Question

1.What is the pH of a solution containing 0.042 M NaH2PO4 and 0.058 M Na2HPO4 ? The pKa of sodium phosphate is 6.86. pH = 6.86 + log [0.058] / [0.042]; pH = 6.86 + 0.14; pH = 7.00
That is my answer to the above question. The following question is what I am having trouble with.
2.What is the pH of the sodium phosphate buffer described in problem #1 if 1 ml of 10M NaOH is added? What is the pH of 1 liter of water if 1ml of NaOH is added?

Answer 1

2a. Adding 1 mL of 10M NaOH to HOW MUCH of the buffer. Since the problem gives molarities, I will assume 1L.
millimols HPO4^2- = 1000 mL x 0.058 = 58
mmols H2PO4 = 1000 mL x 0.042 = 42
mmols NaOH added = 1 mL x 10M = 10

.......H2PO4^- + OH^- ==> HPO4^2- + H2O
I......42........0........58.........0
added...........10.................
C......-10.....-10........+10.....+10
E.......32......0.........68

pH = 6.86 + log(68/32) = about 7.19

2b.
10M NaOH x (1/1001) = 0.00999M
pOH = 2
pH = 10
Check my work for typos. The chemistry is ok.