Answer to Question #184721 in Chemistry for leshan

Question

What is the pH of of a solution of 0.60 M ammonium chloride ?

Answer 1

Solution:

NH₄Cl ⇌ NH₄⁺ + Cl⁻

1) NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, K_H = ???

2) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, K_B = 1.8×10⁻⁵

3) H₂O ⇌ H⁺ + OH⁻, K_W = 1.0×10⁻¹⁴

Hence,

K_H = K_W / K_B = (1.0×10⁻¹⁴) / (1.8×10⁻⁵) = 5.56×10⁻¹⁰

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

From this, we can construct an ICE(Initial, Change, Equilibrium) Table:

From the ICE Table,

[NH₃] = [H₃O⁺] = x

[NH₄⁺] = 0.60 - x

K_H = [NH₃] × [H₃O⁺] / [NH₄⁺]

5.56×10⁻¹⁰ = x² / (0.60 - x)

We can assume that (0.60 - x) = 0.60

5.56×10⁻¹⁰ = x² / 0.60

x² = 3.336×10⁻¹⁰

x = 1.827×10⁻⁵

Hence,

x = [H₃O⁺] = 1.827×10⁻⁵

We can convert between [H₃O⁺] and pH using the following equation:

pH = - log[H₃O⁺]

pH = - log(1.827×10⁻⁵) = 4.74

pH = 4.74

Answer: pH = 4.74