals your Ka, or your pH equals your pKa, okay? So in this case, it would be 1.8. That's 10 to the negative 5 molar, okay? And then to figure out the pH, the negative log of that number, okay, wonderful. We're going to now see what happens when we add some base to this thing, okay? So we'll kill it, this one, and then redo this problem with base being added, okay? Any questions on this one?