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If 1 volume of 0.1 KH2PO4 is mixed with 2 volumes of 0.1 OHEtNH2 (pka=9.44), what will be the pH of the mixture?

I know I know:

HH equation

2 answers

I would do it this way.
.tNH2+H2PO4^-==>tNH3^+ + HPO4^-2

I assume the units you omitted on the concns are M. Make up a volume, say 100 mL for the phosphate and 200 mL for the amine. That gives, in millimoles.
begin 20.....10......0......0
change -10...-10.....+10.....+10
final..10.....0.....+10......+10

Now substitute into the HH equation. Won't that be a pH of 9.44?
pH = 9.44 + log(10/10)
Ur the BEST DR.BOB!