What is the pH of a 0.820 M aqueous ammonia solution? Ka? Kb (NH3) = 1.8 × 10^-5

• A. 2.42
• B. 13.91
• C. 11.58
• D. 9.25

Answer: (C)

When a weak base like ammonia is dissolved in water, it undergoes hydrolysis: NH3 + H2O ⇌ NH4+ + OH−. The base dissociation constant Kb relates the concentrations of the species at equilibrium. For a solution with initial base concentration c, you can estimate the hydroxide concentration by [OH−] ≈ sqrt(Kb × c) when the amount that reacts is small compared with c.

Here, c = 0.820 M and Kb = 1.8 × 10^−5. So [OH−] ≈ sqrt(1.8 × 10^−5 × 0.820) = sqrt(1.476 × 10^−5) ≈ 3.85 × 10^−3 M.

Next, pOH = −log10([OH−]) ≈ −log10(3.85 × 10^−3) ≈ 2.41. Therefore, pH = 14 − pOH ≈ 14 − 2.41 = 11.59.

As a quick cross-check, the conjugate acid NH4+ would have Ka = Kw/Kb ≈ 1.0×10^−14 / 1.8×10^−5 ≈ 5.6×10^−10, giving pKa ≈ 9.25, which is consistent with the solution being strongly basic. The pH is about 11.58.