Weak-Acid Titration: Equivalence-Point pH, Buffer Point, and Indicator Choice
A 25.0 mL sample of 0.100 M HA (Ka = 1.8 × 10⁻⁵ at 298 K) is titrated with 0.100 M NaOH. The measured initial pH is 2.87, and at 12.5 mL of added base the pH is 4.74. A student writes that at the equivalence point the solution will be neutral (pH = 7.00), since equal moles of acid and base have reacted, and selects bromothymol blue as the indicator. Available indicators: methyl orange (transition 3.1–4.4), bromothymol blue (6.0–7.6), phenolphthalein (8.2–10.0).
1. Calculate the pH at the equivalence point and explain qualitatively why it is not 7.00. Then choose the best indicator from the list and explain the consequence of the student's choice.
Read the worked answer
Moles of HA = (0.100 M)(0.0250 L) = 2.50 × 10⁻³ mol, so 25.0 mL of NaOH reaches equivalence and the total volume is 50.0 mL. All HA is converted to A⁻: [A⁻] = 2.50 × 10⁻³ mol / 0.0500 L = 0.0500 M.
A⁻ is the conjugate base of a weak acid, so it hydrolyzes: Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. [OH⁻] = √(Kb·C) = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ M → pOH = 5.28 → pH = 8.72.
The equivalence point is basic because the only species present that reacts with water is A⁻, which produces OH⁻; Na⁺ is a spectator. "Equal moles reacted" means the amounts are stoichiometrically equivalent, not that the solution is neutral.
Phenolphthalein (8.2–10.0) brackets 8.72 and is the correct choice. Bromothymol blue would signal the endpoint well before the true equivalence point, giving a systematically low calculated acid concentration.
2. Explain why the measured pH at 12.5 mL equals 4.74, and why this point is described as having maximum buffer capacity. Predict qualitatively how the pH–volume curve behaves past the equivalence point.
Read the worked answer
At 12.5 mL, exactly half of the HA has been neutralized: n(A⁻) = n(HA), and the Henderson–Hasselbalch equation gives pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa. Since Ka = 1.8 × 10⁻⁵, pKa = 4.74, which matches the measured value. This is the half-equivalence point.
A buffer resists pH change best when the two conjugate partners are present in comparable amounts, and capacity is maximal when their ratio is 1:1 - exactly this point. At 12.5 mL the solution contains both HA and A⁻ in equal concentration, so added strong acid is consumed by A⁻ and added strong base by HA. Once the ratio departs far from 1 (say beyond 10:1), one component is nearly exhausted and the pH changes rapidly.
Past the equivalence point, HA is gone and the pH is governed by excess OH⁻: pH climbs steeply, then flattens as added base dilutes and the curve approaches the pH of the titrant (~12.7 for 0.100 M NaOH).
3. Calculate ΔG° for the dissociation HA ⇌ H⁺ + A⁻ at 298 K. Then judge the claim: "If a student measured a larger Ka for a different acid, that acid must dissociate faster."
Read the worked answer
ΔG° = −RT ln Ka = −(8.314 J mol⁻¹ K⁻¹)(298 K)(ln 1.8 × 10⁻⁵). ln(1.8 × 10⁻⁵) = −10.93. ΔG° = −(2477.6 J/mol)(−10.93) ≈ +2.7 × 10⁴ J/mol = +27 kJ/mol.
The positive sign is consistent with a weak acid: dissociation is non-spontaneous under standard conditions, and Ka < 1. The magnitude, not just the sign, matters - a larger positive ΔG° would correspond to a smaller Ka.
The claim is not valid. Ka is a thermodynamic quantity, the ratio of the forward and reverse rate constants (Ka = k_f/k_r). A large Ka requires only that dissociation be favorable relative to recombination; it says nothing about the absolute speed of either process. An acid could dissociate very rapidly yet recombine even faster, giving a small Ka - and therefore a small rate of net dissociation. Reaction rate is governed by Ea; equilibrium position is governed by ΔG°.