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A Level H1 Chemistry Practice Paper 2

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TuitionGoWhere Exam Practice (AI) - Chemistry H1 A-Level

Marking Scheme & Answer Key (Version 2)

Subject: Chemistry H1
Paper: Practice Paper 2

Section A: Structured Questions

1
(a) A weak acid is an acid that partially dissociates (or ionizes) in water. [1]
(b) CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)

  • Must use reversible arrow (⇌\rightleftharpoons). [1]
  • State symbols correct. [0 if missing, but usually part of the mark for the equation in this context]
    (c)
  1. Expression: Ka=[H+][CH3COO−][CH3COOH]K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}
  2. Assumption: [H+]=[CH3COO−][H^+] = [CH_3COO^-] and [CH3COOH]eq≈[CH3COOH]initial[CH_3COOH]_{eq} \approx [CH_3COOH]_{initial}
  3. [H+]=Ka×[CH3COOH]=1.7×10−5×0.10[H^+] = \sqrt{K_a \times [CH_3COOH]} = \sqrt{1.7 \times 10^{-5} \times 0.10}
  4. [H+]=1.7×10−6=1.30×10−3 mol dm−3[H^+] = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}
  5. pH=−log⁡(1.30×10−3)=2.88pH = -\log(1.30 \times 10^{-3}) = 2.88 (or 2.89) [3]
    • 1 mark for substitution into correct formula.
    • 1 mark for correct [H+][H^+].
    • 1 mark for correct pH.

2
(a) C6H5COOH(aq)+NaOH(aq)→C6H5COONa(aq)+H2O(l)C_6H_5COOH(aq) + NaOH(aq) \rightarrow C_6H_5COONa(aq) + H_2O(l) [1]
(b)

  1. Volume in dm3=22.40/1000=0.0224 dm3dm^3 = 22.40 / 1000 = 0.0224 \text{ dm}^3
  2. n(NaOH)=c×V=0.050×0.0224=1.12×10−3 moln(NaOH) = c \times V = 0.050 \times 0.0224 = 1.12 \times 10^{-3} \text{ mol} [1]
    (c)
  3. Ratio is 1:1, so n(acid)=1.12×10−3 moln(\text{acid}) = 1.12 \times 10^{-3} \text{ mol}
  4. Volume of acid = 25.0 cm3=0.025 dm325.0 \text{ cm}^3 = 0.025 \text{ dm}^3
  5. c(acid)=1.12×10−30.025=0.0448 mol dm−3c(\text{acid}) = \frac{1.12 \times 10^{-3}}{0.025} = 0.0448 \text{ mol dm}^{-3} [2]
    • 1 mark for moles of acid.
    • 1 mark for concentration.

3
(a) Ka1=[HCO3−][H+][H2CO3]K_{a1} = \frac{[HCO_3^-][H^+]}{[H_2CO_3]} [1]
(b)

  1. It is harder to remove a positive proton (H+H^+) from a negatively charged ion (HCO3−HCO_3^-) than from a neutral molecule (H2CO3H_2CO_3).
  2. Electrostatic attraction between H+H^+ and HCO3−HCO_3^- is stronger. [2]
    (c) [H+]=10−pH=10−5.6=2.5×10−6 mol dm−3[H^+] = 10^{-pH} = 10^{-5.6} = 2.5 \times 10^{-6} \text{ mol dm}^{-3} [1]

4
(a) An amphoteric substance can act as both an acid and a base (reacts with both acids and bases). [1]
(b)
(i) Al2O3(s)+6H+(aq)→2Al3+(aq)+3H2O(l)Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l) [1]
(ii) Al2O3(s)+2OH−(aq)+3H2O(l)→2[Al(OH)4]−(aq)Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq) [1]
* Accept Al2O3+2NaOH→2NaAlO2+H2OAl_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O if balanced correctly, but complex ion form is preferred in modern syllabi.

5
(a) A weak acid and its salt (conjugate base). [1]
(b)

  1. pH=pKa+log⁡([salt][acid])pH = pK_a + \log \left( \frac{[\text{salt}]}{[\text{acid}]} \right)
  2. Since [salt]=[acid]=0.10[\text{salt}] = [\text{acid}] = 0.10, the log term is log⁡(1)=0\log(1) = 0.
  3. pH=pKa=−log⁡(1.7×10−5)=4.77pH = pK_a = -\log(1.7 \times 10^{-5}) = 4.77 [2]
    (c)
  4. Added H+H^+ ions react with the conjugate base (CH3COO−CH_3COO^-).
  5. Equation: CH3COO−(aq)+H+(aq)→CH3COOH(aq)CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)
  6. This removes most of the added H+H^+, keeping pH relatively constant. [2]

6
(a) Ksp=[Mg2+][OH−]2K_{sp} = [Mg^{2+}][OH^-]^2 [1]
(b)

  1. Let solubility be s mol dm−3s \text{ mol dm}^{-3}.
  2. [Mg2+]=s[Mg^{2+}] = s, [OH−]=2s[OH^-] = 2s.
  3. Ksp=(s)(2s)2=4s3K_{sp} = (s)(2s)^2 = 4s^3.
  4. 1.8×10−11=4s31.8 \times 10^{-11} = 4s^3.
  5. s3=4.5×10−12s^3 = 4.5 \times 10^{-12}.
  6. s=4.5×10−123=1.65×10−4 mol dm−3s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3} [3]
    (c)
  7. NaOHNaOH provides a high concentration of OH−OH^- ions (Common Ion Effect).
  8. According to Le Chatelier’s Principle / KspK_{sp} expression, increasing [OH−][OH^-] shifts equilibrium to the left (precipitation), decreasing solubility. [2]

7
(a) Ka=10−pKa=10−4.87=1.35×10−5 mol dm−3K_a = 10^{-pK_a} = 10^{-4.87} = 1.35 \times 10^{-5} \text{ mol dm}^{-3} [1]
(b)

  • Start pH: Weak acid, so pH > 1 (approx 2.9).
  • Buffer region: Gradual rise.
  • Equivalence point: pH > 7 (approx 8-9) because salt is basic.
  • Final pH: Approaches pH of NaOH (approx 13).
  • Shape: Sigmoidal. [3]
    • 1 mark for start pH > 1.
    • 1 mark for equivalence point pH > 7.
    • 1 mark for general shape.

8
(a) NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq) [1]
(b) pOH=14−11.0=3.0pOH = 14 - 11.0 = 3.0.
[OH−]=10−3.0=1.0×10−3 mol dm−3[OH^-] = 10^{-3.0} = 1.0 \times 10^{-3} \text{ mol dm}^{-3} [1]
(c)

  1. Kb=[NH4+][OH−][NH3]K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}
  2. Assume [NH4+]=[OH−]=1.0×10−3[NH_4^+] = [OH^-] = 1.0 \times 10^{-3}.
  3. 1.8×10−5=(1.0×10−3)2[NH3]eq1.8 \times 10^{-5} = \frac{(1.0 \times 10^{-3})^2}{[NH_3]_{eq}}
  4. [NH3]eq=1.0×10−61.8×10−5=0.0556 mol dm−3[NH_3]_{eq} = \frac{1.0 \times 10^{-6}}{1.8 \times 10^{-5}} = 0.0556 \text{ mol dm}^{-3}
  5. Initial [NH3]≈0.0556 mol dm−3[NH_3] \approx 0.0556 \text{ mol dm}^{-3} (since dissociation is small). [3]

Section B: Data-Based and Application Questions

9
(a)

  1. pH=2.44⇒[H+]=10−2.44=3.63×10−3 mol dm−3pH = 2.44 \Rightarrow [H^+] = 10^{-2.44} = 3.63 \times 10^{-3} \text{ mol dm}^{-3}
  2. Ka=[H+]2[HA]=(3.63×10−3)20.10K_a = \frac{[H^+]^2}{[HA]} = \frac{(3.63 \times 10^{-3})^2}{0.10}
  3. Ka=1.32×10−50.10=1.32×10−4 mol dm−3K_a = \frac{1.32 \times 10^{-5}}{0.10} = 1.32 \times 10^{-4} \text{ mol dm}^{-3} [3]
    (b)
  4. If it were strong, [H+][H^+] would equal 0.10 mol dm−30.10 \text{ mol dm}^{-3} (pH 1.0).
  5. The actual pH is 2.44 (much higher), indicating partial dissociation. [2]
    (c)
  6. Enzymes are proteins with specific 3D structures (tertiary structure) maintained by hydrogen/ionic bonds.
  7. Excess H+H^+ disrupts these bonds, causing denaturation. The active site changes shape, and the substrate no longer fits. [2]

10
(a)

  1. Adding F−F^- shifts the equilibrium: Ca5(PO4)3OH(s)+F−(aq)⇌Ca5(PO4)3F(s)+OH−(aq)Ca_5(PO_4)_3OH(s) + F^-(aq) \rightleftharpoons Ca_5(PO_4)_3F(s) + OH^-(aq).
  2. Since fluoroapatite is less soluble (smaller KspK_{sp}), the equilibrium lies far to the right, forming a more resistant layer on the tooth. [3]
    (b)
  3. H+H^+ from the acid reacts with OH−OH^- and PO43−PO_4^{3-} ions.
  4. This decreases the concentration of products, shifting the dissolution equilibrium to the right (Le Chatelier), causing more enamel to dissolve. [2]

11
(a) [H+]=10−2.30=5.01×10−3 mol dm−3[H^+] = 10^{-2.30} = 5.01 \times 10^{-3} \text{ mol dm}^{-3} [1]
(b)

  1. % dissociation=[H+]eq[HA]initial×100\% \text{ dissociation} = \frac{[H^+]_{eq}}{[HA]_{initial}} \times 100
  2. =5.01×10−30.050×100=10.0%= \frac{5.01 \times 10^{-3}}{0.050} \times 100 = 10.0\% [2]
    (c) Weak acid. Strong acids dissociate ~100%. 10% is significantly less than 100%. [1]

12
(a)

  1. Chlorine is electronegative and exerts a negative inductive effect (-I effect).
  2. This withdraws electron density from the O-H bond, making the H more positive and easier to lose.
  3. It also stabilizes the resulting carboxylate anion by dispersing the negative charge. More Cl atoms = stronger effect = lower pKapK_a. [3]
    (b)
  4. Ka=10−2.86=1.38×10−3K_a = 10^{-2.86} = 1.38 \times 10^{-3}.
  5. [H+]=Ka×c=1.38×10−3×0.10=1.38×10−4=0.0117[H^+] = \sqrt{K_a \times c} = \sqrt{1.38 \times 10^{-3} \times 0.10} = \sqrt{1.38 \times 10^{-4}} = 0.0117.
  6. pH=−log⁡(0.0117)=1.93pH = -\log(0.0117) = 1.93 [3]

13
(a) When pH=pKapH = pK_a, the ratio [salt][acid]=1\frac{[\text{salt}]}{[\text{acid}]} = 1. [1]
(b)

  1. Initial moles: Acid = 0.10, Salt = 0.10.
  2. Add 0.01 mol NaOH. NaOH reacts with Acid.
  3. New moles: Acid = 0.10−0.01=0.090.10 - 0.01 = 0.09. Salt = 0.10+0.01=0.110.10 + 0.01 = 0.11.
  4. pH=4.76+log⁡(0.110.09)pH = 4.76 + \log(\frac{0.11}{0.09}).
  5. pH=4.76+log⁡(1.22)=4.76+0.087=4.85pH = 4.76 + \log(1.22) = 4.76 + 0.087 = 4.85 [4]

14
(a) Ksp=[Ca2+][OH−]2K_{sp} = [Ca^{2+}][OH^-]^2 [1]
(b)

  1. Solubility decreases as T increases.
  2. This implies the reverse reaction (precipitation) is favored by heat, or the forward reaction (dissolution) is favored by cold.
  3. Therefore, dissolution is exothermic (ΔH<0\Delta H < 0). [2]

15
(a)

  1. Kind=[H+][In−][HIn]K_{ind} = \frac{[H^+][In^-]}{[HIn]}
  2. [H+]=Kind[HIn][In−][H^+] = K_{ind} \frac{[HIn]}{[In^-]}
  3. −log⁡[H+]=−log⁡Kind−log⁡([HIn][In−])-\log[H^+] = -\log K_{ind} - \log(\frac{[HIn]}{[In^-]})
  4. pH=pKind+log⁡([In−][HIn])pH = pK_{ind} + \log(\frac{[In^-]}{[HIn]}) [2]
    (b) When [HIn]=[In−][HIn] = [In^-], log⁡(1)=0\log(1) = 0, so pH=pKind=3.7pH = pK_{ind} = 3.7. [1]
    (c)
  5. Strong Acid + Weak Base titration has an equivalence point in the acidic range (pH < 7).
  6. Methyl orange changes color in the acidic range (3.1–4.4), matching the steep part of the curve.
  7. Weak Acid + Strong Base has equivalence point in basic range (pH > 7), where methyl orange has already changed color. [2]

Section C: Extended Response

16
(a) pH=−log⁡10[H+]pH = -\log_{10}[H^+] [1]
(b)

  1. H2SO4H_2SO_4 is diprotic. [H+]=2×0.050=0.10 mol dm−3[H^+] = 2 \times 0.050 = 0.10 \text{ mol dm}^{-3}.
  2. pH=−log⁡(0.10)=1.0pH = -\log(0.10) = 1.0 [2]
    (c)
  3. The second dissociation is incomplete, so fewer H+H^+ ions are produced than assumed in (b).
  4. Lower [H+][H^+] means higher pH. [2]

17
(a)

  1. Rate is faster with HCl (Acid A).
  2. HCl is a strong acid (fully dissociated), so [H+][H^+] is much higher (1.0 M1.0 \text{ M}) compared to ethanoic acid (weak, partial dissociation, low [H+][H^+]).
  3. Collision frequency between H+H^+ and Mg is higher. [3]
    (b)
  4. The total volume of gas produced is the same.
  5. Both acids have the same volume and concentration, so they contain the same total number of moles of potential H+H^+ (stoichiometrically). Excess Mg ensures all acid reacts. [2]

18
(a)

  1. Ksp=s2K_{sp} = s^2.
  2. s=1.8×10−10=1.34×10−5 mol dm−3s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5} \text{ mol dm}^{-3} [2]
    (b)
  3. In 0.10 M NaCl, [Cl−]=0.10 M[Cl^-] = 0.10 \text{ M}.
  4. Ksp=[Ag+][Cl−]⇒1.8×10−10=[Ag+](0.10)K_{sp} = [Ag^+][Cl^-] \Rightarrow 1.8 \times 10^{-10} = [Ag^+](0.10).
  5. [Ag+]=1.8×10−9 mol dm−3[Ag^+] = 1.8 \times 10^{-9} \text{ mol dm}^{-3}.
  6. Solubility is 1.8×10−9 mol dm−31.8 \times 10^{-9} \text{ mol dm}^{-3}. [3]
    (c)
  7. High [Cl−][Cl^-] from NaCl shifts equilibrium AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq) to the left.
  8. This reduces the concentration of dissolved Ag+Ag^+, lowering solubility. [2]

19
(a)

  • NaCl: Neutral (pH 7).
  • NH4ClNH_4Cl: Acidic (pH < 7). [2]
    (b)
  1. NH4+NH_4^+ is the conjugate acid of a weak base (NH3NH_3).
  2. It hydrolyzes in water: NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq).
  3. Production of H3O+H_3O^+ makes the solution acidic. [2]

20
(a)

  • Start pH: 1.0 (0.1 M HCl0.1 \text{ M } HCl).
  • Equivalence point: pH 7.0 (vertical section).
  • End pH: ~13.0 (0.1 M NaOH0.1 \text{ M } NaOH).
  • Curve starts low, stays low, rises sharply at 25 cm³, levels off high. [3]
    (b) Phenolphthalein (colorless to pink) or Methyl Orange (red to yellow). Both work for Strong/Strong. [1]
    (c) The pH change at the equivalence point is very gradual (no steep vertical section), making it hard for an indicator to show a sharp color change. [1]