NH4OH is 4.3% ionised at 298 K in 0.01 M solution. Calculate the ioniz

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

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

Answer»

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