A Level H1 Chemistry Practice Paper 4

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

Answer Key and Marking Scheme
Paper: Practice Paper (Version 4 of 5)
Topic: Acids, Bases, and Salts

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 for equation, 1 for state symbols and reversible arrow] (c) Ethanoic acid forms dimers via strong hydrogen bonding between two molecules (two H-bonds per dimer). Ethanol forms hydrogen bonds but not dimers to the same extent. More energy is required to break the intermolecular forces in ethanoic acid. [1 for dimer/H-bonding detail, 1 for comparison of energy/forces]

2. (a) $n(NaOH) = c \times V = 0.050 \times \frac{22.4}{1000} = 1.12 \times 10^{-3} \ mol$ [1] (b) Ratio is 1:1. So $n(acid) = 1.12 \times 10^{-3} \ mol$. $c(acid) = \frac{n}{V} = \frac{1.12 \times 10^{-3}}{0.025} = 0.0448 \ mol \ dm^{-3}$ [1 for moles, 1 for conc] (c) Benzoic acid is a weak acid. In acidic conditions, the equilibrium $C_6H_5COOH \rightleftharpoons C_6H_5COO^- + H^+$ shifts to the left, favoring the uncharged molecular form ($C_6H_5COOH$). The uncharged molecule is more lipid-soluble and can penetrate bacterial cell membranes more effectively than the ionized form. [1 for reference to equilibrium/uncharged form]

3. (a) An amphoteric substance can act as both an acid and a base. [1] (b)(i) $Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l)$ [1 for species, 1 for balancing] (b)(ii) $Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq)$ [1 for species, 1 for balancing] (Note: Accept $Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaAl(OH)_4$)

4. (a) $[H^+] = 10^{-pH} = 10^{-2.38} = 4.17 \times 10^{-3} \ mol \ dm^{-3}$ [1] (b) $K_a = \frac{[HCOO^-][H^+]}{[HCOOH]}$ [1] (c) Assume $[H^+] = [HCOO^-]$ and $[HCOOH]_{eq} \approx [HCOOH]_{initial}$ (since dissociation is small). $K_a = \frac{(4.17 \times 10^{-3})^2}{0.10} = \frac{1.74 \times 10^{-5}}{0.10} = 1.74 \times 10^{-4} \ mol \ dm^{-3}$ [1 for substitution, 1 for answer, 1 for units]

5. (a) The buffer contains high concentrations of $CH_3COOH$ and $CH_3COO^-$. When $H^+$ is added, it reacts with the conjugate base $CH_3COO^-$ to form $CH_3COOH$: $CH_3COO^- + H^+ \rightarrow CH_3COOH$. This removes most of the added $H^+$, keeping the pH relatively constant. [1 for identifying reaction with conjugate base, 1 for equation, 1 for explanation of pH stability] (b) $pH = pK_a + \log_{10} \left( \frac{[salt]}{[acid]} \right)$ $pK_a = -\log(1.7 \times 10^{-5}) = 4.77$ $pH = 4.77 + \log_{10} \left( \frac{0.20}{0.10} \right) = 4.77 + \log_{10}(2) = 4.77 + 0.30 = 5.07$ [1 for pKa, 1 for substitution, 1 for final answer]

Section B: Data Interpretation and Application

6. (a) $HCO_3^-$ (Hydrogencarbonate ion) [1] (b) $pH = pK_a + \log_{10} \left( \frac{[HCO_3^-]}{[H_2CO_3]} \right)$ $pH = 6.35 + \log_{10} \left( \frac{2.4 \times 10^{-2}}{1.2 \times 10^{-3}} \right) = 6.35 + \log_{10}(20) = 6.35 + 1.30 = 7.65$ [1 for formula, 1 for substitution, 1 for answer] (c) Loss of $CO_2$ shifts the equilibrium $CO_2 + H_2O \rightleftharpoons H^+ + HCO_3^-$ to the left to restore $CO_2$. This consumes $H^+$ ions, causing the pH to increase (become more alkaline). [1 for shift direction, 1 for effect on pH]

7. (a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1] (b) Let solubility be $s \ mol \ dm^{-3}$. Then $[Mg^{2+}] = s$ and $[OH^-] = 2s$. $K_{sp} = (s)(2s)^2 = 4s^3$ $2.0 \times 10^{-11} = 4s^3$ $s^3 = 5.0 \times 10^{-12}$ $s = \sqrt[3]{5.0 \times 10^{-12}} = 1.71 \times 10^{-4} \ mol \ dm^{-3}$ [1 for expression in s, 1 for calculation, 1 for answer] (c) Common ion effect. Adding $NaOH$ increases $[OH^-]$. To maintain constant $K_{sp}$, the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ shifts to the left, precipitating more solid and decreasing solubility. [1 for common ion effect, 1 for shift explanation]

8. (a) Propyl propanoate [1] (b) 1. Use excess alcohol or acid. 2. Remove water/ester as it forms (distillation). [1 for each] (c) The ethyl group ($C_2H_5-$) in propanoic acid is more electron-releasing (positive inductive effect) than the methyl group ($CH_3-$) in ethanoic acid. This increases electron density on the carboxylate anion, destabilizing it slightly and making the O-H bond harder to break, thus making it a weaker acid. [1 for inductive effect comparison, 1 for link to stability/acidity]

9. (a) $SiO_2$ (Silicon dioxide). It has a giant covalent structure with strong covalent bonds throughout the lattice requiring large amounts of energy to break. The others are simple molecular (weak van der Waals) or ionic (strong but generally lower MP than giant covalent network like silica). [1 for SiO2, 1 for giant covalent, 1 for strong bonds] (b) $P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$ [1] (c) Acidic. $SO_2 + H_2O \rightleftharpoons H_2SO_3$ (sulfurous acid). The solution contains $H^+$ ions. [1 for acidic, 1 for equation]

10. (a) Carboxylic acid [1] (b) e.g., $CH_3CH_2CH_2COOH$ (Butanoic acid) or $(CH_3)_2CHCOOH$ (2-methylpropanoic acid) [1] (c) Ester [1]

Section C: Extended Response and Synthesis

11. (a) $K_a = \frac{[H^+]^2}{[HA]}$ $[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.4 \times 10^{-4} \times 0.050} = \sqrt{7.0 \times 10^{-6}} = 2.65 \times 10^{-3} \ mol \ dm^{-3}$ $pH = -\log(2.65 \times 10^{-3}) = 2.58$ [1 for formula rearrangement, 1 for [H+], 1 for pH, 1 for sig figs/units] (b) Assumption: The dissociation of water is negligible / $[H^+]_{eq} \approx [A^-]_{eq}$. Or: The change in $[HA]$ is negligible ($[HA]_{eq} \approx [HA]_{initial}$). Justification: $K_a$ is small ($10^{-4}$), so the degree of dissociation is low (<5%). [1 for assumption, 1 for justification]

12. (a) Ethanoic < Chloroethanoic < Fluoroethanoic [1] (b) Chlorine is more electronegative than hydrogen. It exerts a negative inductive effect (-I), withdrawing electron density from the carboxylate group. This disperses the negative charge on the conjugate base ($ClCH_2COO^-$), stabilizing it. A more stable conjugate base means the acid is stronger. [1 for electronegativity, 1 for inductive effect, 1 for stabilization] (c) Fluorine is more electronegative than chlorine. It has a stronger -I effect, stabilizing the conjugate base more effectively. Thus, fluoroethanoic acid is stronger. [1 for prediction, 1 for reasoning]

13. (a) $pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)$ $4.00 = 4.76 + \log \left( \frac{[A^-]}{[HA]} \right)$ $\log \left( \frac{[A^-]}{[HA]} \right) = -0.76$ $\frac{[A^-]}{[HA]} = 10^{-0.76} = 0.174$ [1 for substitution, 1 for log calc, 1 for ratio] (b) Mixing equal volumes of equal concentration gives a ratio of 1:1. $pH = pK_a + \log(1) = 4.76$. 4.76 is higher than 4.00. [1 for identifying pH 4.76, 1 for comparison]

14. (a) $MgO$: Blue/Purple (Alkaline). $SO_2$: Red/Orange (Acidic). [1 for each] (b) Mg is a metal with low electronegativity. The Mg-O bond is ionic. $O^{2-}$ ions react with water to form $OH^-$. S is a non-metal with high electronegativity. The S-O bonds are covalent and polar. $SO_2$ reacts with water to form covalent bonds with O, releasing $H^+$. The high oxidation state and electronegativity of S draw electron density, weakening O-H bonds in the resulting oxoacid. [1 for Mg ionic/basic, 1 for S covalent/acidic, 1 for electronegativity/polarity link, 1 for mechanism]

15. (a) Benzene ring with -COOH at position 1 and -OCOCH3 at position 2. [1 for ring/substituents, 1 for correct positions] (b) Products: Sodium salicylate (2-hydroxybenzoate ion) and Sodium ethanoate (acetate ion). Structures: Benzene ring with -COO- Na+ and -OH; and CH3COO- Na+. [1 for salicylate, 1 for ethanoate, 1 for correct ionic forms] (c) In NaOH, the carboxylic acid group reacts to form a soluble ionic salt ($COO^- Na^+$). Ionic compounds are generally more soluble in polar solvents like water due to ion-dipole interactions. [1 for salt formation, 1 for solubility explanation]

16. (a) $K_{sp} = [Ca^{2+}][SO_4^{2-}]$. In pure water, $[Ca^{2+}] = [SO_4^{2-}] = s$. $s^2 = 2.4 \times 10^{-5}$ $s = \sqrt{2.4 \times 10^{-5}} = 4.90 \times 10^{-3} \ mol \ dm^{-3}$ [1 for expression, 1 for answer] (b) Total volume = 200 $cm^3$. Concentrations halve. $[Ca^{2+}] = 0.005 \ mol \ dm^{-3}$ $[SO_4^{2-}] = 0.005 \ mol \ dm^{-3}$ Ionic Product (IP) = $0.005 \times 0.005 = 2.5 \times 10^{-5}$ $IP (2.5 \times 10^{-5}) > K_{sp} (2.4 \times 10^{-5})$. Yes, a precipitate will form. [1 for new concs, 1 for IP calc, 1 for comparison, 1 for conclusion]

17. (a) $H_2NCH_2COO^-$ [1] (b) $^+H_3NCH_2COOH$ [1] (c) Glycine exists as a zwitterion ($^+H_3NCH_2COO^-$) in the solid state. This results in strong electrostatic forces (ionic bonding) between ions in the lattice. Ethanoic acid exists as simple molecules held by weaker hydrogen bonds. Ionic forces are stronger than H-bonds. [1 for zwitterion/ionic nature, 1 for comparison of forces]

18. (a) $HNO_2(aq) \rightleftharpoons H^+(aq) + NO_2^-(aq)$ [1] (b) Nitric acid is strong (fully dissociated), so $[HNO_3] = [H^+] = 0.1 \ mol \ dm^{-3}$. Nitrous acid is weak (partially dissociated), so to achieve the same $[H^+]$ of 0.1, the initial concentration of $HNO_2$ must be much higher than 0.1 $mol \ dm^{-3}$. [1 for strong/weak distinction, 1 for conc comparison] (c) In $HNO_2$: H is +1, O is -2. $1 + N + 2(-2) = 0 \Rightarrow N = +3$. In $NO$: O is -2. $N - 2 = 0 \Rightarrow N = +2$. [1 for each]

19. (a) $HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)$ [1] (b) At low pH (high $[H^+]$), equilibrium shifts left, favoring $HIn$ (yellow). As pH increases, $[H^+]$ decreases, equilibrium shifts right, favoring $In^-$ (blue). At $pH = pK_{In} = 7$, $[HIn] = [In^-]$ (green/mix). [1 for shift left/right, 1 for species color, 1 for equilibrium principle] (c) No. The titration of a weak acid (ethanoic) with a strong base (NaOH) has an equivalence point at pH > 7 (approx 8-9). Bromothymol blue changes color at pH 7, which is before the equivalence point, leading to an inaccurate endpoint. Phenolphthalein (range 8-10) would be better. [1 for No, 1 for explanation of eq point pH]

20. (a) 1. $H_2A \rightleftharpoons H^+ + HA^-$ 2. $HA^- \rightleftharpoons H^+ + A^{2-}$ [1 for each] (b) It is harder to remove a positive proton ($H^+$) from a negatively charged ion ($HA^-$) than from a neutral molecule ($H_2A$) due to electrostatic attraction. [1 for charge argument, 1 for electrostatic attraction] (c) Use $K_{a1}$. $[H^+] = \sqrt{K_{a1} \times [H_2A]} = \sqrt{1.0 \times 10^{-3} \times 0.10} = \sqrt{1.0 \times 10^{-4}} = 1.0 \times 10^{-2} \ mol \ dm^{-3}$ $pH = -\log(1.0 \times 10^{-2}) = 2.00$ [1 for formula, 1 for [H+], 1 for pH]