Weak-acid titration: which pH should be trusted?
A 25.00 mL sample of a monoprotic weak acid HA is titrated with 0.1000 mol/L sodium hydroxide using a calibrated pH meter. The equivalence point is reached after 22.40 mL of base. At half of this volume, 11.20 mL, the measured pH is 4.20. At the equivalence point the measured pH is 8.32. The student uses phenolphthalein, whose colour-change interval is approximately pH 8.2-10.0, and stops at 22.55 mL when a faint pink colour persists. The acid is known to be pure, the ionic strength is low, and all solutions are at 25 degC. The burette readings have an uncertainty of about 0.05 mL, and the electrode was calibrated with buffers at pH 4.00 and 7.00. The student must decide whether to calculate Ka from the half-equivalence point or from the equivalence-point pH, because the two values do not agree exactly.
1. Calculate the concentration of the acid and its Ka from the half-equivalence data. Explain why the half-equivalence pH can be used in this way.
Read the worked answer
At equivalence, moles of NaOH = 0.1000 mol/L x 0.02240 L = 2.240 x 10^-3 mol. Since HA is monoprotic, moles of HA in 25.00 mL are also 2.240 x 10^-3 mol, so
[HA] = 2.240 x 10^-3 / 0.02500 = 0.08960 mol/L.
At half-equivalence, half of the HA has been converted to A-, so [HA] is approximately equal to [A-]. The Henderson-Hasselbalch equation gives pH = pKa + log([A-]/[HA]). When [A-] = [HA], the logarithm is zero and pH = pKa. Therefore pKa = 4.20 and
Ka = 10^-4.20 = 6.31 x 10^-5.
This is a valid approximation because the acid is weak and the amounts of HA and A- are both much larger than the hydrogen-ion concentration.
2. Using the Ka from the half-equivalence point, calculate the expected pH at the equivalence point and compare it with the measured 8.32. Comment on the agreement and on the suitability of phenolphthalein.
Read the worked answer
At equivalence, the total volume is 25.00 + 22.40 = 47.40 mL, and all HA has been converted to A-. Thus
[A-] = 2.240 x 10^-3 / 0.04740 = 0.04726 mol/L.
For the conjugate base, Kb = Kw/Ka = 1.00 x 10^-14 / 6.31 x 10^-5 = 1.58 x 10^-10.
[OH-] = sqrt(Kb[A-]) = sqrt[(1.58 x 10^-10)(0.04726)] = 2.74 x 10^-6 mol/L.
pOH = -log(2.74 x 10^-6) = 5.56, so pH = 14.00 - 5.56 = 8.44.
The measured value 8.32 is 0.12 pH unit lower than expected. This small discrepancy could come from dissolved CO2, electrode drift, or activity effects, but it does not change the main conclusion. Phenolphthalein begins to show colour near pH 8.2, so the equivalence-point pH of about 8.4 lies within its useful range. The endpoint at 22.55 mL is 0.15 mL past the equivalence volume, an overshoot of about 0.67%, which is acceptable for a school titration but explains a slightly high result.
3. The equivalence-point pH of 8.32, if taken alone, would imply a different Ka. Show this briefly and decide which Ka value is more reliable.
Read the worked answer
If pH = 8.32 at equivalence, then pOH = 5.68 and [OH-] = 10^-5.68 = 2.09 x 10^-6 mol/L. With [A-] = 0.04726 mol/L,
Kb = [OH-]^2/[A-] = (2.09 x 10^-6)^2/0.04726 = 9.25 x 10^-11.
Then Ka = Kw/Kb = 1.00 x 10^-14/9.25 x 10^-11 = 1.08 x 10^-4, which corresponds to pKa = 3.97.
This differs from pKa = 4.20 by 0.23 unit. The half-equivalence value is more reliable here because it uses the pH directly at a point where the buffer ratio is fixed at 1:1 and is less sensitive to small errors in concentration. The equivalence-point pH depends on the square root of Kb and on the total volume, so it amplifies errors and is more affected by CO2 absorption. The student should report Ka = 6.3 x 10^-5 and treat the equivalence-point discrepancy as experimental evidence, not as a reason to discard the half-equivalence result.