Zadanie 3a
Wzorowane na: Chemistry HL Paper 1B, May 2026, TZ3, pytanie 3(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 3a-3c.
A student estimated the acid dissociation constant, Ka, of propanoic acid, CH3CH2COOH, by measuring the pH of a 0.0200 mol dm−3 standard solution of the acid with a pH probe that had been calibrated beforehand. Three trials gave the pH values below.
| Trial | pH (± 0.1) |
|---|---|
| 1 | 3.4 |
| 2 | 3.2 |
| 3 | 3.3 |
Polecenie
Complete the table for Trial 1 and so determine its Ka value. Use section 1 of the data booklet.
| Trial | pH (± 0.1) | [H3O+] / mol dm−3 | [CH3CH2COOH] / mol dm−3 | Ka |
|---|---|---|---|---|
| 1 | 3.4 |
Sprawdź rozwiązanieUkryj rozwiązanieKlucz CKE i punktacja
Odpowiedź z klucza
[H3O+] = 10−3.4 = 3.98 × 10−4 «mol dm−3» ✔
[CH3CH2COOH] = «0.0200 − 3.98 × 10⁻⁴ =» 1.96 × 10−2 «mol dm−3» ✔
Ka = « =» 8.1 × 10−6 ✔
Schemat punktowania
M1 hydronium concentration; M2 concentration of the undissociated acid (do not award M2 for 0.0200, the initial concentration); M3 Ka (accept 8.0 to 8.2 × 10⁻⁶). Award [3] for the correct final answer. ECF for M3 from M1 and M2: using 0.0200 for the acid concentration gives 7.9 × 10−6 and scores M3 only.
Komentarz
The dissociation removes only about 2 % of the acid, so [CH₃CH₂COOH] differs from 0.0200 only in the third significant figure; the mark requires the subtraction to be shown.