Ionic Equilibrium: Conceptual Doubts Cleared (6)

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Question

The pH of a 0.10.1 M solution of acetic acid is 2.872.87. Find the dissociation constant KaK_a and the degree of dissociation α\alpha.

Solution — Step by Step

[H+]=102.87[\text{H}^+] = 10^{-2.87}. We compute: 102.871.349×10310^{-2.87} \approx 1.349 \times 10^{-3} M.

CH3_3COOH \rightleftharpoons CH3_3COO^- + H+^+. Initial concentrations: [[CH3_3COOH]=0.1] = 0.1 M, others 0\approx 0. At equilibrium: [[CH3_3COOH]=0.1x] = 0.1 - x, [[CH3_3COO]=x^-] = x, [[H+]=x^+] = x.

So x=[x = [H+]=1.349×103^+] = 1.349 \times 10^{-3} M.

Ka=[CH3COO][H+][CH3COOH]=x20.1xK_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} = \frac{x^2}{0.1 - x}

Since x0.1x \ll 0.1, 0.1x0.10.1 - x \approx 0.1:

Ka(1.349×103)20.11.82×105K_a \approx \frac{(1.349 \times 10^{-3})^2}{0.1} \approx 1.82 \times 10^{-5}
α=x[CH3COOH]0=1.349×1030.1=0.013491.35%\alpha = \frac{x}{[\text{CH}_3\text{COOH}]_0} = \frac{1.349 \times 10^{-3}}{0.1} = 0.01349 \approx 1.35\%

Final answers: Ka1.82×105K_a \approx 1.82 \times 10^{-5}, α1.35%\alpha \approx 1.35\%.

Why This Works

For weak acids at low concentration, the approximation 0.1x0.10.1 - x \approx 0.1 is excellent because α1\alpha \ll 1. This avoids solving a quadratic. The key check: α<5%\alpha < 5\% means the approximation is safe.

The textbook value of KaK_a for acetic acid at 25°25°C is 1.8×1051.8 \times 10^{-5} — our answer matches almost exactly. NEET often quotes this number as a sanity check.

Alternative Method

Use Ostwald’s dilution law: Ka=cα2/(1α)K_a = c \alpha^2 / (1 - \alpha). With c=0.1c = 0.1 M and α=0.01349\alpha = 0.01349: Ka=0.1×(0.01349)2/(10.01349)1.82×105K_a = 0.1 \times (0.01349)^2 / (1 - 0.01349) \approx 1.82 \times 10^{-5}. Same answer.

Ostwald’s dilution law in shortcut form: α=Ka/c\alpha = \sqrt{K_a / c} for weak acids when α<5%\alpha < 5\%. Memorise — saves a quadratic.

Common Mistake

Solving the full quadratic when the approximation is fine. For weak acids (KacK_a \ll c), the linear approximation gives the same answer to 3 significant figures and saves time.

Forgetting to take antilog of pH correctly. pH=2.87\text{pH} = 2.87[H+]=102.87[\text{H}^+] = 10^{-2.87}, not 10310^{-3} or 0.2870.287. Use a log table or memorise 100.870.13510^{-0.87} \approx 0.135.

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