Ionic Equilibrium: Real-World Scenarios (4)

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Question

A buffer solution is prepared by mixing 0.2 M acetic acid and 0.3 M sodium acetate in water. Find the pH of the buffer. Given KaK_a of acetic acid =1.8×105= 1.8 \times 10^{-5}. (Real-world: this is the pH range of vinegar-based buffers used in food preservation.)

Solution — Step by Step

Acetic acid (weak acid) + sodium acetate (its conjugate base salt) — classic acid buffer. Use the Henderson-Hasselbalch equation:

pH=pKa+log([salt][acid])\text{pH} = \text{p}K_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)

pKa=log(1.8×105)=5log1.850.255=4.745\text{p}K_a = -\log(1.8 \times 10^{-5}) = 5 - \log 1.8 \approx 5 - 0.255 = 4.745

pH=4.745+log(0.30.2)=4.745+log1.5=4.745+0.1764.92\text{pH} = 4.745 + \log\left(\frac{0.3}{0.2}\right) = 4.745 + \log 1.5 = 4.745 + 0.176 \approx 4.92

pH of the buffer ≈ 4.92.

Why This Works

A buffer resists pH changes because it has both an acid (to neutralize added base) and a conjugate base (to neutralize added acid). The Henderson-Hasselbalch equation comes from rearranging the KaK_a expression assuming the salt and acid concentrations don’t change much during the buffering action.

For maximum buffering capacity, the salt:acid ratio should be 1:1, giving pH = pKaK_a. Our buffer at 1.5:1 is still effective and within the useful range (typically pKa±1K_a \pm 1).

For basic buffers (weak base + its salt), use pOH=pKb+log([salt]/[base])\text{pOH} = \text{p}K_b + \log([\text{salt}]/[\text{base}]), then pH = 14 - pOH. Don’t directly apply the acid version.

Alternative Method

From first principles: write the equilibrium expression Ka=[H+][A]/[HA]K_a = [H^+][A^-]/[HA]. Salt fully dissociates, so [A]=0.3[A^-] = 0.3 M. Acid is weak, so [HA]0.2[HA] \approx 0.2 M (negligible dissociation suppressed by the salt — common ion effect). Solving: [H+]=Ka[HA]/[A]=1.8×1050.2/0.3=1.2×105[H^+] = K_a \cdot [HA]/[A^-] = 1.8 \times 10^{-5} \cdot 0.2/0.3 = 1.2 \times 10^{-5}. pH =log(1.2×105)4.92= -\log(1.2 \times 10^{-5}) \approx 4.92. Same answer.

Common Mistake

Students forget that the Henderson-Hasselbalch equation uses log to base 10, not natural log. Also, for a basic buffer, applying the acid version directly gives the wrong answer — use pOH first.

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