Preparing an Acetate Buffer Without Sodium Acetate
A biochemistry practical needs 250.0 mL of acetate buffer at pH 5.00 with a total acetate concentration ([CH3COOH] + [CH3COO⁻]) of 0.100 M, because higher ionic strength inhibits the enzyme being studied. The shelf holds glacial acetic acid (17.4 M; density 1.05 g mL⁻¹), NaOH pellets (M = 40.00 g mol⁻¹) and distilled water, but the sodium acetate bottle is empty. The protocol states: pH must be 5.00 ± 0.05, and the buffer must stay inside that window after 1.00 mL of 1.00 M HCl is injected during the run. A colleague suggests simply mixing equal volumes of 0.100 M acetic acid and 0.100 M sodium acetate 'since that gives pH = pKa = 4.76, close enough'. The student has a calibrated pH meter with buffers at 4.00 and 7.00. Take pKa(CH3COOH) = 4.76 at 25 °C, ignore activity coefficients, and note that the total acetate figure refers to the final 250.0 mL.
1. Design the preparation using only glacial acetic acid and NaOH. Give the moles of each component, the mass of NaOH and the volume of glacial acid needed.
Read the worked answer
Total acetate: 0.100 M × 0.2500 L = 0.0250 mol.
Required ratio from Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]) 5.00 = 4.76 + log r → log r = 0.24 → r = 10^0.24 = 1.738.
Let [A⁻] = 1.738x and [HA] = x with 2.738x = 0.0250 mol, so x = 0.00913 mol.
- CH3COO⁻ required: 0.01587 mol
- CH3COOH required: 0.00913 mol
Because the conjugate base is generated in situ, all the acid must first be measured out and then partly neutralised:
- Total acid to pipette: 0.0250 mol ÷ 17.4 mol L⁻¹ = 1.44 mL of glacial acetic acid
- NaOH pellets: 0.01587 mol × 40.00 g mol⁻¹ = 0.635 g
Procedure: dissolve the NaOH in about 150 mL of water, add the glacial acid slowly with swirling and cooling, allow to reach room temperature, then make up to 250.0 mL. Adding acid to concentrated base is avoided because the neutralisation is strongly exothermic and can volatilise acetic acid. Check with the pH meter and adjust by no more than a few drops.
2. The experiment injects 1.00 mL of 1.00 M HCl. Calculate the new pH and decide whether the buffer meets the ±0.05 specification. If it fails, what single change would fix it without altering the target pH or the total acetate?
Read the worked answer
Added acid: 1.00 × 10⁻³ L × 1.00 mol L⁻¹ = 1.00 × 10⁻³ mol. This converts an equal amount of acetate into acetic acid.
- CH3COO⁻: 0.01587 − 0.00100 = 0.01487 mol
- CH3COOH: 0.00913 + 0.00100 = 0.01013 mol
Since both are in the same 250 mL, the ratio alone matters: r = 0.01487/0.01013 = 1.468 pH = 4.76 + log 1.468 = 4.76 + 0.167 = 4.93.
Drift = 0.07 pH unit, which exceeds the ±0.05 window. The buffer fails.
Diagnosis: the buffer is too dilute in total acetate relative to the acid insult (0.001 mol is 4% of total acetate). Raising total acetate to 0.150 M without changing the target pH keeps the same ratio r = 1.738 but increases the reservoir:
- A⁻ = 0.150 × 0.2500 × 0.6348 = 0.02381 mol; HA = 0.01369 mol
- After HCl: A⁻ = 0.02281, HA = 0.01469, r = 1.553, pH = 4.95 (drift 0.05).
So a total acetate of 0.150 M just satisfies the specification. Increasing concentration is the correct lever; changing the set pH would move the working point away from the enzyme optimum.
3. Evaluate the colleague's equal-volume mixing proposal, and state whether diluting the correctly prepared buffer to 500 mL with water would change its pH or its ability to absorb the injected acid.
Read the worked answer
Colleague's proposal: equal volumes of 0.100 M acid and 0.100 M salt give equal concentrations of HA and A⁻ (0.050 M each after mixing), so pH = pKa = 4.76. That is 0.24 units below the target - nearly five times the permitted tolerance - and would sit on the steep part of the titration curve only if the ratio were near 1 while the enzyme needs the ratio at 1.738. Simple dilution from 0.100 M each to a 1.738 : 1 ratio would be needed, i.e. the colleague's recipe is not adjustable without adding more base. It also halves the total acetate, cutting capacity.
Dilution of the correct buffer to 500 mL: the ratio [A⁻]/[HA] is unchanged, so the calculated pH remains 5.00 (ignoring small activity changes). The moles of acetate present are also unchanged, so adding the same 1.00 × 10⁻³ mol of HCl produces the same ratio shift and the same pH drop of 0.07. Dilution therefore does not rescue the specification. It does lower the buffer index β (capacity per unit volume), so a larger volume of dilute acid would be needed to do the same damage - but the experiment specifies 1.00 mL, so the failure stands.