Preparing and Stress-Testing an Acetate Buffer of pH 5.00
A technician must prepare 1.00 L of an acetate buffer for a teaching laboratory. She measures 0.100 mol of pure acetic acid (Ka = 1.8×10⁻⁵, pKa = 4.74) into a 1.00 L volumetric flask, adds 0.0500 mol of solid NaOH, swirls until it dissolves, and makes up to the mark with distilled water. The calibrated pH meter reads 4.74. She expected 'half-neutralised' to mean pH 7 and is convinced the meter is faulty.
The protocol actually requires pH 5.00. She is permitted to add more solid NaOH (assume the volume stays 1.00 L) but is not permitted to add any strong acid, because the lab forbids handling concentrated HCl at the bench. Later, 0.0200 mol of HCl(g) is bubbled in with stirring and she must judge whether the buffer has failed and whether the same amount of HCl added to 1.00 L of distilled water would have 'basically the same' effect.
1. Explain the 4.74 reading, and calculate the additional mass of solid NaOH needed to bring the buffer to pH 5.00.
Read the worked answer
Why 4.74 is correct, not a fault. Adding 0.0500 mol of NaOH to 0.100 mol of acetic acid converts exactly half of the acid into acetate:
CH3COOH + OH⁻ → CH3COO⁻ + H2O
leaving n(CH3COOH) = 0.0500 mol and n(CH3COO⁻) = 0.0500 mol in 1.00 L. The ratio is 1, so by the Henderson–Hasselbalch equation
pH = pKa + log([A⁻]/[HA]) = 4.74 + log 1 = 4.74
A half-neutralised weak acid is not neutral: it is the point where pH = pKa. The meter is right.
Reaching pH 5.00. The required ratio is
[A⁻]/[HA] = 10^(5.00 − 4.74) = 10^0.26 = 1.82
Let y = extra moles of NaOH. Then (0.0500 + y)/(0.0500 − y) = 1.82
0.0500 + y = 0.0910 − 1.82y → 2.82y = 0.0410 → y = 1.45×10⁻² mol
Mass = 1.45×10⁻² mol × 40.00 g mol⁻¹ = 0.58 g of NaOH.
Check: base 0.0645 mol, acid 0.0355 mol, ratio 1.82, pH = 4.74 + 0.26 = 5.00 ✓. The added base is well under the 0.0500 mol of acid still available, so the buffer is not destroyed in the process.
2. After 0.0200 mol of HCl(g) is bubbled in, calculate the new pH. Decide whether the buffer has failed, and compare with adding the same amount of HCl to 1.00 L of pure water.
Read the worked answer
Stoichiometry first, equilibrium second. The strong acid converts acetate into acetic acid, essentially completely:
CH3COO⁻ + H⁺ → CH3COOH
n(CH3COO⁻) = 0.0645 − 0.0200 = 0.0445 mol n(CH3COOH) = 0.0355 + 0.0200 = 0.0555 mol
pH = 4.74 + log(0.0445/0.0555) = 4.74 + log(0.802) = 4.74 − 0.096 = 4.64
The buffer moved by only 0.36 pH units for 0.0200 mol of strong acid.
Has it failed? The usual working criterion is that a buffer is useful while 0.1 ≤ [A⁻]/[HA] ≤ 10. Here the ratio is 0.80, comfortably inside that window, and neither component has been used up (both remain above 0.04 mol). The buffer has not failed; it has only been partly consumed, and its capacity towards further base is now larger than its capacity towards more acid.
Comparison with water. In 1.00 L of pure water, 0.0200 mol of HCl gives [H⁺] = 0.0200 mol L⁻¹ and pH = −log(0.0200) = 1.70, a change of about 5.3 units from pH 7. The two situations are therefore not 'basically the same' - they differ by roughly 3 pH units and by a factor of about 900 in [H⁺].
This is the key judgement: the buffer does not neutralise the acid, it absorbs it by protonating acetate, which is why the pH barely moves.
3. The technician now suggests adding a little more acetic acid to push the pH up to exactly 5.00. Evaluate this proposal and state what it reveals about the permitted-operation constraint.
Read the worked answer
The proposal is backwards. Acetic acid is the weak acid component (HA). Adding it increases the denominator of the ratio [A⁻]/[HA], so the logarithm becomes more negative and the pH falls, moving away from 5.00. From the current pH 4.64, adding HA would push the mixture towards pH 4.3 or lower, never towards 5.00.
To raise the pH of an acetate buffer one must increase the conjugate base: either add a strong base (NaOH, which is permitted) or add solid sodium acetate. The lab's rule forbidding strong acid is therefore not the binding constraint here at all - strong acid would only ever lower the pH, and it is the wrong reagent in this direction. What the technician actually needed, both before and after the HCl, was the conjugate base.
What this reveals about the chemistry. pH is set by the ratio of the two buffer components, not by the total amount. A useful habit is to state, before picking a reagent: 'which direction must the ratio move, and which component appears in the numerator?' Here the numerator is [A⁻], so the ratio must rise, so base must be added.
A further point worth making: diluting the buffer with water would not have solved the problem either. Dilution leaves the ratio, and hence the pH, almost unchanged (it mainly reduces buffer capacity and slightly changes activity coefficients) - the common exam trap of 'dilute the acid to raise the pH' does not apply to a buffered solution.