pH of 0.10 M NH₃ (Kb = 1.8×10⁻⁵) ≈

11.1 [OH⁻] ≈ √(1.8×10⁻⁶) ≈ 1.34×10⁻³ → pOH ≈ 2.87 → pH ≈ 11.1.

Weak Acid and Base Equilibria

AP Chemistry· difficulty 4/5

pH of 0.10 M NH₃ (K_b = 1.8\times10⁻⁵) \approx

Explanation

[OH⁻] \approx \sqrt(1.8\times10⁻⁶) \approx 1.34\times10⁻³ \to pOH \approx 2.87 \to pH \approx 11.1.

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