A Level H2 Chemistry Practice Paper 5

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key and Marking Scheme (Version 5)

Topic: Acids, Bases and Salts
Total Marks: 60

Section A: Structured Questions

1 (a) $\text{pH} = -\log_{10}[\text{H}^+]$ [1] (b) $K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$ Assume $[\text{H}^+] = [\text{CH}_3\text{COO}^-]$ and $[\text{CH}_3\text{COOH}]_{eq} \approx 0.10$. $[\text{H}^+]^2 = K_a \times 0.10 = 1.7 \times 10^{-5} \times 0.10 = 1.7 \times 10^{-6}$ $[\text{H}^+] = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}$ $\text{pH} = -\log(1.30 \times 10^{-3}) = 2.88$ [2] (c) (i) In a buffer where $[\text{acid}] = [\text{salt}]$, $\text{pH} = pK_a$. $pK_a = -\log(1.7 \times 10^{-5}) = 4.77$. $\text{pH} = 4.77$ [2] (ii) $\text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq) \rightarrow \text{CH}_3\text{COOH}(aq)$ The ethanoate ions remove added $\text{H}^+$, minimizing pH change. [2]

2 (a) HA is the strong acid. [1] For a $0.10 \text{ M}$ strong monoprotic acid, $[\text{H}^+] = 0.10 \text{ M}$, so $\text{pH} = -\log(0.10) = 1.0$. This matches HA. [1] (b) For HB, $\text{pH} = 2.9 \Rightarrow [\text{H}^+] = 10^{-2.9} = 1.26 \times 10^{-3} \text{ M}$. $K_a = \frac{[\text{H}^+]^2}{[\text{HB}]} = \frac{(1.26 \times 10^{-3})^2}{0.10} = 1.59 \times 10^{-5} \text{ mol dm}^{-3}$ [3] (c) (i) Sketch: Start pH ~4.5, gradual rise, vertical jump at $25.0 \text{ cm}^3$ (equivalence), final pH ~12-13. Equivalence point pH > 7 (basic). Buffer region around half-equivalence ($12.5 \text{ cm}^3$). [3] (ii) Phenolphthalein. [1] The equivalence point is in the basic range (pH 8-10), which falls within the color change range of phenolphthalein (8.3-10.0). Methyl orange changes in acidic range. [1]

3 (a) $K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2$ [1] (b) Let solubility be $s \text{ mol dm}^{-3}$. $[\text{Mg}^{2+}] = s$, $[\text{OH}^-] = 2s$. $K_{sp} = (s)(2s)^2 = 4s^3$ $1.8 \times 10^{-11} = 4s^3$ $s^3 = 4.5 \times 10^{-12}$ $s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$ [3] (c) Common ion effect. [1] Adding NaOH increases $[\text{OH}^-]$. To maintain constant $K_{sp}$, $[\text{Mg}^{2+}]$ must decrease, causing precipitation of $\text{Mg(OH)}_2$. [1]

4 (a) $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$ [1] (b) $K_b = 10^{-4.75} = 1.78 \times 10^{-5}$. $[\text{OH}^-] = \sqrt{K_b \times [\text{NH}_3]} = \sqrt{1.78 \times 10^{-5} \times 0.050} = \sqrt{8.9 \times 10^{-7}} = 9.43 \times 10^{-4}$. $\text{pOH} = -\log(9.43 \times 10^{-4}) = 3.03$. $\text{pH} = 14 - 3.03 = 10.97$ [3] (c) (i) Acidic [1] (ii) $\text{NH}_4^+$ is the conjugate acid of a weak base and hydrolyzes: $\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+$. $\text{Cl}^-$ is the conjugate base of a strong acid and does not hydrolyze. Net production of $\text{H}_3\text{O}^+$ makes solution acidic. [2]

5 (a) Methyl propanoate. [1] $\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{H}_2\text{O}$ [1] (b) Use excess alcohol or remove water/ester as it forms. [1] (c) (i) Saponification (or alkaline hydrolysis). [1] (ii) $\text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+ + \text{CH}_3\text{OH}$ [2] (iii) The carboxylate ion ($\text{RCOO}^-$) formed is stable and does not react with the alcohol to reform the ester. The reaction is effectively irreversible. [2]

Section B: Data-Based and Application Questions

6 (a) At equivalence, moles acid = moles base. Moles NaOH = $0.100 \times \frac{25.0}{1000} = 0.0025 \text{ mol}$. Moles HX = $0.0025 \text{ mol}$. $[\text{HX}] = \frac{0.0025}{0.025} = 0.10 \text{ mol dm}^{-3}$ [3] (b) At half-equivalence, $\text{pH} = pK_a$. $pK_a = 4.75$. $K_a = 10^{-4.75} = 1.78 \times 10^{-5} \text{ mol dm}^{-3}$ [2] (c) At equivalence, solution contains salt NaX. $[\text{X}^-] = \frac{0.0025 \text{ mol}}{0.050 \text{ dm}^3} = 0.050 \text{ M}$. Hydrolysis: $\text{X}^- + \text{H}_2\text{O} \rightleftharpoons \text{HX} + \text{OH}^-$. $K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.78 \times 10^{-5}} = 5.62 \times 10^{-10}$. $[\text{OH}^-] = \sqrt{K_b \times [\text{X}^-]} = \sqrt{5.62 \times 10^{-10} \times 0.050} = \sqrt{2.81 \times 10^{-11}} = 5.30 \times 10^{-6}$. $\text{pOH} = 5.28$. $\text{pH} = 14 - 5.28 = 8.72$ [4]

7 (a) Acids produce $\text{H}^+$. $\text{H}^+$ reacts with $\text{OH}^-$ in the equilibrium to form water. [1] This decreases $[\text{OH}^-]$, shifting equilibrium to the right (Le Chatelier). [1] More hydroxyapatite dissolves, causing demineralization. [1] (b) (i) $\text{Ca}_5(\text{PO}_4)_3\text{F}(s) \rightleftharpoons 5\text{Ca}^{2+}(aq) + 3\text{PO}_4^{3-}(aq) + \text{F}^-(aq)$ [1] (ii) $\text{F}^-$ is a weaker base than $\text{OH}^-$. [1] It reacts less readily with $\text{H}^+$ to form HF (weak acid) compared to $\text{OH}^-$ forming water. [1] Thus, the equilibrium position is less disturbed by acid, maintaining solid structure. [1] (Alternative: $K_{sp}$ of fluoroapatite is lower, so it is less soluble generally.)

8 (a)

1. $\text{H}_2\text{A} \rightleftharpoons \text{H}^+ + \text{HA}^-$ [1]
2. $\text{HA}^- \rightleftharpoons \text{H}^+ + \text{A}^{2-}$ [1] (b) It is harder to remove a positive proton ($\text{H}^+$) from a negatively charged ion ($\text{HA}^-$) than from a neutral molecule ($\text{H}_2\text{A}$) due to electrostatic attraction. [2] (c) Use $K_{a1}$. $[\text{H}^+] = \sqrt{K_{a1} \times [\text{H}_2\text{A}]} = \sqrt{1.0 \times 10^{-3} \times 0.10} = \sqrt{1.0 \times 10^{-4}} = 0.010 \text{ M}$. $\text{pH} = -\log(0.010) = 2.0$ [3]

9 (a) $\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-$ [1] (b) $K_{In} = \frac{[\text{H}^+][\text{In}^-]}{[\text{HIn}]}$. $[\text{H}^+] = K_{In} \frac{[\text{HIn}]}{[\text{In}^-]}$. $-\log[\text{H}^+] = -\log K_{In} - \log \left( \frac{[\text{HIn}]}{[\text{In}^-]} \right)$. $\text{pH} = pK_{In} + \log \left( \frac{[\text{In}^-]}{[\text{HIn}]} \right)$ [2] (c) $7.6 = 7.0 + \log \left( \frac{[\text{In}^-]}{[\text{HIn}]} \right)$. $\log \left( \frac{[\text{In}^-]}{[\text{HIn}]} \right) = 0.6$. Ratio = $10^{0.6} \approx 3.98$ [2] (d) Blue. [1] Since ratio $> 1$, $[\text{In}^-] > [\text{HIn}]$, so the base colour (blue) dominates. [1]

10 (a) $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$ [1] (b) Moles NaOH = $0.100 \times 0.0245 = 0.00245 \text{ mol}$. Moles $\text{H}_2\text{SO}_4 = \frac{1}{2} \times 0.00245 = 0.001225 \text{ mol}$. $[\text{H}_2\text{SO}_4] = \frac{0.001225}{0.025} = 0.049 \text{ mol dm}^{-3}$ [3] (c) (i) Red to Yellow (or Orange). [1] (ii) No significant difference. [1] Sulfuric acid is strong; the pH change at equivalence is very sharp, covering both indicator ranges. Both indicators will change colour at the equivalence point volume. [1]

Section C: Long Structured Questions

11 (a) Stability increases down the group. [1] Larger cation size (e.g., $\text{Ba}^{2+}$ vs $\text{Mg}^{2+}$) has lower charge density. [1] Lower polarizing power on the carbonate ion. [1] Less distortion of the C-O bond, making it harder to decompose into oxide and $\text{CO}_2$. [1] (b) (i) Amphoteric substances can act as both an acid and a base. [1] (ii)

1. $\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}$ [1]
2. $\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-$ [1] (c) (i) $\text{BaSO}_4$ is very insoluble ($K_{sp}$ is very low). [1] Concentration of toxic $\text{Ba}^{2+}$ ions in solution is negligible. [1] (ii) Stomach contains HCl (acid). [1] $\text{BaCO}_3 + 2\text{H}^+ \rightarrow \text{Ba}^{2+} + \text{H}_2\text{O} + \text{CO}_2$. [1] The reaction removes carbonate ions, shifting equilibrium to dissolve more $\text{BaCO}_3$, releasing toxic $\text{Ba}^{2+}$ ions. [1]

12 (a) Aspirin has a -COOH group. [1] In NaOH: $\text{RCOOH} + \text{OH}^- \rightarrow \text{RCOO}^- + \text{H}_2\text{O}$. [1] The ionic salt ($\text{RCOO}^-\text{Na}^+$) is soluble in water due to ion-dipole interactions. [1] Unionized aspirin is non-polar/hydrophobic and poorly soluble. [1] (b) (i) Aspirin hydrolyzes slowly in water/aqueous base during titration, leading to inaccurate results. [2] (ii) Back titration. [1] Add excess standard NaOH, heat to ensure complete hydrolysis/reaction, then titrate remaining NaOH with standard acid. [2] (c) (i) $\text{C}_9\text{H}_8\text{O}_4 + \text{H}_2\text{O} \rightarrow \text{C}_7\text{H}_6\text{O}_3 \text{ (salicylic)} + \text{CH}_3\text{COOH}$ [2] (ii) Add neutral $\text{FeCl}_3$ solution. [1] Salicylic acid (phenol) gives a violet/purple complex. [1] Aspirin (no free phenol group) does not react (or gives no colour). [1]

13 (a) $\text{pH} = pK_a + \log \left( \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \right)$. $7.4 = 6.1 + \log (\text{ratio})$. $\log (\text{ratio}) = 1.3$. Ratio = $10^{1.3} \approx 20$ [3] (b) (i) Lactic acid adds $\text{H}^+$. [1] $\text{H}^+$ reacts with $\text{HCO}_3^-$ to form $\text{H}_2\text{CO}_3$. [1] Ratio changes slightly, but pH remains stable due to log relationship/high buffer capacity. [1] (ii) Increased $\text{H}_2\text{CO}_3$ decomposes to $\text{CO}_2$ and $\text{H}_2\text{O}$. [1] Increased breathing rate removes $\text{CO}_2$, shifting equilibrium to reduce $\text{H}^+$. [1] (c) Ratio = $0.050 / 0.025 = 2$. $\text{pH} = 6.1 + \log(2) = 6.1 + 0.30 = 6.40$ [2]

14 (a) (i) $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ [1] (ii) $s^2 = 1.8 \times 10^{-10} \Rightarrow s = 1.34 \times 10^{-5} \text{ mol dm}^{-3}$ [2] (b) (i) $\text{AgCl}(s) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag(NH}_3)_2]^+(aq) + \text{Cl}^-(aq)$ [1] (ii) $\text{NH}_3$ reacts with $\text{Ag}^+$ to form complex ion. [1] This lowers $[\text{Ag}^+]$. [1] Equilibrium $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$ shifts right to restore $K_{sp}$, dissolving precipitate. [1] (c) (i) $K_{sp}(\text{AgI}) \ll K_{sp}(\text{AgCl})$. [1] (ii) The solubility of AgI is so low that even with complex formation, the product of $[\text{Ag}^+][\text{I}^-]$ cannot exceed $K_{sp}$ sufficiently to dissolve significant amounts. The $K_{stab}$ of the complex is not large enough to overcome the very low $K_{sp}$ of AgI. [2]

15 (a) Temperature: ~450°C. [1] Reaction is exothermic; low T favors yield, but high T favors rate. 450°C is a compromise. [1] Pressure: ~1-2 atm. [1] High P favors yield (fewer moles gas), but high P is expensive/dangerous. Conversion is already high at low P. [1] (b) (i) Reaction is highly exothermic and produces a mist of sulfuric acid that is hard to condense. [2] (ii) $\text{SO}_3$ is dissolved in conc. $\text{H}_2\text{SO}_4$ to form oleum ($\text{H}_2\text{S}_2\text{O}_7$). [1] Oleum is then diluted with water to form conc. $\text{H}_2\text{SO}_4$. [1] (c) (i) $\text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^-$ [1] $\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}$ [1] (ii) Removing $\text{H}^+$ from a neutral molecule is easier than removing a positive $\text{H}^+$ from a negative ion ($\text{HSO}_4^-$) due to electrostatic forces. [2]

16 (a) (i) $^+\text{H}_3\text{NCH}_2\text{COOH}$ [1] (ii) $\text{H}_2\text{NCH}_2\text{COO}^-$ [1] (iii) $^+\text{H}_3\text{NCH}_2\text{COO}^-$ [1] (b) Solubility is lowest at isoelectric point. [1] Zwitterion has no net charge, so ion-dipole interactions with water are weaker than for fully charged ions at extreme pH. [1] (c) (i) Alanine has a chiral carbon (attached to H, CH3, NH2, COOH). [1] Glycine's central carbon is attached to two H atoms (not 4 different groups). [1] (ii) Correct 3D drawings showing mirror images. [2]

17 (a) $\text{SO}_2$ emitted from combustion. [1] Oxidized in atmosphere: $2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$. [1] $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$ (sulfuric acid). [1] (b) (i) Acid rain lowers soil pH. [1] $\text{Al}_2\text{O}_3$ (amphoteric) reacts with acid: $\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}$. [1] $\text{Al}^{3+}$ leaches into water. [1] (ii) Add limestone ($\text{CaCO}_3$) or lime ($\text{CaO}$) to the lake. [1] Neutralizes acid and precipitates aluminum as hydroxide. [1]

18 (a) Cl is electronegative. [1] Exerts electron-withdrawing inductive effect (-I). [1] Stabilizes the carboxylate anion ($\text{ClCH}_2\text{COO}^-$) by dispersing negative charge, favoring dissociation. [1] (b) Three Cl atoms exert a stronger -I effect than one. [1] Further stabilizes the anion, increasing acidity. [1] (c) $K_a = 10^{-2.86} = 1.38 \times 10^{-3}$. $[\text{H}^+] = \sqrt{1.38 \times 10^{-3} \times 0.010} = \sqrt{1.38 \times 10^{-5}} = 3.71 \times 10^{-3}$. $\text{pH} = -\log(3.71 \times 10^{-3}) = 2.43$ [3]

19 (a) Withdraw samples at time intervals. [1] Quench reaction (e.g., add ice/cold acid). [1] Titrate remaining $\text{OH}^-$ with standard acid. [1] (b) (i) Rate $= k[\text{ester}][\text{OH}^-]$ [1] (ii) Pseudo-first order. [1] $[\text{OH}^-]$ is effectively constant. Rate depends only on [ester]. [1] (c) $k = A e^{-E_a/RT}$. [1] As T increases, $e^{-E_a/RT}$ increases. [1] More molecules have energy $\ge E_a$, so rate constant increases. [1]

20 (a) (i) Lewis Acid: $\text{BF}_3$ (electron pair acceptor). [1] Lewis Base: $\text{NH}_3$ (electron pair donor). [1] (ii) Dative covalent (coordinate) bond. [1] N donates lone pair to empty orbital on B. [1] (b) (i) $\text{H}_2\text{O} + \text{NH}_3 \rightleftharpoons \text{OH}^- + \text{NH}_4^+$ (Water donates $\text{H}^+$). [1] (ii) $\text{H}_2\text{O} + \text{HCl} \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^-$ (Water accepts $\text{H}^+$). [1] (c) (i) Endothermic. [1] $K_w$ increases with T, so equilibrium shifts right with heat (Le Chatelier). [1] (ii) pH decreases. [1] $[\text{H}^+]$ increases. [1] Water is still neutral because $[\text{H}^+] = [\text{OH}^-]$. [1]