A Level H1 Chemistry Practice Paper 5

TuitionGoWhere Practice Paper - Chemistry H1 A-Level

Answer Key and Marking Scheme (Version 5)

Section A: Structured Questions

1.
(a) A weak acid is an acid that partially dissociates (or ionizes) in water. [1]
(b) $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$ [1]
(Must use reversible arrow $\rightleftharpoons$ and state symbols)
(c) Ethanoic acid forms dimers via strong hydrogen bonds between two molecules (due to the carbonyl and hydroxyl group). 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]
(1 mark for H-bonds/dimers, 1 mark for comparison/energy)

2.
(a) Sketch: Starts at high pH (~13), gradual drop, steep vertical drop at 25.0 cm³ through pH 7, ends at low pH (~1). Equivalence point labeled at pH 7 and 25.0 cm³. [2]
(b) Indicator: Methyl Orange or Bromothymol Blue or Phenolphthalein (any suitable for strong acid-strong base). [1]
Colour change: e.g., Yellow to Red (Methyl Orange) or Pink to Colourless (Phenolphthalein). [1]
(c) Remain the same. [1] The stoichiometry of the neutralization reaction ($1:1$ mole ratio) depends on the number of moles of acid and base, not the strength of the acid. Since concentration and volume are the same, moles are the same. [1]

3.
(a) $[H^+] = K_a \times \frac{[acid]}{[salt]}$ [1]
$[H^+] = 1.7 \times 10^{-5} \times \frac{0.20}{0.10} = 3.4 \times 10^{-5}$ mol dm⁻³ [1]
$pH = -\log(3.4 \times 10^{-5}) = 4.47$ [1]
(b) The added $H^+$ ions react with the conjugate base ($CH_3COO^-$) to form undissociated ethanoic acid ($CH_3COOH$). [1] This removes most of the added $H^+$, keeping the pH relatively constant. [1]

4.
(a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1]
(b) Let solubility be $s$ mol dm⁻³.
$[Mg^{2+}] = s$, $[OH^-] = 2s$ [1]
$K_{sp} = (s)(2s)^2 = 4s^3$ [1]
$1.8 \times 10^{-11} = 4s^3 \Rightarrow s^3 = 4.5 \times 10^{-12}$
$s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4}$ mol dm⁻³ [1]
(c) Common ion effect. [1] Adding $NaOH$ increases $[OH^-]$. To maintain constant $K_{sp}$, the equilibrium shifts to the left (precipitate forms), decreasing the solubility of $Mg(OH)_2$. [1]

5.
(a) An amphoteric substance can act as both an acid and a base. [1]
(b) (i) $Al_2O_3(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2O(l)$ [2]
(ii) $Al_2O_3(s) + 2NaOH(aq) + 3H_2O(l) \rightarrow 2Na[Al(OH)_4](aq)$ [2]
(Accept $NaAlO_2$ if balanced correctly, but tetrahydroxoaluminate is preferred in aqueous context)

6.
(a) Methyl propanoate. [1]
Structure: $CH_3CH_2COOCH_3$ (Displayed formula showing C=O and O-CH3 linkage). [1]
(b) Remove water (distillation) or use excess alcohol/acid. [1]
(c) $K_a = 10^{-pK_a} = 10^{-4.87} = 1.35 \times 10^{-5}$ mol dm⁻³. [1]

7.
(a) Pair 1: $NH_3$ (base) and $NH_4^+$ (acid). [1]
Pair 2: $H_2O$ (acid) and $OH^-$ (base). [1]
(b) Ammonia accepts a proton ($H^+$) from water. [1]

8.
(a) $[H^+] = 2 \times 0.050 = 0.10$ mol dm⁻³. [1]
$pH = -\log(0.10) = 1.0$. [1]
(b) Higher. [1] The second dissociation is incomplete, so the actual $[H^+]$ is less than 0.10 mol dm⁻³. Lower $[H^+]$ means higher pH. [1]

9.
(a) The chlorine atom is electronegative and exerts an electron-withdrawing inductive effect. [1] This withdraws electron density from the carboxylate group, stabilizing the negative charge on the conjugate base ($CH_2ClCOO^-$). [1] A more stable conjugate base means the acid dissociates more readily, making it stronger. [1]
(b) Larger. [1]

10.
(a) $CaCO_3(s) + 2H^+(aq) \rightarrow Ca^{2+}(aq) + H_2O(l) + CO_2(g)$ [2]
(1 mark for correct species, 1 mark for balancing/states)
(b) 1. Effervescence / Bubbles of gas. [1]
2. Solid calcium carbonate dissolves / disappears. [1]

Section B: Data Analysis and Application

11.
(a) pH = 6.4 (This is $pK_{a1}$). [1]
(b) $HCO_3^-$ (Hydrogencarbonate ion). [1]
(c) $K_{a2} = \frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}$ [1]

12.
(a) Moles NaOH = $0.100 \times \frac{20.0}{1000} = 0.0020$ mol. [1]
Moles HA = 0.0020 mol (1:1 ratio).
$[HA] = \frac{0.0020}{0.025} = 0.080$ mol dm⁻³. [1]
(b) At half-equivalence, $pH = pK_a$. [1]
$pK_a = 4.75 \Rightarrow K_a = 10^{-4.75} = 1.78 \times 10^{-5}$ mol dm⁻³. [1]
(c) At equivalence, we have a solution of NaA (salt of weak acid/strong base).
Total volume = 45.0 cm³.
$[A^-] = \frac{0.0020}{0.045} = 0.0444$ mol dm⁻³.
Hydrolysis: $A^- + H_2O \rightleftharpoons HA + OH^-$
$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} = 5.62 \times 10^{-10}$.
$[OH^-] = \sqrt{K_b \times [A^-]} = \sqrt{5.62 \times 10^{-10} \times 0.0444} = 4.99 \times 10^{-6}$.
$pOH = -\log(4.99 \times 10^{-6}) = 5.30$.
$pH = 14 - 5.30 = 8.70$. [3]
(1 mark for salt conc, 1 mark for Kb or expression, 1 mark for final pH)

13.
(a) Giant ionic lattice. [1] Strong electrostatic forces of attraction between $Mg^{2+}$ and $O^{2-}$ ions. [1]
(b) $MgO$ is a giant ionic structure with strong electrostatic forces requiring much energy to break. [1] $P_4O_{10}$ is a simple molecular structure. [1] The intermolecular forces (van der Waals) between $P_4O_{10}$ molecules are weak and require little energy to overcome. [1]

14.
(a) $K_{sp} = s^2 \Rightarrow s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}$ mol dm⁻³. [2]
(b) In 0.10 M NaCl, $[Cl^-] = 0.10$ M.
$K_{sp} = [Ag^+][Cl^-] \Rightarrow 1.8 \times 10^{-10} = [Ag^+](0.10)$.
$[Ag^+] = 1.8 \times 10^{-9}$ mol dm⁻³.
Solubility = $1.8 \times 10^{-9}$ mol dm⁻³. [3]
(1 mark for expression, 1 mark for substitution, 1 mark for answer)
(c) Common ion effect reduces solubility. [1]

15.
(a) $pH = pK_a + \log \left( \frac{[salt]}{[acid]} \right)$
$5.00 = 4.76 + \log \left( \frac{[salt]}{[acid]} \right)$
$0.24 = \log \left( \frac{[salt]}{[acid]} \right)$
Ratio = $10^{0.24} = 1.74$. [3]
(b) Mixing equal volumes of equal concentration gives a ratio of 1:1.
$\log(1) = 0$, so $pH = pK_a = 4.76$.
4.76 is lower than 5.00. [1]

Section C: Extended Response

16.
Marking Points:

1. Strength Definition: Stronger acids have larger $K_a$ / lower $pK_a$ / dissociate more.
2. Chloroethanoic vs Ethanoic: Cl is electronegative, withdraws electrons (inductive effect), stabilizes conjugate base ($COO^-$), makes acid stronger.
3. Phenol vs Ethanoic: Phenol is weaker. The phenoxide ion is stabilized by resonance delocalization into the ring, but the charge is delocalized over carbon atoms (less electronegative than oxygen in carboxylate). In ethanoate, negative charge is delocalized over two oxygen atoms (more stable). Also, O-H bond in phenol is stronger/harder to break than in carboxylic acid due to partial double bond character with ring.
4. Comparison: Chloroethanoic > Ethanoic > Phenol.
   [6 marks: 2 for Cl effect, 2 for Phenol/Ethanoic comparison, 1 for order, 1 for clarity/coherence]

17.
Marking Points:

1. Dynamic Equilibrium: Rate of forward reaction (dissociation) equals rate of reverse reaction (recombination). Concentrations of species remain constant.
2. Le Chatelier: $HA \rightleftharpoons H^+ + A^-$.
3. Adding Water: Dilution decreases concentration of all species. System shifts to the side with more particles (right) to oppose the change.
4. Result: Degree of dissociation ($\alpha$) increases, but $[H^+]$ decreases overall (so pH increases).
   [4 marks: 1 for definition, 1 for principle application, 1 for direction of shift, 1 for outcome]

18.
Marking Points:

1. Procedure: Pipette known volume of weak acid into beaker. Add NaOH from burette in small increments. Measure pH with calibrated pH meter after each addition. Stir.
2. Plotting: Plot pH (y-axis) vs Volume NaOH (x-axis).
3. Analysis: Identify equivalence point (steepest part of curve). Determine volume at equivalence ($V_{eq}$).
4. Finding pKa: Find volume $\frac{1}{2} V_{eq}$. Read pH at this volume. $pH = pK_a$ at half-equivalence.
   [5 marks: 1 for setup, 1 for measurement, 1 for plot, 1 for eq point, 1 for half-eq method]

19.
(a) $NH_4^+$ is a weak acid. $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$. Produces $H_3O^+$, so acidic. [2]
(b) $CH_3COO^-$ is a weak base. $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$. Produces $OH^-$, so alkaline. [2]
(c) pH $\approx$ 7 (Neutral). [1] Since $K_a \approx K_b$, the extent of acid hydrolysis of $NH_4^+$ equals the extent of base hydrolysis of $CH_3COO^-$. $[H^+] \approx [OH^-]$. [1]

20.
(a) Propanoic < Ethanoic < Methanoic. [1]
(b) Alkyl groups (methyl, ethyl) are electron-releasing (positive inductive effect). [1] This destabilizes the carboxylate anion by intensifying the negative charge, making dissociation harder. Longer chain = slightly more releasing, but effect is small; H in methanoic has no releasing effect, so it is strongest. [1]
(c) $[H^+] = \sqrt{K_a \times [acid]} = \sqrt{1.8 \times 10^{-4} \times 0.10} = \sqrt{1.8 \times 10^{-5}} = 4.24 \times 10^{-3}$.
$pH = -\log(4.24 \times 10^{-3}) = 2.37$. [3]