A Level H2 Chemistry Practice Paper 2

TuitionGoWhere Practice Paper - Chemistry H2 A-Level (Answers)

TuitionGoWhere Practice Paper (AI)

Subject: Chemistry H2
Level: A-Level
Paper: Practice Paper (Version 2 of 5)
Topic Focus: Acids, Bases & 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 ethanoic acid is small, so $[CH_3COOH]_{eq} \approx [CH_3COOH]_{initial}$. Also, $[H^+] \approx [CH_3COO^-]$. $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \approx \frac{[H^+]^2}{0.10}$ $[H^+]^2 = 1.7 \times 10^{-5} \times 0.10 = 1.7 \times 10^{-6}$ $[H^+] = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}$ $\text{pH} = -\log_{10}(1.30 \times 10^{-3}) = 2.89$ [3]

(c) Moles of $CH_3COOH$ = $0.10 \times 0.500 = 0.050 \text{ mol}$. Moles of $CH_3COO^-$ added = $0.05 \text{ mol}$. Using the Henderson-Hasselbalch equation or $K_a$ expression: $[H^+] = K_a \times \frac{[Acid]}{[Salt]}$ Since the volume is the same for both, the ratio of concentrations equals the ratio of moles. $[H^+] = 1.7 \times 10^{-5} \times \frac{0.050}{0.05} = 1.7 \times 10^{-5} \text{ mol dm}^{-3}$ $\text{pH} = -\log_{10}(1.7 \times 10^{-5}) = 4.77$ [3]

(d) The buffer contains significant amounts of weak acid ($CH_3COOH$) and its conjugate base ($CH_3COO^-$). When $OH^-$ is added, it reacts with the weak acid: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$ This removes the added $OH^-$, preventing a significant rise in pH. [2]

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

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

(c) In $0.10 \text{ mol dm}^{-3}$ NaOH, $[OH^-] = 0.10 \text{ mol dm}^{-3}$. Let $s'$ be the new solubility. $[Mg^{2+}] = s', \quad [OH^-] \approx 0.10 \text{ (since } 2s' \ll 0.10)$ $K_{sp} = [Mg^{2+}][OH^-]^2$ $1.8 \times 10^{-11} = (s')(0.10)^2$ $s' = \frac{1.8 \times 10^{-11}}{0.01} = 1.8 \times 10^{-9} \text{ mol dm}^{-3}$ [3]

(d) This is due to the common ion effect. The presence of a high concentration of $OH^-$ ions from the strong base NaOH shifts the equilibrium position of the dissolution of $Mg(OH)_2$ to the left (Le Chatelier's Principle), thereby decreasing its solubility. [2]

3 (a) Name: Methyl propanoate Structure: $CH_3CH_2-C(=O)-O-CH_3$ [2]

(b) Let $x$ be the moles of ester formed at equilibrium. Initial: Acid = 1.0, Alcohol = 1.0, Ester = 0, Water = 0 Equilibrium: Acid = $1.0-x$, Alcohol = $1.0-x$, Ester = $x$, Water = $x$ $K_c = \frac{[Ester][Water]}{[Acid][Alcohol]} = \frac{(x/V)(x/V)}{((1.0-x)/V)((1.0-x)/V)} = \frac{x^2}{(1.0-x)^2}$ $4.0 = \left( \frac{x}{1.0-x} \right)^2$ Taking square root: $2.0 = \frac{x}{1.0-x}$ $2.0(1.0-x) = x \Rightarrow 2.0 - 2.0x = x \Rightarrow 3.0x = 2.0$ $x = \frac{2}{3} = 0.67 \text{ mol}$ [4]

(c) Remove one of the products (e.g., distill off the ester or water) as it forms. According to Le Chatelier’s principle, removing a product shifts the equilibrium position to the right to oppose the change, thus increasing the yield of the ester. [2]

4 (a)

• Start pH: Weak base ($NH_3$), pH $\approx$ 11.
• Equivalence point: Acidic pH (approx 5-6) because the salt formed ($NH_4Cl$) is acidic due to hydrolysis of $NH_4^+$.
• Shape: Gradual decrease, steep drop around equivalence point, then levels off at low pH. [3]

(b) Methyl orange. The equivalence point occurs in the acidic range (pH 3.1 – 4.4 covers the steep part of the curve for Weak Base-Strong Acid titration). Phenolphthalein changes color in the basic range, which is not near the equivalence point. [2]

(c) The titration of ethanoic acid (weak acid) with NaOH (strong base) has an equivalence point in the basic region (pH > 7, typically ~8-9). Methyl orange changes color in the acidic range (3.1-4.4), which is far from the equivalence point, leading to a large titration error. [2]

5 (a) $[Al(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H_3O^+$ [2]

(b) $Al^{3+}$ has a high charge density (small size, high charge), which polarizes the O-H bonds in the coordinated water molecules, weakening them and allowing $H^+$ to be released. $Na^+$ has a low charge density and does not polarize water molecules significantly, so no $H^+$ is released. [2]

(c) Structure of $Al_2Cl_6$: Two Al atoms bridged by two Cl atoms. Each Al is bonded to 4 Cl atoms (2 terminal, 2 bridging). The bridging Cl atoms form dative bonds from their lone pairs to the empty orbitals of the Al atoms. (Diagram should show Cl-Al-Cl bridges). [2]

(d) $AlCl_3$ monomer: $sp^2$ (trigonal planar). $Al_2Cl_6$ dimer: $sp^3$ (tetrahedral around each Al). [2]

6 (a) $H_2A + 2NaOH \rightarrow Na_2A + 2H_2O$ [1]

(b) Moles of NaOH = $0.100 \times \frac{25.0}{1000} = 0.0025 \text{ mol}$. From stoichiometry, moles of $H_2A$ = $\frac{1}{2} \times$ moles of NaOH = $0.00125 \text{ mol}$. Volume of $H_2A$ = $12.5 \text{ cm}^3 = 0.0125 \text{ dm}^3$. Concentration of $H_2A$ = $\frac{0.00125}{0.0125} = 0.100 \text{ mol dm}^{-3}$. [2]

(c) It is harder to remove a positively charged proton ($H^+$) from a negatively charged ion ($HA^-$) than from a neutral molecule ($H_2A$) due to electrostatic attraction. Therefore, the second dissociation is less extensive, resulting in a smaller $K_a$ value. [2]

(d)

• Low pH: $H_2A$ dominant.
• pH $\approx pK_{a1}$ (3): $[H_2A] = [HA^-]$.
• Intermediate pH: $HA^-$ dominant.
• pH $\approx pK_{a2}$ (7): $[HA^-] = [A^{2-}]$.
• High pH: $A^{2-}$ dominant. (Sketch should show two sigmoidal curves crossing at the pKa values). [3]

Section B: Data-Based & Application Questions

7 (a) As we move from Na to Al, the charge on the cation increases (+1 to +3) and the ionic radius decreases. This leads to an increase in charge density. Higher charge density polarizes the surrounding water molecules more strongly, weakening the O-H bonds and facilitating the release of $H^+$ ions, thus lowering the pH. [3]

(b) $SiCl_4 + 2H_2O \rightarrow SiO_2 + 4HCl$ (Or $SiCl_4 + 4H_2O \rightarrow Si(OH)_4 + 4HCl$) The reaction produces HCl, a strong acid, which fully dissociates to give a high concentration of $H^+$ ions, resulting in a very low pH. [2]

(c) $PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl$ [2]

(d) Carbon is a Period 2 element and does not have available d-orbitals in its valence shell to expand its octet. Therefore, it cannot accept a lone pair from a water molecule to initiate hydrolysis. Silicon (Period 3) has empty d-orbitals and can expand its octet, allowing nucleophilic attack by water. [2]

8 (a) Ester group and Carboxylic acid group (though usually ester hydrolysis is the focus for stability, both can hydrolyze under specific conditions. In the context of aspirin stability, the ester group hydrolyzes to salicylic acid and ethanoic acid). Accept: Ester group. [1]

(b) Suitable weak base: Sodium hydrogen carbonate ($NaHCO_3$) or Sodium hydroxide (strong, but forms salt). Better: Magnesium hydroxide or Aluminum hydroxide (antacids). Explanation: The salt (e.g., sodium acetylsalicylate) is ionic. Ionic compounds are generally more soluble in polar solvents like water due to ion-dipole interactions compared to the covalent aspirin molecule which relies on weaker intermolecular forces. [2]

(c) $pH = pK_a + \log_{10} \left( \frac{[A^-]}{[HA]} \right)$ $1.5 = 3.5 + \log_{10} \left( \frac{[A^-]}{[HA]} \right)$ $-2.0 = \log_{10} \left( \frac{[A^-]}{[HA]} \right)$ $\frac{[A^-]}{[HA]} = 10^{-2} = 0.01$ [2]

(d) Unionized form ($HA$) is dominant (ratio 1:100). The stomach lining is lipid-based (non-polar). The unionized form is non-polar and lipid-soluble, allowing it to diffuse easily through the cell membranes of the stomach lining into the bloodstream. [2]

9 (a) $^+H_3N-CH_2-COO^-$ [1]

(b) $pI = \frac{pK_{a1} + pK_{a2}}{2} = \frac{2.34 + 9.60}{2} = \frac{11.94}{2} = 5.97$ [1]

(c)

• Start at low pH (fully protonated $^+H_3NCH_2COOH$).
• First buffer region around pH 2.34 ($pK_{a1}$).
• First equivalence point (zwitterion dominant) around pH 5.97.
• Second buffer region around pH 9.60 ($pK_{a2}$).
• Final high pH (fully deprotonated $H_2NCH_2COO^-$).
• Two steep rises in pH. [3]

(d) The adjacent positively charged ammonium group ($-NH_3^+$) in glycine exerts an electron-withdrawing inductive effect. This withdraws electron density from the carboxyl group, weakening the O-H bond and stabilizing the resulting carboxylate anion ($-COO^-$) through electrostatic attraction. This makes the proton easier to lose, lowering the $pK_a$. [2]