Calculate the value of the ionization constant of HCN.

Question

A 0.0010 M solution of KCN is hydrolyzed to the extent of 14.0 %. Calculate the value of the ionization constant of HCN.

Answer 1

Potassium cyanide (KCN) dissociates fully in water:

KCN →K⁺ + CN⁻

The cyanide ion then hydrolyzes in water:

CN⁻ + H₂O= HCN + OH⁻

^{As we know}

K_a ^x K_b⁼ K_w

_Where K_a ⁼ ionization constant (K_a​) of HCN

K_b= hydrolysis constant of CN^-

Kw=1.0×10⁻¹⁴

The initial concentration of CN⁻=0.0010M.

The extent of hydrolysis is 14.0%, so [OH-]=0.14 x 0.0010= 0.00014 M

K_b= [HCN][OH^-]/[CN^-]_eq

At equilibrium: [HCN]=[OH⁻]=0.00014 M.

[CN^-]=0.0010−0.00014=0.00086 M.

On putting the values on K_b =[(0.00014)(0.00014)​]/ (0.00086)

=2.279 x 10^-5 M

Calculate K_a​ for HCN:

K_a ^{x (}2.279 x 10^-5)= 1.0×10⁻¹⁴

K_a = 4.39 x 10^-10

The ionization constant K_a​ of HCN is approximately 4.39×10⁻¹⁰