 So, let's try to do this problem, preparing a buffer. So how would you prepare a benzoic acid, sodium benzoate buffer with pH equals 4.25, starting with 5 liters of .050 whole or sodium benzoate solution, and adding the acidic component. So it wants you to add the benzoic acid, okay, so, and then it gives you the Ka for benzoic acid, which is 6.3 times 10 to the length of 5. So, first things first, we're going to have to figure out what the concentration of C6H5C00H. So, you know the Henderson-Hasselbach equation, so pH equals pKa plus the log of the conjugate base is up here, so that's going to be this benzoate, so C6H5C00 minus or Na, whichever you prefer, on the bottom C6H5C00H. Okay, does everybody understand that's the Henderson-Hasselbach equation? So, first thing we need to look for is that concentration value there, okay. So, in order to do that, let's isolate this thing here, okay, so we get the log of, are you okay with doing that? So, we're going to raise both sides 10 to that to get rid of the log, okay, so 10 to the log and 10 to the pH minus pKa like that, okay. When we do that, that cancels with that and brings this whole quantity down in front there, okay. Is everybody okay with doing that? Okay, wonderful. I'm just going to erase this so I have some more room. So, concentration of C6H5C00 equals the concentration of the benzoate divided by 10 to the pH line. Is everybody okay with getting to that point? So, now we just want to plug in, so we have this value, 0.05010, well, to the pH is 4.25 minus, well, the pKa, we can figure that out from the kA, right. So, the pKa, negative log of the kA, the negative log of 6.3 times 10 to the negative 5, which is 4.20, 0.50 divided by 10 to the 0.05. So, the concentration that I get of the benzoate acid is 4.5 times 10 to the negative 2 whole. Is everybody okay with that? Yeah, but there's more than just the 5, there's a 6 after that too. So, everybody get to here? Okay, I'm going to have to erase everything else. Is that all right? Can I erase all the other stuff? So, the volume is 5 liters, right. So, the number of moles of benzoate acid, C6H5C00H, is 4.5 times 10 to the negative 2 moles per 1 liter, right. We can multiply that by 5.0 liters, cancel that out. Is everybody okay with doing that? So, it doesn't tell us, it says how would you prepare. I'm assuming they want you to get to mass, okay. So, somebody figure out the molar mass of benzoate acid. I'll figure it out. So, this is the number of moles of benzoate acid. And for every 122.12 grams of benzoate acid, you have one mole. So, cancel, cancel. I'll give you times 27, oh, that's it, right. So, 27 grams of benzoate acid. So, that's how you would prepare that buff, okay. So, you have to add 27 grams of benzoate acid to the 5 liters of the sodium benzoate solution that you already have. Does everybody understand? Okay, so, make sure you can use Henderson-Hasselbock equation, okay. Questions before I kill it?