By using the ICE table :
initial 0.2 M 0 0
change -X + X +X
Equ (0.2 -X) X X
when Ka = (X) (X) / (0.2-X)
so by substitution:
4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X
∴X ≈ 10^-6
∴[HCN] = 10^-6
and PH = -㏒[H+]
= -㏒ 10^-6
= 6
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