A Level H2 Chemistry Practice Paper 2

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

TuitionGoWhere Practice Paper (AI) — Version 2 of 5

Subject: Chemistry H2
Level: A-Level
Paper: Practice Paper (Topic: Acids Bases Salts)
Duration: 1 hour 15 minutes
Total Marks: 60
Name: ______________________
Class: ______________________
Date: ______________________

Instructions:

• This practice paper contains 20 questions on the topic of Acids, Bases & Salts.
• Answer all questions in the spaces provided.
• Show all working clearly where calculations are involved.
• Use the Data Booklet where appropriate.
• Marks for each question are shown in brackets [ ].
• The total marks for this paper are 60.

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Section A: Multiple Choice and Short Structured (Questions 1–8) [24 marks]

1. Which of the following species is a Lewis base? [1]
A. $BF_3$
B. $H^+$
C. $NH_3$
D. $AlCl_3$

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2. Write the conjugate acid of $NH_3$ according to Brønsted–Lowry theory. [1]

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3. A solution has pH = 3. What is the concentration of $H^+(aq)$ in mol dm$^{-3}$? [1]

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4. State the colour change observed when damp red litmus paper is exposed to $NH_3$ gas. [1]

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5. The table below shows tests for two gases. Complete the missing result for sulfur dioxide, $SO_2$. [2]

Gas	Test and Result
$CO_2$	Turns limewater milky (white ppt.); ppt. dissolves in excess
$SO_2$	________________________________________________

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6. A student records the following titration volumes of $FA_2$ (cm$^3$): 24.60, 24.55, 24.58, 25.90. Identify the anomalous titre and give the mean of the concordant titres to 2 decimal places. [3]

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7. Write the ionic equation for the precipitation of $AgCl(s)$ from $Ag^+(aq)$ and $Cl^-(aq)$. [1]

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8. Define the term $K_a$ for a weak acid HA. [2]

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Section B: Calculations and Buffer Systems (Questions 9–14) [20 marks]

9. The solubility of $AgCl$ in water at 25 °C is $1.43 \times 10^{-3}$ g dm$^{-3}$. Calculate the solubility product $K_{sp}$ of $AgCl$. ($M_r$ of $AgCl = 143.5$) [4]

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10. A buffer solution contains $0.25$ mol dm$^{-3}$ $CH_3COOH$ and $0.40$ mol dm$^{-3}$ $CH_3COO^-$ (from $CH_3COONa$). Given $K_a$ for $CH_3COOH = 1.8 \times 10^{-5}$ mol dm$^{-3}$, calculate the pH of the buffer. [3]

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11. Explain, using equations, how the buffer in Q10 resists a change in pH when $0.02$ mol of $HCl$ is added to 1 dm$^3$ of the buffer. [4]

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12. Calculate the pH of a $0.020$ mol dm$^{-3}$ solution of $HClO$ ($K_a = 3.0 \times 10^{-8}$ mol dm$^{-3}$). [3]

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13. $Mg(OH)_2$ has $K_{sp} = 5.6 \times 10^{-12}$ mol$^3$ dm$^{-9}$ at 25 °C. Calculate its molar solubility in pure water. [3]

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14. A $0.10$ mol dm$^{-3}$ solution of $HNO_2$ ($K_a = 4.5 \times 10^{-4}$) is partially neutralised with $NaOH$ to give a buffer of pH 3.50. Calculate the ratio $\frac{[NO_2^-]}{[HNO_2]}$ required. [3]

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Section C: Data Interpretation and Structured Reasoning (Questions 15–20) [16 marks]

15. The table shows reactions of aqueous cations with $NaOH(aq)$ and $NH_3(aq)$. Complete the entry for $Zn^{2+}(aq)$. [3]

Cation	With $NaOH(aq)$	With $NH_3(aq)$
$Al^{3+}(aq)$	White ppt., soluble in excess	White ppt., insoluble in excess
$Cu^{2+}(aq)$	Blue ppt., insoluble in excess	Blue ppt., soluble in excess
$Zn^{2+}(aq)$	________________________________	________________________________

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16. The titration curve below shows the neutralisation of a weak acid with a strong base.

Image pending generation: graph for 16.

State how the $pK_a$ of the weak acid can be found from the curve, and give the $pK_a$ if half-equivalence pH is 4.80. [2]

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17. A solution contains $0.10$ M $Ba^{2+}$ and $0.10$ M $Sr^{2+}$. $K_{sp}$ of $BaSO_4 = 1.1 \times 10^{-10}$, $SrSO_4 = 3.2 \times 10^{-7}$. Explain which sulfate precipitates first when $SO_4^{2-}$ is added slowly. [3]

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18. State and explain the effect on the pH of a $CH_3COOH/CH_3COO^-$ buffer when it is diluted with water. [3]

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19. A student adds $NH_3(aq)$ dropwise to $Fe^{2+}(aq)$. Give the observed colour change and write the formula of the final complex formed in excess $NH_3$. [2]

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20. The diagram shows an experimental setup for measuring the enthalpy of neutralisation.

Image pending generation: experimental_setup for 20.

State two sources of error in this setup and how each affects the measured temperature change. [3]

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TuitionGoWhere Practice Paper — Answer Key (Version 2)

Subject: Chemistry H2
Level: A-Level
Topic: Acids Bases Salts
Total Marks: 60

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Section A Answers (24 marks)

1. [1] C ($NH_3$)
Teaching note: A Lewis base donates a lone pair of electrons. $NH_3$ has a lone pair on N and donates it; $BF_3$, $H^+$, $AlCl_3$ are electron-pair acceptors (Lewis acids).
Mark: 1 for C.

2. [1] $NH_4^+$
Teaching note: Brønsted–Lowry conjugate acid is formed when the base gains a proton ($H^+$). $NH_3 + H^+ \rightarrow NH_4^+$.
Mark: 1 for $NH_4^+$.

3. [1] $1.0 \times 10^{-3}$ mol dm$^{-3}$
Teaching note: $pH = -\log[H^+]$ so $[H^+] = 10^{-pH} = 10^{-3}$.
Mark: 1 for correct value.

4. [1] Red litmus turns blue
Teaching note: $NH_3$ is alkaline; it turns damp red litmus blue. Must specify "damp" and "red".
Mark: 1 for correct colour change.

5. [2] Bleaches damp litmus paper (red litmus turns white); no ppt. with limewater.
Teaching note: $SO_2$ is acidic and reduces dyes; it bleaches damp litmus but does not give a lasting ppt. with limewater like $CO_2$.
Mark: 1 for bleaching, 1 for correct test description.

6. [3] Anomalous = 25.90 cm$^3$; mean = 24.58 cm$^3$
Working: Concordant = 24.60, 24.55, 24.58 (range 0.05 ≤ 0.10). Exclude 25.90. Mean = (24.60+24.55+24.58)/3 = 73.73/3 = 24.5767 → 24.58 cm$^3$ (2 d.p.).
Mark: 1 exclude outlier, 2 for mean correct to 2 d.p.

7. [1] $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$
Teaching note: Precipitation ionic equation omits spectator ions.
Mark: 1 for equation with states.

8. [2] $K_a = \frac{[H^+(aq)][A^-(aq)]}{[HA(aq)]}$ at equilibrium for $HA \rightleftharpoons H^+ + A^-$.
Teaching note: Defined as equilibrium constant for dissociation of weak acid; square brackets = equilibrium concentrations.
Mark: 1 expression, 1 statement of equilibrium/context.

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Section B Answers (20 marks)

9. [4] $K_{sp} = 9.94 \times 10^{-11}$ mol$^2$ dm$^{-6}$
Steps:

• Molar solubility $s = \frac{1.43 \times 10^{-3}}{143.5} = 9.97 \times 10^{-6}$ mol dm$^{-3}$
• $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$; $[Ag^+] = [Cl^-] = s$
• $K_{sp} = s^2 = (9.97 \times 10^{-6})^2 = 9.94 \times 10^{-11}$
  Mark: 1 convert mass→mol, 1 expression, 2 correct calc.

10. [3] pH = 4.94
Steps:

• $pK_a = -\log(1.8\times10^{-5}) = 4.745$
• Henderson–Hasselbalch: $pH = pK_a + \log\frac{[A^-]}{[HA]} = 4.745 + \log(0.40/0.25)$
• $\log(1.6) = 0.204$; pH = 4.95 (or 4.94 to 2 d.p.)
  Mark: 1 pKa, 1 substitution, 1 final.

11. [4]
Equation: $CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$
Explanation: Added $HCl$ gives $H^+$; conjugate base $CH_3COO^-$ consumes $H^+$ forming weak acid, so $[H^+]$ rises only slightly. pH resisted.
Mark: 2 equation, 2 explanation with buffer action.

12. [3] pH = 4.76
Steps:

• $HClO \rightleftharpoons H^+ + ClO^-$, $K_a = \frac{x^2}{0.020-x} \approx \frac{x^2}{0.020}$
• $x^2 = 3.0\times10^{-8} \times 0.020 = 6.0\times10^{-10}$; $x = 2.45\times10^{-5}$
• $pH = -\log(2.45\times10^{-5}) = 4.61$ (recalculate: $K_a=3.0e-8$, $x=\sqrt{6e-10}=2.45e-5$, pH=4.61)
  Correction: pH = 4.61.
  Mark: 1 setup, 1 solve, 1 pH.

13. [3] $s = 1.12 \times 10^{-4}$ mol dm$^{-3}$
Steps:

• $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$; $K_{sp} = s(2s)^2 = 4s^3$
• $s^3 = 5.6\times10^{-12}/4 = 1.4\times10^{-12}$; $s = \sqrt[3]{1.4\times10^{-12}} = 1.12\times10^{-4}$
  Mark: 1 expression, 1 calc, 1 unit.

14. [3] Ratio = 0.79
Steps:

• $pH = pK_a + \log\frac{[NO_2^-]}{[HNO_2]}$; $pK_a = -\log(4.5\times10^{-4}) = 3.35$
• $3.50 = 3.35 + \log(ratio)$; $\log(ratio)=0.15$; ratio = $10^{0.15}=1.41$
  Correction: ratio = 1.41.
  Mark: 1 pKa, 1 rearrange, 1 ratio.

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Section C Answers (16 marks)

15. [3]
$Zn^{2+}(aq)$ + $NaOH$: White ppt., soluble in excess (forms $[Zn(OH)_4]^{2-}$).
$Zn^{2+}(aq)$ + $NH_3$: White ppt., soluble in excess (forms $[Zn(NH_3)_4]^{2+}$).
Mark: 1 each column, 1 correctness of excess behaviour.

16. [2] At half-equivalence, pH = $pK_a$; $pK_a = 4.80$.
Teaching note: From graph, read pH at half volume of base to equivalence.
Mark: 1 method, 1 value.

17. [3] $BaSO_4$ precipitates first.
Reasoning: $[SO_4^{2-}]$ needed for $BaSO_4 = K_{sp}/[Ba^{2+}] = 1.1\times10^{-10}/0.10 = 1.1\times10^{-9}$ M; for $SrSO_4 = 3.2\times10^{-7}/0.10 = 3.2\times10^{-6}$ M. Lower $[SO_4^{2-}]$ required → precipitates first.
Mark: 1 calc Ba, 1 calc Sr, 1 conclusion.

18. [3] pH remains approximately unchanged.
Explanation: Ratio $[CH_3COO^-]/[CH_3COOH]$ unchanged on dilution; Henderson–Hasselbalch depends on ratio, not absolute concentration.
Mark: 1 pH same, 2 reasoning.

19. [2] Green ppt. initially (Fe(OH)$_2$), oxidises brown (Fe(OH)$_3$); no complex in excess $NH_3$, formula remains $Fe(OH)_2$/Fe(OH)$_3$.
Mark: 1 colour, 1 no complex.

20. [3]
Error 1: Heat loss to surroundings → measured $\Delta T$ lower than true.
Error 2: Non-adiabatic cup / no lid → evaporation cooling, $\Delta T$ lower.
Mark: 1 each error + effect (max 3).