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A Level H2 Chemistry Practice Paper 2

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Chemistry H2
Level: A-Level
Paper: Practice Paper (Version 2 of 5)
Topic Focus: Acids, Bases & Salts
Duration: 1 hour 15 minutes
Total Marks: 60
Name: __________________________
Class: __________________________
Date: __________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
Write your answers in the spaces provided on this question paper.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a scientific calculator.
A Data Booklet is provided.

Section A: Structured Questions

Answer all questions in this section.

1 Ethanoic acid, $CH_3COOH$ , is a weak acid with a $K_a$ value of $1.7 \times 10^{-5} \text{ mol dm}^{-3}$ at 298 K.

(a) Define the term pH. [1]

(b) Calculate the pH of a $0.10 \text{ mol dm}^{-3}$ solution of ethanoic acid. State any assumptions made in your calculation. [3]

(c) A student adds $0.05 \text{ mol}$ of sodium ethanoate, $CH_3COONa$ , to $500 \text{ cm}^3$ of the $0.10 \text{ mol dm}^{-3}$ ethanoic acid solution from part (b). Calculate the new pH of the resulting buffer solution. [3]

(d) Explain, with the aid of an equation, how this buffer solution resists a change in pH when a small amount of strong base ( $OH^-$ ) is added. [2]

...... [10]

2 The solubility product constant, $K_{sp}$ , for magnesium hydroxide, $Mg(OH)_2$ , is $1.8 \times 10^{-11} \text{ mol}^3 \text{ dm}^{-9}$ at 298 K.

(a) Write the expression for the solubility product, $K_{sp}$ , of $Mg(OH)_2$ . [1]

(b) Calculate the solubility of $Mg(OH)_2$ in pure water in $\text{mol dm}^{-3}$ . [2]

(d) Explain why the solubility of $Mg(OH)_2$ decreases in the presence of $NaOH$ compared to pure water. [2]

3 Propanoic acid ( $CH_3CH_2COOH$ ) reacts with methanol ( $CH_3OH$ ) in the presence of an acid catalyst to form an ester.

(a) Name the ester formed and draw its structural formula. [2]

(b) The equilibrium constant, $K_c$ , for this esterification reaction is 4.0 at a specific temperature. $CH_3CH_2COOH + CH_3OH \rightleftharpoons CH_3CH_2COOCH_3 + H_2O$ If 1.0 mol of propanoic acid and 1.0 mol of methanol are mixed and allowed to reach equilibrium, calculate the amount (in moles) of ester present at equilibrium. [4]

(c) Suggest one method, other than adding more reactants, to increase the yield of the ester. Explain your answer using Le Chatelier’s principle. [2]

4 Consider the following indicators and their pH ranges:

Methyl orange: 3.1 – 4.4
Bromothymol blue: 6.0 – 7.6
Phenolphthalein: 8.3 – 10.0

(a) Sketch the pH curve for the titration of $25.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ ammonia ( $NH_3$ ) with $0.10 \text{ mol dm}^{-3}$ hydrochloric acid ( $HCl$ ). Label the equivalence point. [3]

(b) Select the most suitable indicator for this titration from the list above. Justify your choice. [2]

(c) Explain why methyl orange would not be a suitable indicator for the titration of ethanoic acid with sodium hydroxide. [2]

5 Aluminum chloride ( $AlCl_3$ ) behaves differently in aqueous solution compared to the gas phase.

(a) In aqueous solution, aluminum ions exist as the complex ion $[Al(H_2O)_6]^{3+}$ . This ion acts as a Brønsted-Lowry acid. Write an equation to show how this ion generates acidity in water. [2]

(b) Explain why a solution of $AlCl_3$ is acidic, whereas a solution of $NaCl$ is neutral. [2]

(c) In the gas phase at high temperatures, $AlCl_3$ exists as a dimer, $Al_2Cl_6$ . Draw the structure of $Al_2Cl_6$ , clearly showing the coordinate (dative covalent) bonds. [2]

(d) State the hybridization of the aluminum atom in the monomer $AlCl_3$ and in the dimer $Al_2Cl_6$ . [2]

6 A student performs a titration to determine the concentration of a diprotic acid, $H_2A$ . $25.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ $NaOH$ requires $12.5 \text{ cm}^3$ of the acid $H_2A$ for complete neutralization.

(a) Write the balanced chemical equation for the reaction between $H_2A$ and $NaOH$ . [1]

(b) Calculate the concentration of the acid $H_2A$ in $\text{mol dm}^{-3}$ . [2]

(c) The first dissociation constant $K_{a1}$ of $H_2A$ is $1.0 \times 10^{-3}$ and the second dissociation constant $K_{a2}$ is $1.0 \times 10^{-7}$ . Explain why $K_{a1} > K_{a2}$ . [2]

(d) Sketch the distribution diagram for the species $H_2A$ , $HA^-$ , and $A^{2-}$ as a function of pH. Label the regions where each species is dominant. [3]

Section B: Data-Based & Application Questions

Answer all questions in this section.

7 The table below shows the pH values of $0.1 \text{ mol dm}^{-3}$ solutions of various chlorides.

Chloride	pH of $0.1 \text{ mol dm}^{-3}$ solution
$NaCl$	7.0
$MgCl_2$	6.5
$AlCl_3$	3.0
$SiCl_4$	< 1.0 (reacts violently)
$PCl_5$	< 1.0 (reacts violently)

(a) Explain the trend in pH from $NaCl$ to $AlCl_3$ . Refer to the charge density of the cations in your answer. [3]

(b) Write the equation for the reaction of $SiCl_4$ with water. Explain why the resulting solution is strongly acidic. [2]

(c) $PCl_5$ reacts with water to form phosphoric acid ( $H_3PO_4$ ) and hydrogen chloride. Write the balanced equation for this reaction. [2]

(d) Why does $CCl_4$ not react with water, despite carbon being in the same group as silicon? [2]

8 Aspirin (acetylsalicylic acid) is a weak acid with the formula $C_9H_8O_4$ . It contains a carboxylic acid group and an ester group.

(a) Identify the functional groups in aspirin that can undergo hydrolysis. [1]

(b) Aspirin is often formulated with a buffer to reduce stomach irritation. Suggest a suitable weak base that could be used to form a salt with aspirin, and explain why this salt is more soluble in water than pure aspirin. [2]

(c) The $pK_a$ of the carboxylic acid group in aspirin is 3.5. Calculate the ratio of $[A^-]/[HA]$ in the stomach at pH 1.5. [2]

(d) Based on your answer in (c), is aspirin primarily in its ionized or unionized form in the stomach? Explain how this affects its absorption through the stomach lining (which is lipid-based). [2]

9 The amino acid glycine ( $H_2NCH_2COOH$ ) exists as a zwitterion in solid state and in neutral aqueous solution.

(a) Draw the structure of the zwitterion of glycine. [1]

(b) Glycine has two $pK_a$ values: $pK_{a1} = 2.34$ (carboxyl group) and $pK_{a2} = 9.60$ (amino group). Calculate the isoelectric point (pI) of glycine. [1]

(c) Sketch the titration curve for the addition of $NaOH$ to a solution of glycine hydrochloride ( $^+H_3NCH_2COOH$ ). Label the points corresponding to $pK_{a1}$ , $pK_{a2}$ , and the isoelectric point. [3]

(d) Explain why the $pK_a$ of the carboxyl group in glycine (2.34) is lower than that of ethanoic acid (4.76). [2]

Answers

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]

(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]