[Rutherford]

Calculate the pH of a 0.7M solution of NH₄CN. Kb_NH3=1.79*10^-5, Ka_HCN=7.9*10^-10.

The reactions are:
NH₄⁺+H₂O NH₃+H₃O⁺ Ka=Kh=Kw/Kb=5.59*10^-10
0.7                  x      x
CN^-+H₂O HCN+OH^- Kb=Kw/Ka=1.27*10^-5
0.7                y     y

[H⁺]=x=1.98*10^-5
[OH^-]=y=2.98*10^-3
There are more OH^- ions y-x≈2.98*10^-3
pOH=2.52
pH=14-2.52=11.48 but in my book it is 9.18.
What did I do wrong?

[Borek]

You can't assume there are two different, separate reactions taking place. You ended with OH^- being neutralized by H⁺ - once they react, they are removed from the solution, so the equilibrium shifts.

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified

[Rutherford]

Understood, thanks.