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A Level H2 Chemistry Acids Bases Salts Quiz

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Questions

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 where calculations are required. Use the Data Booklet if needed. Section A is short structured questions; Section B is calculations and explanations; Section C is data interpretation and applied reasoning.

Section A: Short Structured Questions (Questions 1–5)

1. [2 marks] State the Brønsted–Lowry definitions of an acid and a base.
Acid: ________________________________________________________
Base: ________________________________________________________

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

3. [2 marks] A student adds aqueous sodium hydroxide dropwise to a solution containing $Al^{3+}(aq)$ . Describe the observation and write the formula of the complex ion formed in excess NaOH.
Observation: __________________________________________________
Complex ion: __________________________________________________

4. [1 mark] Define the term $K_a$ for a weak acid HA.

5. [2 marks] Complete the gas test table below.

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

Section B: Calculations and Explanations (Questions 6–15)

6. [3 marks] The solubility of AgCl in water at 25 °C is $0.00143\text{ g dm}^{-3}$ . Calculate the solubility product $K_{sp}$ of AgCl. ( $M_r$ of AgCl = 143.5)
Working:

7. [4 marks] A buffer solution contains $0.20\text{ mol dm}^{-3}$ $CH_3COOH$ and $0.30\text{ mol dm}^{-3}$ $CH_3COONa$ . Calculate the pH of the buffer. ( $K_a$ for $CH_3COOH = 1.8 \times 10^{-5}\text{ mol dm}^{-3}$ )
Working:

8. [3 marks] Explain, using equations, how the buffer in Q7 resists a change in pH when a small amount of $HCl(aq)$ is added.

9. [2 marks] Calculate the pH of a $0.010\text{ mol dm}^{-3}$ solution of $HCl$ (assume complete dissociation).
Working: ________________________________________________________

10. [3 marks] $0.0250\text{ dm}^3$ of $0.100\text{ mol dm}^{-3}$ $NaOH$ is exactly neutralised by $0.0200\text{ dm}^3$ of $H_2SO_4$ . Calculate the concentration of the $H_2SO_4$ solution.
Working:

11. [2 marks] Write the expression for $K_{sp}$ of $Mg(OH)_2$ and state its units.
Expression: __________________________________________________
Units: ______________________________________________________

12. [4 marks] $25.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $CH_3COOH$ is titrated with $0.100\text{ mol dm}^{-3}$ $NaOH$ . Calculate the pH at the equivalence point. ( $K_a$ for $CH_3COOH = 1.8 \times 10^{-5}$ ; $K_w = 1.0 \times 10^{-14}$ at 25 °C)
Working:

13. [2 marks] State the Lewis definition of a base and give one example of a Lewis base.
Definition: __________________________________________________
Example: ____________________________________________________

14. [3 marks] A solution has $[H^+] = 3.2 \times 10^{-4}\text{ mol dm}^{-3}$ . Calculate the pH and state whether the solution is acidic, neutral, or basic.
Working: ____________________________________________________
Classification: ______________________________________________

15. [3 marks] Explain why $CH_3COOH$ is a weak acid whereas $HCl$ is a strong acid, in terms of extent of dissociation.

Section C: Data Interpretation and Applied Reasoning (Questions 16–20)

16. [3 marks] The table below shows titration volumes from a practical session.

Titration	Volume / cm³
Rough	24.80
1	24.35
2	24.40
3	24.38

Obtain a suitable mean titre for calculations and show your working.
Mean: ______________________________________________________
Working: ____________________________________________________

17. [3 marks] The following titration curve was generated for a weak acid–strong base titration.

Graph for 17 (ALEVEL Chemistry H2)

Generated graph for 17.

Using the graph, state the pH at the half-equivalence point and deduce the $pK_a$ of the acid.
pH at half-equivalence: ________
$pK_a$ : ________

18. [3 marks] A student prepares a buffer by mixing $0.050\text{ mol dm}^{-3}$ $NH_3$ and $0.030\text{ mol dm}^{-3}$ $NH_4Cl$ . Given $K_b$ for $NH_3 = 1.8 \times 10^{-5}$ , calculate the pH. (Use $K_a = K_w / K_b$ )
Working:

19. [2 marks] State two assumptions made when using the Henderson–Hasselbalch equation for buffer pH calculations.

20. [3 marks] A water sample contains $Ca^{2+}(aq)$ at $2.0 \times 10^{-4}\text{ mol dm}^{-3}$ . If $K_{sp}$ of $CaF_2 = 3.9 \times 10^{-11}\text{ mol}^3\text{ dm}^{-9}$ , calculate the maximum $[F^-]$ before precipitation occurs.
Working:

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Answers

A-Level Chemistry H2 Quiz - Acids Bases Salts (Answer Key)

Total Marks: 40
Topic: Acids, Bases & Salts (syllabus-first generated content; not claimed as past-year derived)

1. [2 marks]

Acid: Proton ( $H^+$ ) donor. [1]
Base: Proton ( $H^+$ ) acceptor. [1]
Teaching note: Brønsted–Lowry theory extends Arrhenius by not requiring water. A acid donates $H^+$ ; a base accepts it.

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

3. [2 marks]
Observation: White precipitate forms, dissolves in excess NaOH to give a colourless solution. [1]
Complex ion: $[Al(OH)_4]^-$ [1]
Teaching note: $Al^{3+}$ is amphoteric; with excess $OH^-$ , $Al(OH)_3 + OH^- \rightarrow [Al(OH)_4]^-$ .

4. [1 mark]
$K_a = \frac{[H^+][A^-]}{[HA]}$ for $HA \rightleftharpoons H^+ + A^-$ [1]

5. [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]

6. [3 marks]
$M_r$ AgCl = 143.5 g mol⁻¹ [1]
Solubility in mol dm⁻³ = $0.00143 / 143.5 = 9.97 \times 10^{-6}$ [1]
$AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ ; $K_{sp} = [Ag^+][Cl^-] = (9.97\times10^{-6})^2 = 9.94\times10^{-11}\text{ mol}^2\text{ dm}^{-6}$ [1]

7. [4 marks]
$pK_a = -\log(1.8\times10^{-5}) = 4.74$ [1]
$pH = pK_a + \log\frac{[A^-]}{[HA]} = 4.74 + \log(0.30/0.20)$ [1]
$= 4.74 + \log(1.5) = 4.74 + 0.18$ [1]
$= 4.92$ [1]

8. [3 marks]
Added $H^+$ reacts with $CH_3COO^-$ : $CH_3COO^- + H^+ \rightarrow CH_3COOH$ [1]
Equilibrium $CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$ shifts left (Le Chatelier) [1]
Free $[H^+]$ does not rise significantly; pH stable. [1]

9. [2 marks]
$[H^+] = 0.010$ ; $pH = -\log(0.010) = 2.00$ [2]

10. [3 marks]
Moles $NaOH = 0.0250 \times 0.100 = 2.50\times10^{-3}$ mol [1]
$2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$ ; moles $H_2SO_4 = 1.25\times10^{-3}$ [1]
Conc $H_2SO_4 = 1.25\times10^{-3} / 0.0200 = 0.0625\text{ mol dm}^{-3}$ [1]

11. [2 marks]
$K_{sp} = [Mg^{2+}][OH^-]^2$ [1]
Units: $\text{mol}^3\text{ dm}^{-9}$ [1]

12. [4 marks]
At eq, all $CH_3COOH$ → $CH_3COO^-$ ; total vol = 50 cm³; $[CH_3COO^-] = 0.0500$ M [1]
$K_b = K_w/K_a = 1.0\times10^{-14}/1.8\times10^{-5} = 5.56\times10^{-10}$ [1]
$[OH^-] = \sqrt{K_b \times 0.0500} = 5.27\times10^{-6}$ [1]
$pOH = 5.28$ , $pH = 8.72$ [1]

13. [2 marks]
Base: electron-pair donor. [1]
Example: $NH_3$ (or $H_2O$ , $Cl^-$ ). [1]

14. [3 marks]
$pH = -\log(3.2\times10^{-4}) = 3.49$ [2]
Acidic (pH < 7). [1]

15. [3 marks]
$HCl$ dissociates fully: $HCl \rightarrow H^+ + Cl^-$ [1]
$CH_3COOH$ partially: $CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$ [1]
Weak acid has small $K_a$ , most molecules remain undissociated. [1]

16. [3 marks]
Exclude rough (24.80). Concordant: 24.35, 24.40, 24.38 (range 0.05 < 0.10) [1]
Mean = $(24.35+24.40+24.38)/3 = 24.376...$ [1]
To 2 d.p. = 24.38 cm³ [1]

17. [3 marks]
From graph: pH at half-equivalence = 4.7 [1.5]
At half-equivalence, $pH = pK_a$ , so $pK_a = 4.7$ [1.5]

18. [3 marks]
$K_a = 1.0\times10^{-14} / 1.8\times10^{-5} = 5.56\times10^{-10}$ ; $pK_a = 9.25$ [1]
$pH = 9.25 + \log(0.050/0.030) = 9.25 + 0.22$ [1]
$= 9.47$ [1]

19. [2 marks]

$[HA]$ and $[A^-]$ approx equal to initial concentrations (small dissociation). [1]
Volume change negligible / activity ≈ concentration. [1]

20. [3 marks]
$K_{sp} = [Ca^{2+}][F^-]^2$ [1]
$[F^-]^2 = 3.9\times10^{-11} / 2.0\times10^{-4} = 1.95\times10^{-7}$ [1]
$[F^-] = 4.4\times10^{-4}\text{ mol dm}^{-3}$ [1]