What is the pH of a 0.253 M aqueous solution of ammonium chloride?
K_b (NH₃) = 1.2 × 10^-5

Question

What is the pH of a 0.253 M aqueous solution of ammonium chloride? K_b (NH₃)= 1.2 × 10^-5 A)2.7 B)4.8 C)9.2 D)11.3 E)9.9

Quizplus · Accepted Answer

Ammonium chloride is a salt composed of the ammonium cation (NH4+) and the chloride anion (Cl-). When dissolved in water, the salt undergoes hydrolysis, meaning it reacts with water to form a weak acid (NH4+) and a strong base (OH-):
NH4+ + H2O ↔ NH3 + H3O+
The equilibrium favors the side with the weaker acid and weaker base, so more H3O+ ions are present in the solution. Therefore, the solution is acidic. The pH can be calculated using the equilibrium constant (Kw = 1.0 x 10^-14) and the concentration of H3O+:
Kw = [H3O+][OH-] = 1.0 x 10^-14
[H3O+] = Kw/[OH-] = 1.0 x 10^-14/0.253 = 3.95 x 10^-14
pH = -log[H3O+] = -log(3.95 x 10^-14) = 4.8
Therefore, the pH of the solution is 4.8, and the best choice is B.

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