Describe the preparation of 5 liters of a 0.3 M acetate buffer, pH 4.47, starting from a 2 M solution of acetic acid and a 2.5 M solution of KOH.(pKa 4.77).
I have worked this problem at least half a dozen times in the last two weeks. It is long and tedious. Here is a quick summary. You can ask questions about what you don't understand. The Henderson-Hasselbalch equation is
pH = pKa + log (base)/(acid). The base is acetate (Ac^- or b) and the acid is (HAc or a). So equation 1 is 4.47 = 4.77 + log b/a. Solve for b/a. The second equation is a + b = 0.3 M. Solve these two equation simultaneously and determine (HAc) and (Ac^-) or a and b. Those concentrations are what you need to prepare the solution. Then you go to the actual preparation from the solutions. I would switch to millimols. You want millimols HAc = mL x M = 5000 mL x M from the above calculation. For Ac^- millimoles = 5000 mL x M from the above. Then, using those numbers plug into the below equation to determined millimols you need for HAc and for KOH . Knowing millimoles you need to start with then use M = millimoles/mL. You know the millimols and M you want, solve for mL you need to prepare that. Post your work if you have trouble. I will be available later this evening and I'll get back to you.
.........................HAc + KOH ==> KAc + H2O