Buffer Problem?

Question

How much solid NaCN must be added to 1.0 L of a 0.5 M HCN solution to give a pH of 7.0? (Assume Ka = 6.2 x 10-10 for HCN.)

A. 0.0034 g
B. 11 g
C. 160 g
D. 24 g
E. 0.15 g

Answer 1

CN- ion dissolves in water and some reacts:
CN- + H2O => HCN + OH-
To use this equation for equilibrium, we need a Kb for CN-. It can be shown that KaKb= 1x10-14, so Kb = 10x10-15/6.2x10-10 = 1.6x10-5.
If the amount of CN- at equilibrium is (x), then
[HCN][OH-]/ [CN-]=Kb= [0.5][10-7]/[x]
(we assume that the amount of CN- converted to HCN is << 0.5M. Then...........
x= 0.5x10-7 /1.6x10-5= 3.2x10-3.
Since each mole of NaCN= 23+12+14=49 g/gmole, the amount needed is about 0.15 g. E is closest.