Weak Acid Dilution: Faulty Meter or Normal Equilibrium?
During a practical on weak acids, a student prepares 100.0 cm3 of a 0.100 mol dm-3 solution of a monoprotic weak acid, HA. A calibrated pH meter reads pH 2.85 at 298 K. The student then pipettes 10.0 cm3 of this solution into a 100.0 cm3 volumetric flask, makes up to the mark with distilled water, and measures pH 3.35. They expected the pH to rise by exactly 1.00 because the concentration of hydrogen ions should fall by a factor of ten. They conclude that either the pH meter is faulty or the distilled water is contaminated with acid. The teacher asks the student to check the conclusion using equilibrium reasoning before blaming the apparatus. The acid is known not to be volatile or to react with water other than by simple dissociation.
1. Calculate Ka for HA from the first measurement and state one assumption in your calculation.
Read the worked answer
pH = 2.85, so [H+] = 10^-2.85 = 1.41 × 10^-3 mol dm^-3. For a monoprotic acid HA, [A-] is approximately equal to [H+] if water contributes negligibly. The equilibrium concentration of HA is 0.100 - 1.41 × 10^-3 = 0.0986 mol dm^-3. Ka = [H+][A-]/[HA] = (1.41 × 10^-3)^2 / 0.0986 = 2.02 × 10^-5 mol dm^-3. Assumption: activities can be replaced by concentrations, or dissociation is small enough that the approximation is valid; the calculated Ka is therefore an estimate.
2. Use the Ka value to predict the pH after the tenfold dilution. Does the result support the student's conclusion that the apparatus is faulty? Explain.
Read the worked answer
After dilution the formal concentration is 0.0100 mol dm^-3. Let x = [H+] at equilibrium. Ka = x^2/(0.0100 - x) ≈ x^2/0.0100, so x ≈ sqrt(2.02 × 10^-5 × 0.0100) = 4.49 × 10^-4 mol dm^-3. pH = -log10(4.49 × 10^-4) = 3.35. This is exactly the measured value. Dilution increases the degree of dissociation, so [H+] does not fall by a full factor of ten; the pH rise is about 0.50 rather than 1.00. The observation is normal weak-acid behaviour, not evidence of contamination or a faulty meter.
3. Suggest one experimental check that would help distinguish contamination from normal equilibrium behaviour.
Read the worked answer
Measure the pH of a blank sample of the same distilled water, or dilute a strong acid solution of the same initial pH by the same factor. A strong acid should show a pH rise close to 1.00 because it is fully ionised; a contaminated water blank would show a lower pH than pure water. If the blank is neutral and the weak acid behaves as predicted by Ka, contamination is ruled out. Repeating with a freshly prepared solution and a second pH meter would also test reproducibility.