Calculate the percent ionization of 0.20 M solution of hydrocyanic acid, HCN. Ka for HCN = 4.9 x 10-10.

Equation of dissociation of hydrocyanic acid, HCN will be given as:
HCN < ---------> H+ + CN
Let α is the degree of ionization of HCN
Initial concentration:
[HCN] = 0.2
[H+] = 0
[CN] = 0
Equilibrium concentration:
[HCN] = 0.2 (1 – α)
H+] = 0.2 α
[CN] = 0.2 α
Ka = [H+] [CN] / [HCN]
4.9 x 10-10 = (0.2 α) (0.2 α) / 0.2 (1 – α)
As, 1 – α ≈ 1
Therefore, 4.9 x 10-10 = (0.2 α) 2
α = (4.9 x 10-10 / 0.2)1/2
= 4.95 x 10-5
Hence, percent ionization = 0.00495 %

Category: Ionic Equilibria

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