A pH 5.00 Buffer That Reads 4.62: Calculation or Electrode?
A second-year student must prepare 0.50 L of a pH 5.00 buffer for a restriction digestion; the enzyme is active only between pH 4.5 and 5.5. The laboratory has 0.20 M acetic acid (pKa = 4.76) and 0.20 M sodium acetate. Working quickly, the student mixes 200 mL of the acid with 50 mL of the acetate solution and dilutes to 0.50 L. The bench pH meter was last calibrated two months ago against a single pH 7.00 buffer and had not been checked since; it reads 4.62. A demonstrator insists the calculation must be wrong because the true pH should be much lower than 4.62. The student suspects the electrode instead, pointing out that the bulb had been stored dry for a month. The digest is scheduled in three hours and no fresh reagents are available. The student has to decide which explanation fits the numbers, whether the buffer is usable as it stands, and what volumes would give pH 5.00 at a total acetate concentration of 0.10 M.
1. Calculate the pH the mixture should have and quantify the discrepancy with the meter reading.
Read the worked answer
Moles of acid = 0.200 L × 0.20 mol/L = 0.040 mol. Moles of acetate = 0.050 L × 0.20 mol/L = 0.010 mol.
Henderson–Hasselbalch: pH = 4.76 + log₁₀(0.010/0.040) = 4.76 + log₁₀(0.25) = 4.76 − 0.602 = 4.16.
Dilution to 0.50 L changes both concentrations by the same factor, so the ratio - and therefore the pH - is unaffected; total acetate is only 0.10 M overall, far below any activity-coefficient problem worth worrying about here.
The meter reads 4.62, i.e. 0.46 units higher than predicted. The demonstrator's intuition is right on direction (the true pH should be lower), but 4.62 is the wrong number, not the wrong calculation.
2. Decide between electrode error and a mis-standardised acid, using a quantitative test, and state the one measurement that settles it.
Read the worked answer
Model the failed electrode as an offset at pH 7 with reduced slope s. A true value of 4.16 would read 7 − s(7 − 4.16) = 4.62, giving s = 2.38/2.84 ≈ 0.84 - an 84% slope, precisely the signature of a dried-out bulb left in air.
Test the alternative: for the reading to be genuinely 4.62, the acid would have to be far weaker. Solving 4.62 = 4.76 + log(0.010/n_acid) gives n_acid ≈ 0.0069 mol, i.e. 0.034 M instead of 0.20 M - a six-fold dilution of a sealed stock, which is implausible.
The decisive measurement is a fresh two-point calibration with pH 4.00 and 7.00 buffers. If the meter then reads ~4.16 for the same solution, the electrode was at fault; if it still reads 4.62, titrate the acid against standard NaOH to check its molarity. Do the calibration first - it takes five minutes, whereas the titration takes thirty, and the digest cannot wait.
3. Judge whether the prepared buffer can be used, and compute the volumes that would give pH 5.00 at 0.10 M total acetate.
Read the worked answer
Usability: if the electrode story holds, the true pH is 4.16, which is below the enzyme's 4.5–5.5 window - adding the digest would simply fail. The buffer must not be used as it stands; the enzyme needs to be held until the buffer is remade and verified.
Correct recipe at total acetate 0.10 M in 0.50 L: Total acetate = 0.10 × 0.50 = 0.050 mol. Required ratio = 10^(5.00 − 4.76) = 10^0.24 = 1.74, so n_acetate = 1.74 × n_acid and n_acid(1 + 1.74) = 0.050.
n_acid = 0.050/2.74 = 0.0183 mol → 0.0183/0.20 = 0.091 L = 91 mL of 0.20 M acetic acid. n_acetate = 0.0317 mol → 0.0317/0.20 = 0.159 L = 159 mL of 0.20 M sodium acetate.
Combine and dilute to 500 mL. Note that the acetic acid is the limiting reagent here: only 91 mL is needed, so the 200 mL used earlier was nearly three-fold excess - the arithmetic error, not just the electrode, is what pushed the pH down.