QUESTION 1 (11 marks)
Phosphoric acid (H3PO4) is a common triprotic acid that dissociates fully in three stages. The dissociation equations are shown in the table.
Stage | Dissociation equation | Ka |
1 | H3PO4(aq) + H2O(l) ⇌ H2PO4–(aq) + H3O+(aq) | 7.1 × 10–3 |
2 | 2–(aq) + H3O+(aq) H2PO4 –(aq) + H2O(l) ⇌ HPO4 | 6.5 × 10–8 |
3 | 2–(aq) + H2O(l) ⇌ PO4 3–(aq) + H3O+(aq) HPO4 | 4.5 × 10–13 |
a) Use the information to determine the strongest Brønsted-Lowry acid and its conjugate base. Explain your reasoning.
[3 marks]
Acid:
Conjugate base:
Reasoning:
b) Identify an amphiprotic species from the dissociation reactions. Explain your reasoning.
[2 marks]
c) Determine the Kb value for the strongest conjugate base formed when H3PO4 has fully dissociated. Show your working.
[2 marks]
d) Calculate the pH of a 0.05 M solution of dihydrogen phosphate (H2PO4
–).
Show your working and state any assumptions made.
[4 marks]
QCAA guide · typeset solution
QCAA sample response and mark allocation
1a) | (aq) Acid: H3PO4 Conjugate base: H2PO4 −(aq) is the strongest Brønsted-Lowry acid because Reasoning: H3PO4 (aq) donates a proton and it has the largest Ka value. H3PO4 –(aq) accepts a proton. H2PO4 | • identifies H3PO4 as the acid and H2PO4 − as the conjugate base [1 mark] • identifies H3PO4 as the strongest acid due to having the largest Ka [1 mark] • identifies acid as the H+ donor and base as the H+ acceptor [1 mark] | For H+, accept proton. Do not accept hydrogen donor or acceptor. |
1b) | −(aq) is amphiprotic, because it can donate or accept a H2PO4 proton and therefore act as a Brønsted-Lowry acid or a base. | • identifies an amphiprotic species [1 mark] • identifies that this species can accept or donate protons [1 mark] | Acceptable amphiprotic species are: − - H2PO4 2− - HPO4 |
1c) | Kw= Ka × KbKb= KwKa=10−144.5 × 10−13 = 2.2 × 10−2 | • correctly substitutes into formula [1 mark] • determinesKb= 2.2 × 10−2[1 mark] | Allow FT error from substitution. Do not penalise for incorrect decimal places/significant figures. |
1d) | −→HPO4 2− is very As Ka is very small, the dissociation of H2PO4 very small. Assume 0.05 >> x, therefore 0.05 – x ≈ 0.05 −] = 0.05 ‒ x = 0.05 At equilibrium [H2PO4 Let x = [H+] = [HPO4 2−] x2 [x][x] [0.05−x] ≈ 6.5 × 10−8 = 0.05 x2 = 0.05 × 6.5 × 10−8 x = √3.25 × 10−9 x = 5.70 × 10−5 mol L–1 = [H+] pH = ‒log 5.70 × 10−5 pH = 4.2 (to one decimal place) | • indicates assumption 0.05 – x≈0.05[1 mark] • shows substitution correctly performed [1 mark] • correctly determines [H+] = 5.70 × 10−5[1 mark]• determines pH = 4.2[1 mark] | Allow FT error from incorrect substitution of Ka. Accept ICE table or other valid working. Allow FT error from [H+]. Do not penalise for incorrect decimal places/significant figures. |
QCAA sample response and marking criteria reproduced from the official guide.
Compare your working with the guide above.