A Level H2 Chemistry Acids Bases Salts Quiz

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A-Level Chemistry H2 Quiz - Acids Bases Salts

Name: ________________________
Class: ________________________
Date: ________________________
Score: ________ / 40

Duration: 60 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Show all working clearly where calculations are required.
• Use appropriate chemical notation and units.
• Section A: Short structured questions (1–8)
• Section B: Data interpretation and calculations (9–14)
• Section C: Extended response (15–20)

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Section A: Short Structured Questions (1–8)

1. [2 marks]
Define a Brønsted–Lowry acid and a Brønsted–Lowry base.
Acid: ____________________________________________________________
Base: ____________________________________________________________

2. [1 mark]
Write the conjugate base of $HCO_3^-$.
Answer: ________________________

3. [2 marks]
A student recorded the following titration volumes of $0.100\ \text{mol dm}^{-3}$ NaOH required to neutralise $25.0\ \text{cm}^3$ of HCl:

Trial	Volume / cm³
1 (rough)	24.80
2	24.35
3	24.40
4	24.38

From the table, obtain a suitable mean titre for use in calculations. Show your working.
Mean titre = ________________________

4. [2 marks]
Complete the gas test table below.

Gas	Test and Result
$NH_3$	________________________________________________
$CO_2$	________________________________________________

5. [2 marks]
State the observation when $NaOH(aq)$ is added dropwise then in excess to $Al^{3+}(aq)$.
Dropwise: ________________________________________
Excess: __________________________________________

6. [1 mark]
Calculate the pH of $1.0 \times 10^{-3}\ \text{mol dm}^{-3}$ HCl. Assume complete dissociation.
pH = ________

7. [2 marks]
Write the expression for the acid dissociation constant, $K_a$, for ethanoic acid, $CH_3COOH$.
$K_a$ = ________________________________________________

8. [2 marks]
A buffer solution contains $0.10\ \text{mol dm}^{-3}$ $CH_3COOH$ and $0.20\ \text{mol dm}^{-3}$ $CH_3COO^-$. Given $K_a = 1.8 \times 10^{-5}\ \text{mol dm}^{-3}$, calculate its pH.
pH = ________________________

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Section B: Data Interpretation and Calculations (9–14)

9. [3 marks]
$25.0\ \text{cm}^3$ of $0.0800\ \text{mol dm}^{-3}$ $H_2SO_4$ was titrated with $0.100\ \text{mol dm}^{-3}$ KOH.
(a) Write the balanced equation.
(b) Calculate the volume of KOH required for complete neutralisation.

10. [3 marks]
The $K_a$ of benzoic acid, $C_6H_5COOH$, is $6.3 \times 10^{-5}\ \text{mol dm}^{-3}$. Calculate the pH of a $0.050\ \text{mol dm}^{-3}$ solution of benzoic acid.

11. [3 marks]
A solution has $[H^+] = 3.2 \times 10^{-4}\ \text{mol dm}^{-3}$.
(a) Calculate the pH.
(b) State whether the solution is acidic, neutral, or alkaline.
(c) Calculate $[OH^-]$ using $K_w = 1.0 \times 10^{-14}\ \text{mol}^2\ \text{dm}^{-6}$.

12. [4 marks]
A titration curve for a weak acid vs strong base is shown below.

Image pending generation: graph for Q12.

(a) From the graph, state the volume at the equivalence point. [1]
(b) Deduce the $pK_a$ of the acid. [1]
(c) Explain why the pH at equivalence is greater than 7. [2]

13. [3 marks]
$0.200\ \text{g}$ of an impure sample of $CaCO_3$ was dissolved in excess HCl and the solution titrated with $0.0500\ \text{mol dm}^{-3}$ NaOH. $28.6\ \text{cm}^3$ NaOH was required. If the HCl added initially was $50.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$, calculate the percentage purity of $CaCO_3$. ($M_r$ of $CaCO_3 = 100.1$)

14. [3 marks]
State and explain the effect on the pH of a $CH_3COOH$/$CH_3COO^-$ buffer when a small amount of $HCl$ is added.

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Section C: Extended Response (15–20)

15. [3 marks]
Compare the Lewis and Brønsted–Lowry definitions of a base, giving one example of a species that acts as a base under each definition.

16. [3 marks]
Explain, with equations, how a buffer resists changes in pH when a small amount of $NaOH$ is added.

17. [3 marks]
A student adds $NH_3(aq)$ dropwise then in excess to $Cu^{2+}(aq)$. Describe the observations and write the formula of the complex formed in excess.

18. [3 marks]
Describe how you would test for the presence of $SO_4^{2-}$ ions in a solution, and state the observation for a positive test.

19. [3 marks]
Calculate the pH of a solution formed by mixing $50.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ $HCl$ with $50.0\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ $Ba(OH)_2$. Show all steps.

20. [3 marks]
Explain why $Al(OH)_3$ is described as amphoteric, with reference to its reactions with acid and base.

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A-Level Chemistry H2 Quiz - Acids Bases Salts: Answer Key

Total Marks: 40
Question count: 20 (exactly)

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Section A

1. [2 marks]

• Acid: A proton ($H^+$) donor. [1]
• Base: A proton ($H^+$) acceptor. [1]
  Teaching note: Brønsted–Lowry theory extends Arrhenius by not requiring water. A conjugate pair differs by one $H^+$.

2. [1 mark]
$CO_3^{2-}$
Teaching note: Remove one $H^+$ from $HCO_3^-$: $HCO_3^- \rightarrow CO_3^{2-} + H^+$.

3. [2 marks]

• Reject trial 1 (rough). Concordant: 24.35, 24.40, 24.38 (range 0.05 ≤ 0.10). [1]
• Mean = $(24.35 + 24.40 + 24.38)/3 = 24.376... \approx 24.38\ \text{cm}^3$ to 2 d.p. [1]
  Common mistake: Including rough titre or averaging non-concordant values.

4. [2 marks]

• $NH_3$: Turns damp red litmus paper blue. [1]
• $CO_2$: Gives a white precipitate with limewater (ppt. dissolves in excess $CO_2$). [1]
  Marking: Must specify "damp red litmus" and "white ppt." for full credit.

5. [2 marks]

• Dropwise: White precipitate of $Al(OH)_3$. [1]
• Excess: Precipitate dissolves to form colourless $[Al(OH)_4]^-$ solution. [1]
  Note: $Al^{3+}$ is amphoteric; dissolves in excess NaOH.

6. [1 mark]
pH = 3.0
Working: $[H^+] = 1.0 \times 10^{-3}$; $pH = -\log_{10}(1.0 \times 10^{-3}) = 3.00$.

7. [2 marks]
$K_a = \dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}$ [2]
Note: State (aq) optional; exclude water.

8. [2 marks]
$pH = pK_a + \log\left(\dfrac{[CH_3COO^-]}{[CH_3COOH]}\right)$
$pK_a = -\log(1.8 \times 10^{-5}) = 4.74$
$pH = 4.74 + \log(0.20/0.10) = 4.74 + 0.301 = 5.04$ [2]
Common mistake: Reversing ratio.

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Section B

9. [3 marks]
(a) $H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O$ [1]
(b) Moles $H_2SO_4 = 0.0250 \times 0.0800 = 2.00 \times 10^{-3}\ \text{mol}$
Moles KOH needed = $2 \times 2.00 \times 10^{-3} = 4.00 \times 10^{-3}\ \text{mol}$
Vol KOH = $4.00 \times 10^{-3} / 0.100 = 0.0400\ \text{dm}^3 = 40.0\ \text{cm}^3$ [2]

10. [3 marks]
$HA \rightleftharpoons H^+ + A^-$, $K_a = \dfrac{x^2}{0.050 - x} \approx \dfrac{x^2}{0.050}$
$x^2 = 6.3 \times 10^{-5} \times 0.050 = 3.15 \times 10^{-6}$
$x = 1.775 \times 10^{-3}$
$pH = -\log(1.775 \times 10^{-3}) = 2.75$ [3]

11. [3 marks]
(a) $pH = -\log(3.2 \times 10^{-4}) = 3.49$ [1]
(b) Acidic (pH < 7) [1]
(c) $[OH^-] = K_w / [H^+] = 1.0 \times 10^{-14} / 3.2 \times 10^{-4} = 3.1 \times 10^{-11}\ \text{mol dm}^{-3}$ [1]

12. [4 marks]
(a) 25.0 cm³ [1]
(b) At half-equivalence (12.5 cm³), pH = pKa = approx 4.5 (read from graph) [1]
(c) Salt of weak acid + strong base hydrolyses: $A^- + H_2O \rightleftharpoons HA + OH^-$, producing $OH^-$ so pH > 7. [2]

13. [3 marks]
Moles HCl initial = $0.0500 \times 0.100 = 5.00 \times 10^{-3}$
Moles NaOH used = $0.0286 \times 0.0500 = 1.43 \times 10^{-3}$
Moles HCl reacted with $CaCO_3$ = $5.00 \times 10^{-3} - 1.43 \times 10^{-3} = 3.57 \times 10^{-3}$
$CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$
Moles $CaCO_3$ = $3.57 \times 10^{-3}/2 = 1.785 \times 10^{-3}$
Mass pure = $1.785 \times 10^{-3} \times 100.1 = 0.1787\ \text{g}$
% purity = $(0.1787 / 0.200) \times 100 = 89.4\%$ [3]

14. [3 marks]
HCl reacts with $CH_3COO^-$: $CH_3COO^- + H^+ \rightarrow CH_3COOH$. [1]
$[CH_3COO^-]$ decreases slightly, $[CH_3COOH]$ increases; ratio changes little. [1]
Henderson–Hasselbalch shows pH drop is small. [1]

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Section C

15. [3 marks]
Lewis base: electron-pair donor (e.g. $NH_3$). [1]
Brønsted–Lowry base: proton acceptor (e.g. $OH^-$). [1]
Difference: Lewis broader, not requiring $H^+$. [1]

16. [3 marks]
$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$
Added $OH^-$ consumes $H^+$: $H^+ + OH^- \rightarrow H_2O$. [1]
Equilibrium shifts right, $CH_3COOH$ dissociates to replace $H^+$. [1]
$[H^+]$ remains near constant, pH stable. [1]

17. [3 marks]
Dropwise: pale blue precipitate $Cu(OH)_2$. [1]
Excess: precipitate dissolves to deep blue solution. [1]
Complex: $[Cu(NH_3)_4]^{2+}$. [1]

18. [3 marks]
Add $BaCl_2(aq)$ (or $Ba(NO_3)_2$) acidified with HCl. [1]
Positive: white precipitate of $BaSO_4$. [1]
Equation: $Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4(s)$. [1]

19. [3 marks]
Moles $H^+$ = $0.0500 \times 0.100 = 5.00 \times 10^{-3}$
Moles $OH^-$ = $0.0500 \times 0.100 \times 2 = 1.00 \times 10^{-2}$
Excess $OH^-$ = $5.00 \times 10^{-3}$; total vol = 100 cm³ = 0.100 dm³
$[OH^-] = 5.00 \times 10^{-3} / 0.100 = 0.0500\ \text{mol dm}^{-3}$
$pOH = 1.30$, $pH = 12.70$ [3]

20. [3 marks]
Amphoteric: reacts with acid and base. [1]
With acid: $Al(OH)_3 + 3H^+ \rightarrow Al^{3+} + 3H_2O$. [1]
With base: $Al(OH)_3 + OH^- \rightarrow [Al(OH)_4]^-$. [1]