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A Level H1 Chemistry Practice Paper 3

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TuitionGoWhere Exam Practice (AI) - Answer Key

A-Level Chemistry H1 Practice Paper - Version 3

Topic: Acids, Bases and Salts

Total Marks: 40

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)
Marking: Reversible arrow (⇌) required. State symbols required. [1]
(c) Ethanoic acid can form dimers via hydrogen bonding between two molecules (two H-bonds per dimer). Ethanol forms hydrogen bonds but not stable dimers to the same extent. More energy is required to break the intermolecular forces in ethanoic acid. [2]
Note: Accept reference to stronger intermolecular forces due to dimerization.

2.
(a) [H+]=10−pH=10−2.90=1.26×10−3[H^+] = 10^{-pH} = 10^{-2.90} = 1.26 \times 10^{-3} mol dm⁻³. [1]
(b) Moles of NaOH = 0.100×20.01000=0.00200.100 \times \frac{20.0}{1000} = 0.0020 mol.
Ratio of C₆H₅COOH : NaOH is 1:1.
Moles of acid = 0.0020 mol.
Concentration of acid = 0.002025.01000=0.080\frac{0.0020}{\frac{25.0}{1000}} = 0.080 mol dm⁻³. [2]
(c) Sketch:

  • Start pH around 3 (weak acid).
  • Gradual rise (buffer region).
  • Vertical jump at equivalence point (20 cm³).
  • Equivalence point pH > 7 (basic salt).
  • Levels off at high pH (excess strong base).
    Marks: Shape [1], Equivalence point labeled correctly at pH > 7 and Vol = 20 [1]. [2]

3.
(a) A solution that resists changes in pH when small amounts of acid or base are added. [1]
(b) pH=pKa+log⁡10([salt][acid])pH = pK_a + \log_{10} \left( \frac{[salt]}{[acid]} \right)
Since [salt] = [acid], log⁡(1)=0\log(1) = 0.
pH=pKa=−log⁡10(1.7×10−5)=4.77pH = pK_a = -\log_{10}(1.7 \times 10^{-5}) = 4.77. [2]
(c) The added H⁺ ions react with the ethanoate ions (CH₃COO⁻) from the salt:
CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq).
This removes most of the added H⁺, keeping the pH relatively constant. [2]

4.
(a) Amphoteric substances can act as both an acid and a base. [1]
(b) (i) Al₂O₃(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂O(l) [1]
(ii) Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq)
Accept: Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄ [1]

5.
(a) Ksp=[Mg2+][OH−]2K_{sp} = [Mg^{2+}][OH^-]^2 [1]
(b) Let solubility be ss mol dm⁻³.
[Mg2+]=s[Mg^{2+}] = s, [OH−]=2s[OH^-] = 2s.
Ksp=(s)(2s)2=4s3K_{sp} = (s)(2s)^2 = 4s^3.
2.0×10−11=4s32.0 \times 10^{-11} = 4s^3.
s3=5.0×10−12s^3 = 5.0 \times 10^{-12}.
s=5.0×10−123=1.71×10−4s = \sqrt[3]{5.0 \times 10^{-12}} = 1.71 \times 10^{-4} mol dm⁻³. [2]
(c) Common ion effect. MgCl₂ provides Mg²⁺ ions.
According to Le Chatelier’s principle, increasing [Mg²⁺] shifts the equilibrium Mg(OH)2(s)⇌Mg2+(aq)+2OH−(aq)Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq) to the left, causing precipitation and decreasing solubility. [2]

Section B: Data Interpretation and Application

6.
(a) H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq) [1]
(b) Ka=[H+][HCO3−][H2CO3]K_a = \frac{[H^+][HCO_3^-]}{[H_2CO_3]} [1]
(c) H₂SO₃ has a larger KaK_a, meaning it dissociates to a greater extent than H₂CO₃. Therefore, H₂SO₃ is a stronger acid. [1]
(d) CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g) [1]

7.
(a) [H+]=10−2=0.01[H^+] = 10^{-2} = 0.01 mol dm⁻³.
[OH−]=Kw[H+]=1.0×10−140.01=1.0×10−12[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{0.01} = 1.0 \times 10^{-12} mol dm⁻³. [2]
(b) At pH 7, the concentration of H⁺ is much lower than optimal. The change in pH disrupts the ionic and hydrogen bonds maintaining the tertiary structure of the enzyme. This causes denaturation, changing the shape of the active site so the substrate can no longer bind. [2]

8.
(a) Ka=[H+]2[HA]K_a = \frac{[H^+]^2}{[HA]}.
Assumption: [H+]eq≈[H+]initial[H^+]_{eq} \approx [H^+]_{initial} from dissociation, and [HA]eq≈[HA]initial[HA]_{eq} \approx [HA]_{initial} (dissociation is small).
[H+]=Ka×[HA]=1.3×10−5×0.10=1.3×10−6=1.14×10−3[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.3 \times 10^{-5} \times 0.10} = \sqrt{1.3 \times 10^{-6}} = 1.14 \times 10^{-3} mol dm⁻³.
pH=−log⁡(1.14×10−3)=2.94pH = -\log(1.14 \times 10^{-3}) = 2.94. [3]
(b) (i) No change. KaK_a is a constant at a given temperature. [1]
(ii) pH increases (becomes less acidic) because [H⁺] decreases upon dilution. [1]

9.
(a) NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) [1]
(b) Acidic. NH₄Cl dissociates into NH₄⁺ and Cl⁻.
NH₄⁺ is the conjugate acid of a weak base (NH₃) and hydrolyzes:
NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq), producing H⁺ ions.
Cl⁻ is the conjugate base of a strong acid and does not hydrolyze. [2]

10.
(a) Add aqueous NaOH and warm.
Ammonium chloride produces ammonia gas (pungent smell, turns damp red litmus blue).
Sodium chloride shows no reaction/no gas. [2]
(b) Ammonium chloride is thermally unstable and sublimes/decomposes on heating:
NH₄Cl(s) ⇌ NH₃(g) + HCl(g).
Sodium chloride has a giant ionic lattice with strong electrostatic forces, requiring very high temperatures to melt; it does not decompose easily. [2]

11.
(a) Hydroxyl group (-OH) and Carboxyl group (-COOH). [1]
(b) 2CH₃CH(OH)COOH + Na₂CO₃ → 2CH₃CH(OH)COONa + H₂O + CO₂ [2]
(c) (i) Accumulation of acid lowers pH. This disrupts the tertiary structure of the enzyme (denaturation) by breaking ionic/hydrogen bonds, altering the active site shape. [2]
(ii) Calcium hydroxide is a base. It neutralizes the lactic acid, preventing the pH from dropping too low and inhibiting the enzymes. [1]

12.
(a) HCl is a strong acid and fully dissociates, giving a high [H⁺]. Ethanoic acid is weak and partially dissociates, giving a lower [H⁺] and thus higher pH. [2]
(b) The chlorine atom is electronegative. It exerts an electron-withdrawing inductive effect (-I effect). This withdraws electron density from the O-H bond in the carboxyl group, weakening it and making the H⁺ ion easier to lose. It also stabilizes the resulting carboxylate anion by dispersing the negative charge. [3]

13.
(a) Giant ionic lattice. Strong electrostatic forces of attraction between Mg²⁺ and O²⁻ ions. [2]
(b) Giant covalent (macromolecular) structure. Strong covalent bonds between Si and O atoms throughout the lattice. [2]
(c) MgO has a higher melting point. The electrostatic forces in the ionic lattice of MgO (charges +2/-2) are stronger than the covalent bonds in SiO₂?
Correction/Refinement for H1: Actually, comparing lattice energy vs bond energy is complex. Standard A-Level answer: MgO has very high lattice energy due to +2/-2 charges. SiO₂ has strong covalent bonds. MgO mp ~2800°C, SiO₂ ~1700°C.
Explanation: The electrostatic attraction between Mg²⁺ and O²⁻ is extremely strong due to the high charge density, requiring more energy to overcome than the covalent network breaking in SiO₂. [2]

14.
(a) Moles acid = 0.10×501000=0.0050.10 \times \frac{50}{1000} = 0.005 mol.
Moles salt = 0.10×501000=0.0050.10 \times \frac{50}{1000} = 0.005 mol. [1]
(b) Moles HCl added = 0.10×101000=0.0010.10 \times \frac{10}{1000} = 0.001 mol.
H⁺ reacts with CH₃COO⁻:
New moles CH₃COO⁻ = 0.005−0.001=0.0040.005 - 0.001 = 0.004 mol.
New moles CH₃COOH = 0.005+0.001=0.0060.005 + 0.001 = 0.006 mol.
Total volume = 110 cm³ (cancels out in ratio).
pH=pKa+log⁡(0.0040.006)pH = pK_a + \log \left( \frac{0.004}{0.006} \right).
pKa=4.77pK_a = 4.77.
pH=4.77+log⁡(0.667)=4.77−0.176=4.59pH = 4.77 + \log(0.667) = 4.77 - 0.176 = 4.59. [4]

15.
(a) AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) [1]
(b) Ag⁺ ions react with NH₃ to form the complex ion [Ag(NH₃)₂]⁺. This reduces [Ag⁺], shifting the solubility equilibrium to the right, dissolving more AgCl. [2]
(c) AgI has a much smaller KspK_{sp} (is much less soluble) than AgCl. The equilibrium concentration of Ag⁺ is too low to form the complex ion with dilute ammonia effectively. [1]

16.
(a) NH₃(g) + HCl(g) → NH₄Cl(s) [1]
(b) Ionic bonding (in the solid lattice) / Coordinate covalent (dative) bond formation within the ammonium ion.
Accept: Ionic. [1]
(c) Acidic. NH₄Cl is a salt of a weak base and strong acid. NH₄⁺ hydrolyzes to produce H⁺. [2]

17.
(a) Sketch:

  • Start pH ~11 (weak base).
  • Gradual drop.
  • Vertical drop at equivalence point.
  • Equivalence point pH < 7 (acidic salt).
  • Levels off at low pH.
    Marks: Shape [1], Equivalence point pH < 7 [1], Buffer region indicated [1]. [3]
    (b) Methyl orange. The equivalence point is in the acidic range (pH 3-5). Methyl orange changes color in this range (3.1-4.4). Phenolphthalein changes in basic range, so it would change color before the equivalence point. [2]

18.
(a) [H+]=Ka×[HA]=5.6×10−4×0.010=5.6×10−6=2.37×10−3[H^+] = \sqrt{K_a \times [HA]} = \sqrt{5.6 \times 10^{-4} \times 0.010} = \sqrt{5.6 \times 10^{-6}} = 2.37 \times 10^{-3}.
pH=−log⁡(2.37×10−3)=2.63pH = -\log(2.37 \times 10^{-3}) = 2.63. [3]
(b) The H-F bond is very strong (short bond length, high bond energy) due to the small size of F and good orbital overlap. This makes it difficult for the H⁺ to dissociate. [1]

19.
(a) NaCl. Derived from strong acid (HCl) and strong base (NaOH). Neither ion hydrolyzes. [1]
(b) CH₃COONa. CH₃COO⁻ hydrolyzes: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Produces OH⁻, making solution alkaline. [2]
(c) NH₄NO₃. NH₄⁺ hydrolyzes: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. Produces H⁺, making solution acidic. [2]

20.
(a) [H+]=10−3=0.001[H^+] = 10^{-3} = 0.001 mol dm⁻³.
Ka=[H+]2[HA]=(0.001)20.10=10−610−1=1.0×10−5K_a = \frac{[H^+]^2}{[HA]} = \frac{(0.001)^2}{0.10} = \frac{10^{-6}}{10^{-1}} = 1.0 \times 10^{-5} mol dm⁻³. [3]
(b) Weaker. KaK_a of HA (1.0×10−51.0 \times 10^{-5}) is smaller than KaK_a of ethanoic acid (1.7×10−51.7 \times 10^{-5}). Smaller KaK_a means less dissociation. [1]