A Level H2 Chemistry Practice Paper 4

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key and Marking Scheme (Version 4)

Topic Focus: Acids, Bases and Salts

Section A: Structured Questions

1. (a) pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration. $\text{pH} = -\log_{10}[H^+]$ [1]

(b) Assumption: The dissociation of the acid is small, so $[CH_3COOH]_{eq} \approx [CH_3COOH]_{initial}$. Also, $[H^+] = [CH_3COO^-]$. $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \approx \frac{[H^+]^2}{0.10}$ $[H^+]^2 = 1.7 \times 10^{-5} \times 0.10 = 1.7 \times 10^{-6}$ $[H^+] = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}$ $\text{pH} = -\log(1.30 \times 10^{-3}) = 2.89$ [3] (1 mark for expression, 1 mark for calculation of [H+], 1 mark for pH)

(c) HCl is a strong acid and dissociates completely in water, producing a high concentration of $H^+$ ions. Ethanoic acid is a weak acid and dissociates only partially, producing a much lower concentration of $H^+$ ions. Since pH is inversely related to $[H^+]$, the lower $[H^+]$ in ethanoic acid results in a higher pH. [2]

2. (a) In the buffer, $[CH_3COOH] = [CH_3COO^-]$ because equal volumes and concentrations were mixed. $\text{pH} = pK_a + \log\left(\frac{[salt]}{[acid]}\right)$ $\text{pH} = -\log(1.7 \times 10^{-5}) + \log(1)$ $\text{pH} = 4.77 + 0 = 4.77$ [2]

(b) The buffer contains significant amounts of $CH_3COOH$ and $CH_3COO^-$. When $H^+$ (from HCl) is added, it reacts with the conjugate base $CH_3COO^-$: $CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$ This removes the added $H^+$ ions, keeping the pH relatively constant. [3] (1 mark for equation, 1 mark for identifying reacting species, 1 mark for explanation)

(c) Moles of $H^+$ added = $1.0 \times 10^{-3} \text{ dm}^3 \times 1.0 \text{ mol dm}^{-3} = 0.001 \text{ mol}$. Initial moles of $CH_3COOH$ = $0.050 \times 0.20 = 0.010 \text{ mol}$. Initial moles of $CH_3COO^-$ = $0.050 \times 0.20 = 0.010 \text{ mol}$.

After reaction: Moles $CH_3COOH$ = $0.010 + 0.001 = 0.011 \text{ mol}$. Moles $CH_3COO^-$ = $0.010 - 0.001 = 0.009 \text{ mol}$.

$\text{pH} = 4.77 + \log\left(\frac{0.009}{0.011}\right)$ $\text{pH} = 4.77 + \log(0.818)$ $\text{pH} = 4.77 - 0.087 = 4.68$ [3]

3. (a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1]

(b) Let solubility be $s \text{ mol dm}^{-3}$. Then $[Mg^{2+}] = s$ and $[OH^-] = 2s$. $K_{sp} = (s)(2s)^2 = 4s^3$ $1.8 \times 10^{-11} = 4s^3$ $s^3 = 4.5 \times 10^{-12}$ $s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$ [2]

(c) Adding NaOH increases the concentration of $OH^-$ ions. According to Le Chatelier’s principle (or the common ion effect), the position of equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ shifts to the left to reduce the $[OH^-]$. This causes more $Mg(OH)_2$ to precipitate, decreasing its solubility. [2]

4. (a) $K_{In} = \frac{[H^+][In^-]}{[HIn]}$ $[H^+] = K_{In} \frac{[HIn]}{[In^-]}$ Taking $-\log$ of both sides: $\text{pH} = pK_{In} - \log\left(\frac{[HIn]}{[In^-]}\right) = pK_{In} + \log\left(\frac{[In^-]}{[HIn]}\right)$ [2]

(b) For a strong acid-weak base titration, the equivalence point is in the acidic range (pH < 7). An indicator with $pK_{In} = 5.0$ changes colour in the range pH 4–6, which coincides with the steep part of the titration curve. For a weak acid-strong base titration, the equivalence point is in the alkaline range (pH > 7). An indicator with $pK_{In} = 5.0$ would change colour too early (in the buffer region), leading to a large titration error. [3]

5. (a) Methyl propanoate. Structure: $CH_3CH_2COOCH_3$ [2]

(b) The reaction is an equilibrium. To maximize yield, remove one of the products (e.g., distill off the ester or water) or use an excess of one reactant (usually the cheaper alcohol). This shifts the equilibrium position to the right. [2]

(c) Excess methanol shifts the position of equilibrium to the right (increasing yield). However, the value of the equilibrium constant, $K_c$, remains unchanged as it is only dependent on temperature. [2]

Section B: Data-Based and Application Questions

6. (a) HCl is a strong acid, fully dissociated, so $[H^+] = 0.1 \text{ M}$, pH = 1. Ethanoic acid is weak, partially dissociated, so $[H^+] < 0.1 \text{ M}$, pH > 1. [2]

(b) $CH_3COOH + NH_3 \rightleftharpoons CH_3COO^- + NH_4^+$ Conjugate pairs: $CH_3COOH$ (acid) / $CH_3COO^-$ (conjugate base) $NH_3$ (base) / $NH_4^+$ (conjugate acid) [3]

(c) Sketch:

• Start pH ~2.9.
• Gradual rise (buffer region).
• Vertical jump at equivalence point (pH ~8-9, since weak acid + strong base... wait, Q6 says titration of B (weak acid) with A (strong acid)? No, Q6(b) reacts B and C. Q6(c) says titration of B with A. B is weak acid, A is strong acid. You cannot titrate an acid with an acid. Correction in question interpretation: The question asks for titration of Solution B (Weak Acid) with Solution D (Strong Base, NaOH) usually, or Solution A is Strong Acid. Let's assume the question meant titration of Weak Acid (B) with Strong Base (D) as is standard, OR Weak Base (C) with Strong Acid (A). Re-reading Q6(c): "titration of ... Solution B with Solution A". This is chemically invalid (Acid + Acid). Assumption for Marking: The question likely intended Solution B (Weak Acid) with Solution D (Strong Base) OR Solution C (Weak Base) with Solution A (Strong Acid). Given the context of Q6(b) reacting B and C, let's assume the standard exam pattern: Titration of Weak Acid (B) with Strong Base (NaOH, D). If strictly following text "B with A": No reaction/no curve. Let's assume standard "Weak Acid vs Strong Base" for marking purposes as per syllabus templates:
• Start pH ~2.9.
• Equivalence point pH > 7 (approx 8.5).
• Vertical section around pH 7-10.
• Buffer region at half-equivalence (pH = pKa). [3]

7. (a) A Lewis acid is an electron pair acceptor. [1]

(b) Diagram: Al in center, 4 Cl around. One Cl has a coordinate bond (arrow from Cl lone pair to Al). Al has empty octet initially, fills it. Charge -1 on complex. [2]

(c) $[Al(H_2O)_6]^{3+}(aq) + H_2O(l) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H_3O^+(aq)$ The high charge density of $Al^{3+}$ polarizes the O-H bonds in coordinated water, facilitating proton release. [2]

(d) $AlCl_3$ is covalent (simple molecular) with weak van der Waals forces between molecules. $NaCl$ is ionic with strong electrostatic forces between ions requiring much more energy to break. [2]

8. (a) At half-equivalence, $\text{pH} = pK_a$. Therefore, $pK_a = 4.8$. [1]

(b) At equivalence, moles acid = moles base. Moles NaOH = $0.025 \text{ dm}^3 \times 0.10 \text{ mol dm}^{-3} = 0.0025 \text{ mol}$. Concentration HA = $0.0025 \text{ mol} / 0.025 \text{ dm}^3 = 0.10 \text{ mol dm}^{-3}$. [2]

(c) At equivalence, we have a solution of the salt $NaA$. Volume total = $50 \text{ cm}^3 = 0.050 \text{ dm}^3$. $[A^-] = 0.0025 \text{ mol} / 0.050 \text{ dm}^3 = 0.05 \text{ mol dm}^{-3}$. Hydrolysis: $A^- + H_2O \rightleftharpoons HA + OH^-$. $K_b = K_w / K_a = 10^{-14} / 10^{-4.8} = 10^{-9.2}$. $[OH^-] = \sqrt{K_b [A^-]} = \sqrt{10^{-9.2} \times 0.05}$. $10^{-9.2} \approx 6.31 \times 10^{-10}$. $[OH^-] = \sqrt{3.15 \times 10^{-11}} = 5.61 \times 10^{-6}$. $\text{pOH} = -\log(5.61 \times 10^{-6}) = 5.25$. $\text{pH} = 14 - 5.25 = 8.75$. [4]

9. (a) Neutral: NaCl Acidic: $NH_4Cl$, $AlCl_3$ Alkaline: $CH_3COONa$ [3]

(b) $NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$ OR $[Al(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H_3O^+$ [2]

(c) $CH_3COO^-$ is the conjugate base of a weak acid. It hydrolyzes to produce $OH^-$: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ This increases $[OH^-]$, making pH > 7. [2]

10. (a) Moles NaOH = $0.020 \times 0.10 = 0.002 \text{ mol}$. Moles Ethanoic Acid = $0.040 \times 0.10 = 0.004 \text{ mol}$. [2]

(b) NaOH is limiting. It reacts completely with 0.002 mol acid. Moles Ethanoic Acid remaining = $0.004 - 0.002 = 0.002 \text{ mol}$. Moles Ethanoate formed = 0.002 mol. [2]

(c) Ratio [Salt]/[Acid] = $0.002/0.002 = 1$. $\text{pH} = pK_a + \log(1) = 4.76$. [2]

Section C: Long Structured Questions

11. (a) Thermal stability increases down the group. As the cation size increases (Mg to Ba), the charge density decreases. This reduces the polarizing power of the cation on the carbonate ion ($CO_3^{2-}$). Less polarization means the C-O bond is less weakened, requiring more energy to decompose. [3]

(b) MgO has a giant ionic lattice structure. The $Mg^{2+}$ and $O^{2-}$ ions have high charges and small radii, resulting in very strong electrostatic forces of attraction. A large amount of energy is required to overcome these forces. [2]

(c) $BaSO_4$ has a very low $K_{sp}$ ($1 \times 10^{-10}$), meaning it is virtually insoluble. The concentration of toxic $Ba^{2+}$ ions in solution is negligible. $BaCO_3$ is more soluble ($K_{sp} \approx 5 \times 10^{-9}$) and reacts with stomach acid (HCl) to form soluble $BaCl_2$, releasing toxic $Ba^{2+}$ ions: $BaCO_3 + 2H^+ \rightarrow Ba^{2+} + H_2O + CO_2$ [4]

12. (a) $H_3N^+CH_2COO^-$ [1]

(b) Glycine exists as a zwitterion with strong electrostatic forces (ionic bonding) between the positive and negative ends of adjacent molecules. Propanoic acid relies on hydrogen bonding and van der Waals forces, which are weaker than ionic interactions. [2]

(c) (i) $H_2NCH_2COOH + H^+ \rightarrow H_3N^+CH_2COOH$ [1] (ii) $H_2NCH_2COOH + OH^- \rightarrow H_2NCH_2COO^- + H_2O$ [1]

(d) $\text{pI} = \frac{pK_{a1} + pK_{a2}}{2} = \frac{2.34 + 9.60}{2} = 5.97$ [2]

13. (a) $K_w = [H^+][OH^-]$ [1]

(b) $[H^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}$. $\text{pH} = 7.0$. [1]

(c) (i) $\text{pH} = 6.8 \Rightarrow [H^+] = 10^{-6.8}$. In pure water, $[H^+] = [OH^-]$. $K_w = (10^{-6.8})^2 = 10^{-13.6} = 2.51 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$. [2] (ii) The dissociation of water is endothermic. Increasing temperature shifts equilibrium to the right, increasing $[H^+]$ and $[OH^-]$. Thus, pH decreases. However, since $[H^+]$ still equals $[OH^-]$, the water remains neutral. [3]

14. (a) To ensure accurate determination of the amount of aspirin, the titrant concentration must be known precisely. [1]

(b) Colourless to pink (or faint pink). [1]

(c) Aspirin is more soluble in ethanol. However, ethanol is a weaker solvent for ionization than water. If too much ethanol is used, the dissociation of aspirin might be suppressed, or the indicator might not function correctly. Ideally, aspirin is dissolved in minimal ethanol and diluted with water. [2]

(d) $CH_3COOC_6H_4COOH + 2NaOH \rightarrow CH_3COONa + HOC_6H_4COONa + H_2O$ (Hydrolysis of ester and neutralization of acid). [2]

15. (a) Chlorine is electronegative and exerts a negative inductive effect (-I effect). This withdraws electron density from the carboxylate group, stabilizing the negative charge on the conjugate base ($RCOO^-$). More stable conjugate base means stronger acid. Two Cl atoms exert a stronger -I effect than one, making dichloroethanoic acid stronger. [3]

(b) $pK_a$ would be lower (approx 0.7). Three Cl atoms exert an even stronger -I effect, further stabilizing the conjugate base and increasing acidity. [2]

(c) $\text{pH} = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$ $3.0 = 2.86 + \log\left(\frac{[A^-]}{[HA]}\right)$ $0.14 = \log\left(\frac{[A^-]}{[HA]}\right)$ Ratio = $10^{0.14} = 1.38$ [2]

16. (a) For $CaCO_3$: $[CO_3^{2-}] = K_{sp} / [Ca^{2+}] = 3.4 \times 10^{-9} / 0.010 = 3.4 \times 10^{-7} \text{ M}$. For $MgCO_3$: $[CO_3^{2-}] = K_{sp} / [Mg^{2+}] = 1.0 \times 10^{-5} / 0.010 = 1.0 \times 10^{-3} \text{ M}$. Since $3.4 \times 10^{-7} < 1.0 \times 10^{-3}$, $CaCO_3$ precipitates first. [3]

(b) The second carbonate ($MgCO_3$) starts to precipitate when $[CO_3^{2-}]$ reaches $1.0 \times 10^{-3} \text{ mol dm}^{-3}$. [2]

(c) At this $[CO_3^{2-}]$, the remaining $[Ca^{2+}]$ is determined by $K_{sp}$ of $CaCO_3$. $[Ca^{2+}] = K_{sp} / [CO_3^{2-}] = 3.4 \times 10^{-9} / 1.0 \times 10^{-3} = 3.4 \times 10^{-6} \text{ mol dm}^{-3}$. [3]

17. (a) Increasing pressure shifts equilibrium to the side with fewer moles of gas (right, 4 moles to 2 moles). Yield of ammonia increases. [2]

(b) The forward reaction is exothermic. Increasing temperature shifts equilibrium to the left (endothermic direction). The value of $K_p$ decreases. [2]

(c) $NH_4^+ + OH^- \rightarrow NH_3 + H_2O$ [1]

(d) $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$. Conjugate acid is $NH_4^+$. [2]

18. (a) $\text{pH} = 11.0 \Rightarrow \text{pOH} = 3.0$. $[OH^-] = 10^{-3} = 0.001 \text{ mol dm}^{-3}$. [1]

(b) $B + H_2O \rightleftharpoons BH^+ + OH^-$ $K_b = \frac{[BH^+][OH^-]}{[B]} \approx \frac{(10^{-3})^2}{0.1} = 10^{-5}$. $pK_b = -\log(10^{-5}) = 5.0$. [3]

(c) The methyl group is electron-releasing (+I effect). This increases the electron density on the nitrogen atom, making the lone pair more available for donation to a proton. Thus, methylamine is a stronger base. [2]

19. (a) Ligands cause the d-orbitals of the transition metal to split into different energy levels. Electrons can absorb visible light to jump from lower to higher d-orbitals (d-d transition). The colour observed is the complementary colour of the light absorbed. [3]

(b) (i) $[Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O$ [1] (ii) The ligand field strength of $Cl^-$ is different from $H_2O$. This changes the energy gap ($\Delta E$) between the split d-orbitals. Consequently, a different wavelength of light is absorbed, resulting in a different observed colour. [2]

20. (a)

1. Add aqueous $Na_2CO_3$. Benzoic acid reacts to form soluble sodium benzoate and releases $CO_2$ (effervescence). Phenol does not react (too weak). $2C_6H_5COOH + Na_2CO_3 \rightarrow 2C_6H_5COONa + H_2O + CO_2$ Separate the aqueous layer (contains benzoate) from the ether layer (contains phenol).
2. Acidify the aqueous layer with HCl to precipitate benzoic acid.
3. To recover phenol, add NaOH to the ether layer (or extract with NaOH). Phenol forms soluble sodium phenoxide. $C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$ Separate aqueous layer, acidify to recover phenol. [4]

(b) In benzoic acid, the negative charge on the carboxylate ion ($C_6H_5COO^-$) is delocalized over the two oxygen atoms and stabilized by resonance with the benzene ring (though less effectively than in phenoxide, the key is the stability of the carboxylate vs phenoxide). Actually, the standard explanation: The benzoate ion is stabilized by resonance delocalization of the negative charge over the two electronegative oxygen atoms. In the phenoxide ion, the negative charge is delocalized into the ring, but onto carbon atoms which are less electronegative than oxygen. Therefore, benzoate is more stable, making benzoic acid a stronger acid. [3]