A question on how to find pH.

Question

The base dissociation constant of hydrocyanic acid (HCN) at 25 deg C is 4.9 x 10^-10. What is the pH of an aqueous solution of 0.080 M sodium cyanide (NaCN)?

Answer 1

we have the conjugate base to start with:NaCN

Ka=4.9 x 10^-10

Kb=Kw/Ka=10^-14/4.9 x 10^-10=2.04*10^-5

CN^- +HOH>HCN +HO^-

0.08-x.......x....x+10^-7

Kb=x*(x+10^-7)/(0.08-x)=2.04*10^-5

with neglection:x*x/0.08=2.04*10^-5

x=1.27*10^-6=[HO^-] >pOH=-lg(1.27*10^-6)=5.89

pH=14-pOH=8.02