Posted by Blake on Sunday, January 27, 2013 at 11:00pm.
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.
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