Question

Calculate the pH of a 0.46 M solution of NH 4Cl. ( K b for NH 3 = 1.8 × 10 –5)

a.
9.59
b.
4.41
c.
4.80
d.
0.34
e.
9.20

Answer 1

pH of the solution = 4.8

Further explanation

NH₄Cl is a salt of a weak base and a strong acid so that it will be partially hydrolyzed

Reaction:

NH₄Cl -----> NH₄⁺ + Cl⁻

NH₄ will be hydrolyzed in water

NH₄⁺ + H₂O ------> NH₃ + H₃O⁺ (H⁺)

This solution releases H⁺ ions so they are acidic

Formula

`\tt [H^+]=\sqrt{(Kw)/(Kb).M }`

Kw = The water equilibrium constant = 10⁻¹⁴

Kb = the ionization constant of a base = 1.8.10⁻⁵

M = 0.46

`\tt [H^+]=\sqrt{(10^(-14))/(1.8.10^(-5))* 0.46 }`

[H+]=√2.55 x 10⁻¹⁰ = 1.599 x 10⁻⁵

pH = - log [H+]

pH = 5-log 1.599 = 4.796≈ 4.8