Zadanie 1e, Paper 1B, specimen 2025, Chemia IB HL
Wzorowane na: Chemistry HL Paper 1B, specimen 2025, pytanie 1(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
An outdoor public swimming pool (a lido) stays open throughout the year. The water is not heated, and in winter, when the pool is used by cold-water swimmers, its temperature falls to about 5 °C. The water is disinfected with chlorine. When chlorine dissolves in water it reacts to form hypochlorous acid, HOCl, which is the species that actually kills microorganisms.
Cl2(aq) + H2O(l) ⇌ HOCl(aq) + H+(aq) + Cl−(aq)
Polecenie
Hypochlorous acid is a weak acid. In water it is in equilibrium with its conjugate base, the hypochlorite ion, OCl−, which kills microorganisms far less effectively than the acid molecule does.
The graph shows what percentage of the dissolved chlorine is present as HOCl and what percentage as OCl− across a range of pH values, at 5 °C.
Powiększ(i) Deduce the range of pH values over which the disinfecting action of the water is greatest.
(ii) Using the graph, determine the pKa of hypochlorous acid at 5 °C.
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Pokaż rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) pH below «about» 6.5 / pH 0-6.5 «where almost all of the chlorine is present as HOCl» ✔
(ii) pKa = 7.8 ✔
«at the pH where the curves cross» [HOCl] = [OCl−] «so Ka = [H+] and pH = pKa» ✔
OR
[H+] = Ka «when the acid is 50 % ionized» ✔
Schemat punktowania
(i) [1] for any value or range with an upper limit of pH 7.0 or lower. (ii) M1: pKa read from the crossing point of the two curves, accept 7.7-7.9; M2: the justification that at this pH [HOCl] = [OCl−] so that Ka = [H+] (pH = pKa). Award M2 only if a value in the range 7.6-8.0 has been read.
Komentarz. pKa of HOCl rises as the water gets colder, so the answer is not the textbook room-temperature value. Reading the crossing point at 50 % is the whole of the first mark; the second mark needs the reason, not just a number.