Ionic Equilibrium: Numerical Problems Set (5)

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Question

Calculate the pH of a 0.010.01 M acetic acid solution. Ka=1.8×10−5K_a = 1.8 \times 10^{-5}.

Solution — Step by Step

Acetic acid: CH3COOH⇌CH3COO−+H+\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+.

Let α\alpha be the degree of dissociation. Initially C=0.01C = 0.01 M; at equilibrium, dissociated = CαC\alpha; undissociated = C(1−α)C(1-\alpha).

Ka=[H+][CH3COO−][CH3COOH]=Cα⋅CαC(1−α)=Cα21−αK_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} = \frac{C \alpha^2}{1 - \alpha}

For weak acids, α≪1\alpha \ll 1, so 1−α≈11 - \alpha \approx 1:

Ka≈Cα2  ⟹  α=KaCK_a \approx C \alpha^2 \implies \alpha = \sqrt{\frac{K_a}{C}}

α=1.8×10−50.01=1.8×10−3≈0.0424\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.01}} = \sqrt{1.8 \times 10^{-3}} \approx 0.0424

[H+]=Cα=0.01×0.0424=4.24×10−4 M[\text{H}^+] = C \alpha = 0.01 \times 0.0424 = 4.24 \times 10^{-4} \text{ M}

pH=−log⁡(4.24×10−4)≈3.37\text{pH} = -\log(4.24 \times 10^{-4}) \approx 3.37

Final answer: pH ≈3.37\approx 3.37.

Why This Works

The weak acid approximation hinges on α\alpha being small. We can quickly check: α=0.0424\alpha = 0.0424, so 1−α=0.95761 - \alpha = 0.9576 — the approximation introduces about 4%4\% error, which is acceptable for pH calculations.

For stronger acids or more dilute solutions where α\alpha approaches 0.10.1, solve the quadratic instead.

Alternative Method

Solve the full quadratic: Ka(1−α)=Cα2K_a (1 - \alpha) = C \alpha^2, giving Cα2+Kaα−Ka=0C \alpha^2 + K_a \alpha - K_a = 0. Use the quadratic formula. The result is closer to α=0.0415\alpha = 0.0415, very near our approximation.

A quick rule for shorthand: pH of a weak acid solution of concentration CC is 12(pKa−log⁡C)\frac{1}{2}(pK_a - \log C). For acetic acid at 0.010.01 M: 12(4.74−(−2))=12(6.74)=3.37\frac{1}{2}(4.74 - (-2)) = \frac{1}{2}(6.74) = 3.37. Same answer in one line.

Common Mistake

Students sometimes treat acetic acid as a strong acid and write [H+]=0.01[\text{H}^+] = 0.01 M, giving pH =2= 2. That’s the strong-acid limit — wildly off for weak acids. Always identify whether the acid is strong or weak BEFORE choosing the formula.

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