Question

Ammonium chloride, NH4Cl, is a salt formed from the neutralization of the weak base ammonia, NH3, with the strong acid hydrochloric acid. Given that the value of Kb for ammonia is 1.8×10−5, what is the pH of a 0.176 M solution of ammonium chloride at 25∘C?

Answer 1

Answer:

The pH of a 0.176 M solution of ammonium chloride is 4.9902.

Explanation:

Given:

Kb for ammonia = 1.8×10⁻⁵

Since ammonium ions are the conjugate acid of ammonia.

Thus, Ka for ammonium ions;

Ka = Kw/Kb = 10⁻¹⁴ / 1.8×10⁻⁵ = 5.5556×10⁻¹⁰

Given concentration of Ammonium chloride (C) = 0.176 M

Thus, for weak acids,

$\left[H^+ \right]=\sqrt{K_a\times C}$

$\left[H^+ \right]=\sqrt{5.5556\times 10^{-10}\times 0.176}$

$\left[H^+ \right]=0.9777\times 10^{-5}$

pH is:

$pH\ of\ the\ solution=-log\left[H^+ \right]$

$pH\ of\ the\ solution=-log\left(0.9777\times 10^{-5} \right)$

pH = 4.9902

Thus,

The pH of a 0.176 M solution of ammonium chloride is 4.9902.