Weak-Acid Titration: Extracting Ka and Choosing an Indicator
During a titration practical, 25.00 mL of a monoprotic weak acid HA of unknown concentration is placed in a conical flask and titrated with standardized 0.100 M NaOH from a burette. The equivalence point is reached at 25.00 mL of NaOH. The pH measured at the start is 2.88, and at the half-equivalence point it is 4.76. The laboratory has three indicators available: methyl orange (pH 3.1–4.4), bromothymol blue (pH 6.0–7.6) and phenolphthalein (pH 8.2–10.0). The student is asked to report Ka, the initial concentration of HA, and the pH at the equivalence point, and to justify the choice of indicator. A classmate argues that because the acid is weak, the equivalence pH should be acidic; the student must check this claim with a calculation.
1. Calculate Ka and the initial concentration of HA from the pH data.
Read the worked answer
At the half-equivalence point, [HA] ≈ [A^-], so the Henderson–Hasselbalch equation gives pH ≈ pKa. Therefore: pKa = 4.76 Ka = 10^-4.76 = 1.74 × 10^-5.
At the start, pH = 2.88, so [H+] = 10^-2.88 = 1.32 × 10^-3 M. For a weak acid: Ka = [H+]^2 / (C - [H+]). 1.74 × 10^-5 = (1.32 × 10^-3)^2 / (C - 1.32 × 10^-3) C - 1.32 × 10^-3 = 1.74 × 10^-6 / 1.74 × 10^-5 ≈ 0.100 M C ≈ 0.101 M.
The titration confirms this: 25.00 mL of 0.100 M NaOH neutralizes 25.00 mL of acid, so C_acid = (0.100 × 25.00) / 25.00 = 0.100 M, consistent within rounding.
2. Calculate the pH at the equivalence point and select the most suitable indicator. Justify your choice.
Read the worked answer
At equivalence, all HA has been converted to A^-. The total volume is 50.00 mL, and the number of moles of A^- is 0.100 × 0.02500 = 2.50 × 10^-3 mol, so: [A^-] = 2.50 × 10^-3 / 0.05000 = 0.0500 M.
A^- is the conjugate base of a weak acid, so it hydrolyses: A^- + H2O ⇌ HA + OH^- Kb = Kw / Ka = 1.0 × 10^-14 / 1.74 × 10^-5 = 5.75 × 10^-10. [OH^-] ≈ sqrt(Kb × C) = sqrt(5.75 × 10^-10 × 0.0500) [OH^-] = 5.36 × 10^-6 M pOH = 5.27, so pH = 14.00 - 5.27 = 8.73.
The equivalence pH is basic, not acidic. Phenolphthalein changes over pH 8.2–10.0, which brackets 8.73. Methyl orange would change well before equivalence, and bromothymol blue would also change too early, so phenolphthalein is the correct choice.
3. Explain why the pH changes slowly near the half-equivalence point, and evaluate the classmate's claim that the equivalence pH should be acidic.
Read the worked answer
Near half-equivalence the solution contains comparable amounts of HA and A^-, so it is a buffer. The pH is governed by pH = pKa + log([A^-]/[HA]). Adding a small amount of NaOH changes the ratio [A^-]/[HA] only slightly, and the logarithm compresses that change, so the pH rises slowly. Buffer capacity is greatest when [HA] ≈ [A^-], i.e. at the half-equivalence point.
The classmate's claim is incorrect. A weak acid is only acidic before neutralization. At equivalence the acid has been converted into its conjugate base, which reacts with water to produce OH^-. That hydrolysis makes the equivalence pH greater than 7. The calculation gives pH ≈ 8.73, consistent with phenolphthalein.