Question

1. A buffer was prepared by dissolving 11.7 g of KH2PO4 in 400 mL of water, and with the addition of concentrated solution of KOH the pH was adjusted to 7.10. The volume was then adjusted to 500 mL. What are the [PO43-], [HPO42-], [H2PO41-], and [H3PO4]? Use pKa values of 2.12, 7.21, and 12.7 for phosphoric acid. Show details of your calculation.
2. Which of the naturally occurring amino acids side chains are charged (>10%) at pH 2? pH 7? pH 12?

Answer 1

KH2PO4 + KOH ==> K2HPO4 + H2O
mols KH2PO4 = 11.7/136.08 = about 0.086
Therefore, base + acid = 0.086 (but you need to do it more accurately).
7.10 = 7.21 + log (base)/(acid)
B/A = about 0.8
B+A = about 0.086
Solve the two equations simultaneously for B and A. That will give you HPO4^2- and H2PO4^-
Use k1 to solve for H3PO4 and use k3 to solve for PO4^3-.
Post your work if you get stuck.
I don't know enough about biochem to help with #2.

Answer 2

I got to that part where to find the moles of KH2PO4 but why base/ acid would equal the same. this is how i went about it then i got stuck

11.7 g of KH2PO4/136.08g/mol=.085 mol of KH2PO4
then to find M:
500 ml/1000ml=.5L
.085/.5=17 M KH2PO4
then o lost myself

Answer 3

You don't make sense. base/acid would equal the same as what? base/acid you calculate from pH = pK2 + log (base/acid) and
B/A = approx 0.8 and I did that for you which makes base = about 0.8*A

You slipped some decimal points. 0.085/0.5L = 0.17 BUT you make sure that 0.085 is right first.
The next step you must use is to solve the two equation simultaneously.
A/B = 0.776 mols
A+B = 0.085 mols
Solve those two equation to find A and B. Then you can change to molarity if yu wish but it isn't necessary.