Question

NH₄OH is 4.3% ionised at 298 K in 0.01 M solution. Calculate the ionization constant and pH of NH₄OH.

Answer 1

Given : 

Percent dissociation = 4.3, 

C = 0.01 M, 

K_b = ?, 

pH = ? 

The degree of dissociation and dissociation constant of NH₄OH are related to each other by the formula :

K_b = α²C

K_b = Dissociation constant of NH₄OH = ?

α = Degree of dissociation of NH₄OH = 4.3%

= 4.3 × 10^-2

C = Molar concentration of NH₄OH = 0.01 M

∴ K_b = (4.3 × 10^-2)² × 0.01

= 18.49 × 10^-4 × 10^-2

∴ K_b = 1.849 × 10^-5

Since NH₄OH is a monoacidic base,

[OH^-] = αC

= 4.3 × 10^-2 × 0.01

= 4.3 × 10^-4 mol dm^-3

pOH = -log₁₀[OH^-]

= -log₁₀4.3 × 10^-4

= -[0.6335 – 4] = 3.3665

pH + pOH = 14

pH = 14 – 3.3665 = 10.6335

K_b = 1.849 × 10^-5, pH = 10.6335