Wzorowane na: Chemistry HL Paper 2, November 2025, TZ3, pytanie 3(f). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Informacja do zadań 3a-3h.
In industry, benzene can be formed from cyclohexane through a gas-phase equilibrium reaction.
Hexanoic acid, , is a weak acid. The graph shows how the pH changes during the titration of a 20.0 aqueous solution of hexanoic acid with aqueous sodium hydroxide.
(i) Deduce an expression for the equilibrium constant, , of the ionization of hexanoic acid.
(ii) By annotating the graph, find the p of hexanoic acid.
(iii) Using section 18 of the data booklet, suggest a suitable indicator for this titration.
(iv) With reference to acid–base equilibria, explain why the resulting sodium hexanoate solution has a pH greater than 7.
(v) Given that the sodium hydroxide concentration was 0.020 , calculate the concentration, in , of the hexanoic acid solution.
(i) ✔
(ii) horizontal line drawn from the point on the curve at 12.5 of NaOH «half the equivalence volume» to the pH axis, giving p ≈ 4.9 ✔
(iii) phenolphthalein
OR
phenol red
OR
bromothymol blue ✔
(iv) «hexanoate ion» hydrolyses / reacts with water AND forms
OR
(v) « mol
(i) [1]; accept H3O+ for H+ and Ka for K. (ii) [1] for a line from the half-equivalence point (12.5 cm3) to the pH axis; accept an intersection between pH 4.7 and 5.0. (iii) [1] for any indicator whose range lies within the steep part of the curve, roughly pH 6.5–10.5. (iv) [1]; the equation alone scores. (v) [1].
At half the equivalence volume [HA] = [A−], so pH = pKa; the equivalence point of a weak acid with a strong base lies above pH 7, which rules out methyl orange and methyl red.
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