The Buffer That Would Not Reach Its Target pH
A first-year student must prepare 1.00 dm³ of a buffer of pH 4.74 by mixing equal volumes of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵) and 0.100 mol dm⁻³ sodium ethanoate.
She calculates that 0.0500 mol of sodium ethanoate is required and weighs out 4.10 g, taking the molar mass to be 82.0 g mol⁻¹. The reagent bottle, however, contains the trihydrate, CH₃COONa·3H₂O, M = 136.1 g mol⁻¹, and she does not notice. She dissolves the 4.10 g in water, makes the volume to 500 cm³, mixes it with 500 cm³ of the 0.100 mol dm⁻³ acid, and stirs thoroughly.
A pH meter, calibrated that morning against pH 7.00 and pH 4.00 standards, reads 4.55. The student concludes that the meter is faulty and wants to repeat the whole preparation. The technician disagrees and asks her to check her arithmetic first.
1. Calculate the ethanoate concentration she actually prepared, predict the buffer pH, and decide whether the meter is at fault.
Read the worked answer
Moles of sodium ethanoate actually weighed: 4.10 g ÷ 136.1 g mol⁻¹ = 0.0301 mol, not 0.0500 mol. Dissolved to 500 cm³ this is 0.0603 mol dm⁻³; after mixing with an equal volume of acid it is halved to 0.0301 mol dm⁻³. The acid is likewise halved, to 0.0500 mol dm⁻³.
Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.0301/0.0500) = 4.74 + log 0.602 = 4.74 − 0.221 = 4.52.
The meter read 4.55, a difference of 0.03 pH units, which is inside the ±0.05 tolerance of a freshly calibrated school meter. The meter is behaving correctly; the preparation is wrong because the hydrate's water of crystallisation was ignored. Repeating the measurement with the same mass would reproduce 4.55 and teach nothing. Note also that using concentrations in the log term assumes the ratio of activities is close to the ratio of concentrations, which is reasonable here.
2. State the mass she should have weighed and explain, in terms of the buffer ratio, why the error moved the pH in the direction observed.
Read the worked answer
Required: 0.0500 mol × 136.1 g mol⁻¹ = 6.81 g of the trihydrate.
Why the direction: the target buffer has equal moles of acid and conjugate base, so the log term is zero and pH equals pKa. Her mass delivered only 0.0301 mol of ethanoate against 0.0500 mol of acid, so the ratio [A⁻]/[HA] fell to 0.602. A smaller ratio means a more acidic solution, and the log term becomes negative, pulling the pH below pKa. The shift is set by the logarithm, so a 40% shortfall in base produces only a 0.22-unit change; the buffer is doing exactly what buffers do, resisting the disturbance. Had she overshot the mass, the pH would have risen above pKa by a similar logarithmic amount.
3. Using the correctly prepared buffer, calculate the pH after adding 1.0 cm³ of 1.0 mol dm⁻³ HCl to 1.00 dm³ of it, and compare with the same addition to pure water.
Read the worked answer
Correct buffer: 0.0500 mol HA and 0.0500 mol A⁻ in 1.00 dm³. Added HCl = 0.0010 dm³ × 1.0 mol dm⁻³ = 1.0 × 10⁻³ mol, which converts an equal amount of A⁻ into HA.
After addition: HA = 0.0510 mol, A⁻ = 0.0490 mol (volume change of 0.1% is negligible). pH = 4.74 + log(0.0490/0.0510) = 4.74 + log 0.961 = 4.74 − 0.017 = 4.72.
The pH falls by 0.02 units. The same 1.0 × 10⁻³ mol added to 1.00 dm³ of unbuffered water gives [H⁺] ≈ 1.0 × 10⁻³ mol dm⁻³ and pH ≈ 3.0, a fall of four units. The comparison quantifies buffer capacity: the buffer absorbs the strong acid by protonating ethanoate, so free H⁺ barely rises until the reserve of conjugate base is used up.