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Question What is the pH of an ethanoic acid (CH3CO2H) / sodium ethanoate (CH3CO2

–) buffer containing 0.700 mol dm–3 of CH3CO2H and 0.600 mol dm–3 of CH3CO2

–? (Ka of ethanoic acid = 1.80 × 10–5 mol dm–3)

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Method 1: Constructing the ICE table Let x be the concentration of CH3CO2H that has dissociated. This means x mol dm–3 of H+ and x mol dm–3 of CH3CO2

– are formed from the dissociation of x mol dm–3 of CH3CO2H.

CH3CO2H ∏ H+ + CH3CO2

– (I)nitial conc / mol dm–3 0.700 0 0.600 (C)hange in conc / mol dm–3 – x + x + x (E)qm conc / mol dm–3 0.700 – x x 0.600 + x ≈ 0.700 ≈ 0.600

Assumption: CH3CO2H is a weak acid dissociating in the presence of its conjugate base,

CH3CO2–, of similar concentration.

The presence of CH3CO2– suppresses the dissociation of CH3CO2H due to the

Common Ion Effect. It is thus assumed that x is insignificant, such that 0.700 – x ≈ 0.700 and 0.600 + x ≈

0.600.

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[CH3CO2

–][ H+] Ka = [CH3CO2H]

(0.600 + x)(x) 1.80 × 10–5 = (0.700 – x)

0.600 x 1.80 × 10–5 ≈ 0.700

x = 2.10 × 10–5 ∴ pH of buffer = –lg(2.10 × 10–5) = 4.68

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Method 2: Substituting given data into the Henderson-Hasselbalch equation (H Eqn) directly

[CH3CO2–] pH = pKa + lg [CH3CO2H]

The H Eqn is derived from the Ka expression. Like the Ka expression, all concentration terms in the H Eqn are equilibrium

concentrations. If we substitute the equilibrium concentrations from the ICE table in Method 1 into the

H Eqn, we have the expression shaded in yellow (page 5). Normally, this step is skipped, and many students simply substitute the initial concentrations of CH3CO2H and CH3CO2

– given in the question without fully understanding why. Note that initial concentrations of CH3CO2H and CH3CO2

– are substituted directly into the H Eqn as x is assumed to be insignificant compared to the initial concentrations of CH3CO2H and CH3CO2

–.

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[CH3CO2

–] pH = pKa + lg [CH3CO2H]

(0.600 + x) pH = –lg(1.80 × 10–5) + lg (0.700 – x)

0.600 pH ≈ –lg(1.80 × 10–5) + lg 0.700 = 4.68