Preparing and Stress-Testing a pH 5.00 Acetate Buffer
A first-year laboratory must supply 1.00 L of a pH 5.00 buffer for a spectrophotometric enzyme assay; the enzyme denatures below pH 4.5 and above pH 5.6. The stock cupboard holds 0.100 M acetic acid (Ka = 1.8 × 10⁻⁵, pKa = 4.74) and 0.100 M sodium acetate, together with 1.00 M HCl and 1.00 M NaOH. No pH meter is available, so the buffer must be made by calculation. The protocol also requires that, at the end of each assay, 10.0 mL of 1.00 M HCl be added to stop the reaction, and the demonstrator wants to know whether the buffer can absorb that acid load without leaving the safe window. A second student suggests diluting the same recipe ten-fold to save reagents, arguing that 'dilution cannot change the pH of a buffer, so it must still work'. A failed batch costs the laboratory a week of enzyme preparation time.
1. Calculate the volumes of 0.100 M acetic acid and 0.100 M sodium acetate needed, using only the given data.
Read the worked answer
The Henderson–Hasselbalch relation for this pair is pH = pKa + log([CH₃COO⁻]/[CH₃COOH]).
5.00 = 4.74 + log r, so log r = 0.26 and r = 10^0.26 = 1.82.
Because both stock solutions are 0.100 M, the concentration ratio equals the volume ratio: V_salt/V_acid = 1.82, with V_salt + V_acid = 1000 mL.
V_acid = 1000/2.82 = 355 mL and V_salt = 645 mL.
Check: n(HA) = 0.0355 mol and n(A⁻) = 0.0645 mol; log(0.0645/0.0355) = log 1.82 = 0.26, giving pH 5.00. The ratio 1.82 lies well inside the workable 0.1–10 range, so this is a sensible recipe; note, however, that maximum buffer capacity occurs at ratio 1 (pH 4.74), so the enzyme's upper tolerance of 5.6 is not being exploited.
2. The assay adds 10.0 mL of 1.00 M HCl to the 1.00 L buffer. Show whether the pH stays inside the 4.5–5.6 window.
Read the worked answer
Moles of HCl added = 0.0100 L × 1.00 M = 0.0100 mol. Strong acid converts acetate into acetic acid stoichiometrically:
n(HA) = 0.0355 + 0.0100 = 0.0455 mol; n(A⁻) = 0.0645 − 0.0100 = 0.0545 mol.
pH = 4.74 + log(0.0545/0.0455) = 4.74 + log 1.198 = 4.74 + 0.078 = 4.82.
Volume changes do not enter the ratio, so the dilution by 10 mL is irrelevant to the calculation. The pH shifts by only 0.18 unit and stays inside 4.5–5.6, so the assay is safe.
Compare the unbuffered control: 0.0100 mol of HCl in 1.00 L of water gives [H⁺] ≈ 0.0099 M, pH ≈ 2.0, which would destroy the enzyme instantly. The comparison is the point - the buffer is doing real work, and its capacity (0.0645 mol of acetate available to absorb acid) comfortably exceeds the 0.0100 mol load.
3. Assess the ten-fold dilution proposal, distinguishing the initial pH from the behaviour under acid load.
Read the worked answer
The second student is correct about the initial pH and wrong about the assay. Henderson–Hasselbalch depends only on the ratio, so the diluted buffer still starts at pH 5.00: with 35.5 mL and 64.5 mL made up to 1.00 L, n(HA) = 0.00355 mol and n(A⁻) = 0.00645 mol, ratio still 1.82.
Capacity, however, scales with total buffer concentration. Adding 0.0100 mol of HCl to this diluted buffer consumes all 0.00645 mol of acetate and leaves excess strong acid: 0.0100 − 0.00645 = 0.00355 mol in 1.01 L, i.e. [H⁺] ≈ 3.5 × 10⁻³ M, so pH ≈ 2.45 (the 0.0099 M acetic acid formed contributes only about 4 × 10⁻⁴ M H⁺ and can be neglected). That is far below the enzyme's 4.5 floor.
Judgement: diluting a buffer keeps its pH but destroys its ability to resist acid. Since the protocol guarantees a 0.0100 mol acid challenge, the diluted recipe must be rejected. If reagent economy matters, a smaller volume of the full-strength recipe (say 200 mL, which still absorbs 0.0129 mol of acid) is the correct saving.