The pH of 0.10 M KCN solution at 25 degre sol , for HCN , Ka =6.2*10^-10

 Dissociation of KCN in water can be written as :
KCN → K⁺ + CN^-
CN^- is the conjugate base of HCN. 
Equilibrium reaction of CN^- with H₂O can be written as :
CN^- + H₂O ↔​ HCN + OH^-
So we will determine the concentration of OH^- to find out pOH than OH.
At first calculate the K_b for CN^-
K_aK_b = K_w
​K_b = K_w/K_a
= 1.0 x 10^-14/6.2 x 10^-10
= 1.613 x 10^-5
Now use the equilibrium equatuion to calculate equilibrium concentration
CN^- + H₂O ↔​ HCN + OH^-
Initial conc. of CN^- = 0.1 M
Equilibrium concentration of CN^- = 0.1 M - x
Initial concentration of HCN = 0
Equilibrium conc. of HCN = +x
Initial conc. of OH^- = 0
Equilibrium conc. of OH^- = +x
K_b = [HCN][OH^-]/[CN^-] = 1.613 x 10^-5
     x²/0.1-x = 1.613 x 10^-5
Assuming x << 0.1, remove x from denominator and get
x²/0.1 = 1.613 x 10^-5
x = 1.27 x 10^-3
[OH^-] = ​1.27 x 10^-3
pOH = -log[OH^-]
        = - log [1.27 x 10^-3]
       = 2.899 ≈ 3
pH =14 -pOH
    = 14 -3 
   = 11