The Buffer That Tested Below Tolerance
A final-year student in a Tarragona analytical laboratory must prepare 0.500 L of a pH 4.70 buffer with an acceptance window of 4.60–4.80 for a spectrophotometric assay. Only glacial acetic acid (Mr 60.05) and sodium acetate trihydrate (Mr 136.08) are available, and pKa(CH3COOH) = 4.74. The student dissolves 0.100 mol of acetic acid and 0.075 mol of sodium acetate, makes the volume up to 0.500 L, and the bench pH meter reads 4.62. The assay protocol then requires the technician to add 5.00 mL of 2.00 M HCl to the buffer before the sample is finally diluted to 1.00 L. The supervisor must decide whether to accept the batch as it is, adjust it, or discard it, and must also confirm that the batch can survive the acid addition without leaving the window. A second student claims that diluting the buffer from 0.500 L to 1.00 L will raise the pH towards 7 because 'the acid gets weaker'.
1. Calculate the pH of the buffer as prepared and after the 5.00 mL addition of 2.00 M HCl. Does the batch meet the specification at either point?
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As prepared: n(CH3COOH) = 0.100 mol, n(CH3COO⁻) = 0.075 mol in 0.500 L, so pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.075/0.100) = 4.74 − 0.125 = 4.62. This lies inside 4.60–4.80, so the initial reading is consistent and the batch is initially acceptable.
Acid added: n(HCl) = 2.00 mol/L × 0.00500 L = 0.0100 mol. The strong acid converts acetate into acetic acid quantitatively:
- HA: 0.100 + 0.0100 = 0.110 mol
- A⁻: 0.075 − 0.0100 = 0.065 mol
- New volume 0.505 L (the ratio is what matters, not the volume). pH = 4.74 + log(0.065/0.110) = 4.74 − 0.228 = 4.52.
Judgement: the batch is acceptable as prepared but fails the specification after the protocol's acid addition (4.52 < 4.60). It must be reformulated, not simply accepted.
2. How many millilitres of the 2.00 M HCl can be added before the pH falls to 4.60? Use that result to advise the supervisor.
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Set the target ratio from the Henderson–Hasselbalch equation: 4.60 = 4.74 + log(A⁻/HA) ⟹ log(A⁻/HA) = −0.14 ⟹ A⁻/HA = 10^(−0.14) = 0.724.
Let x = moles of HCl added. Then A⁻ = 0.075 − x and HA = 0.100 + x: (0.075 − x)/(0.100 + x) = 0.724 0.075 − x = 0.0724 + 0.724x 0.0026 = 1.724x ⟹ x = 1.51 × 10⁻³ mol. V = n/c = 1.51 × 10⁻³/2.00 = 7.5 × 10⁻⁴ L ≈ 0.75 mL.
So the buffer tolerates only about 0.75 mL of 2.00 M HCl before leaving the window, whereas the protocol adds 5.00 mL - roughly seven times its capacity. The supervisor should discard the batch and prepare a buffer with a higher acetate/acid ratio (for example 0.125 mol acetate and 0.075 mol acid in 0.500 L), which gives pH 4.96 before acid addition but stays near 4.70 after the 0.0100 mol of HCl has been consumed.
3. A classmate proposes adding solid sodium acetate to the already-acidified buffer to bring it back to pH 4.70. Calculate the mass required, and evaluate the claim that diluting the buffer to 1.00 L will drive the pH towards 7.
Read the worked answer
After acidification the flask contains 0.110 mol HA and 0.065 mol A⁻. For pH 4.70 the required ratio is: A⁻/HA = 10^(4.70−4.74) = 10^(−0.04) = 0.912. Required A⁻ = 0.912 × 0.110 = 0.100 mol. Extra acetate needed = 0.100 − 0.065 = 0.035 mol. Using sodium acetate trihydrate: m = 0.035 mol × 136.08 g/mol ≈ 4.8 g.
The correction is chemically valid but changes the ionic strength appreciably, so the calculated value should be checked with the meter before use; adding the solid also slightly increases the volume, which cancels in the ratio and can be neglected to two significant figures.
Dilution: Henderson–Hasselbalch depends only on the ratio n(A⁻)/n(HA), and dilution changes both moles identically (both are diluted by the same factor), so the ratio is unchanged. The pH stays at 4.52, not 7. Only at extreme dilution (below about 10⁻⁴ M, where water autoionisation and activity coefficients matter) would the pH drift towards 7. The classmate's reasoning confuses concentration with amount.