QUESTION 3 (9 marks)
The concentration, pH and dissociation constant (Ka) of aqueous solutions of ethanoic acid and two unknown monoprotic acids, I and II, are shown.
Acid | Concentration (M) | pH | Ka |
CH3COOH(aq) | 0.2 | 1.8 × 10−5 | |
I | 0.2 | 1.9 | 6.6 × 10−4 |
II | 0.1 | 1.1 | 1.3 × 106 |
a) Compare the relative strength of an aqueous solution of acid I and CH3COOH(aq).
[3 marks]
Similarity:
Difference:
Significance:
b) Determine whether an aqueous solution of acid I or acid II would have a higher electrical conductivity. Explain your reasoning.
[3 marks]
c) Calculate the pH of 0.2 M CH3COOH(aq). Show your working.
[3 marks]
QCAA guide · typeset solution
QCAA sample response and mark allocation
3a) | Similarity: both weak acids Difference: ethanoic acid (CH3COOH) is weaker than acid I Significance: Ethanoic acid dissociates less to produce a lower [H+] | • identifies that both acids are weak acids [1 mark] • identifies that ethanoic acid is weaker than acid I [1 mark] • explains that ethanoic acid dissociates less to produce a lower [H+] [1 mark] |
3b) | Acid II has a Ka greater than 1 and is therefore a strong acid, while acid I is a weak acid due to small Ka. Acid II dissociates to produce more ions in an aqueous solution than acid I. Therefore, acid II has a higher electrical conductivity than acid I. | • identifies that acid II is stronger than acid I [1 mark] • explains that acid II dissociates to produce more ions in solution [1 mark] • determines that the electrical conductivity of acid II will be greater than acid I [1 mark] |
3c) | +33[H O ] = [CH COO ] = x−+333a[H O ][CH COO ][CH COOH] =K−25( )0.2 =1.8×10x−+33= 1.910M[H O ]−×103pH =log [H O ]+−pH = 3 | • identifies +33[H O ] = [CH COO ]− [1 mark]• determines +3[H O ] [1 mark]• calculates pH [1 mark] |
QCAA sample response and marking criteria reproduced from the official guide.
Compare your working with the guide above.