A Level H2 Chemistry Practice Paper 3

TuitionGoWhere Exam Practice (AI) - Answer Key

Subject: Chemistry H2 (9476)
Paper: Practice Paper - Acids, Bases & Salts (Version 3 of 5)

Section A: Structured Questions

1 (a) Titres: * Titration 1: $23.80 - 0.00 = 23.80 \, \text{cm}^3$ * Titration 2: $47.30 - 23.80 = 23.50 \, \text{cm}^3$ * Titration 3: $24.10 - 0.30 = 23.80 \, \text{cm}^3$ [1]

(b) Suitable titres are 1 and 3 ($23.80 \, \text{cm}^3$). Titration 2 ($23.50 \, \text{cm}^3$) is an outlier as it differs by more than $0.10 \, \text{cm}^3$ from the others. The rough titration is also excluded. [2]

(c) Mean titre = $\frac{23.80 + 23.80}{2} = 23.80 \, \text{cm}^3$. [1]

(d) * Moles of $\text{NaOH} = 0.100 \times \frac{23.80}{1000} = 2.38 \times 10^{-3} \, \text{mol}$. * Ratio $\text{CH}_3\text{COOH} : \text{NaOH}$ is $1:1$. * Moles of acid = $2.38 \times 10^{-3} \, \text{mol}$. * Concentration of acid = $\frac{2.38 \times 10^{-3}}{25.0/1000} = 0.0952 \, \text{mol dm}^{-3}$. [2]

2 (a) A solution that resists changes in pH when small amounts of acid or base are added. [2]

(b) (i) $[\text{Salt}] = [\text{Acid}]$ since volumes and concentrations are equal. $\text{pH} = \text{p}K_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) = -\log(1.7 \times 10^{-5}) + \log(1) = 4.77$. [2] (ii) $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$. The ethanoate ions remove added $\text{H}^+$ ions to form weak ethanoic acid, minimizing pH change. [2]

(c) Graph should show: * Starting pH approx 2.9. * Gradual rise (buffer region). * Steep vertical section at equivalence point (approx pH 8-9). * Leveling off at high pH (approx 13). * Label equivalence point and buffer region. [3]

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

(b) (i) Let solubility be $s$. $[\text{Mg}^{2+}] = s$, $[\text{OH}^-] = 2s$. $K_{sp} = s(2s)^2 = 4s^3$. $1.2 \times 10^{-11} = 4s^3 \Rightarrow s^3 = 3.0 \times 10^{-12} \Rightarrow s = 1.44 \times 10^{-4} \, \text{mol dm}^{-3}$. [2] (ii) $[\text{OH}^-] = 2s = 2.88 \times 10^{-4} \, \text{mol dm}^{-3}$. $\text{pOH} = -\log(2.88 \times 10^{-4}) = 3.54$. $\text{pH} = 14 - 3.54 = 10.46$. [2]

(c) $\text{H}^+$ ions from the acid react with $\text{OH}^-$ ions to form water. This decreases $[\text{OH}^-]$, shifting the equilibrium $\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq)$ to the right, dissolving more solid. [2]

4 (a) (i) $\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^-$. [1] (ii) As pH increases, $[\text{H}^+]$ decreases. Equilibrium shifts right to restore $[\text{H}^+]$, increasing $[\text{In}^-]$ (blue) and decreasing $[\text{HIn}]$ (yellow). [2]

(b) No. The equivalence point for weak acid-strong base titration is alkaline (pH > 7). Bromothymol blue changes colour around pH 7, which is before the equivalence point, leading to inaccurate end-point detection. Phenolphthalein would be better. [2]

5 (a) HA. pH = $-\log(0.1) = 1.0$. Since the measured pH matches the calculated pH for complete dissociation, it is a strong acid. [2]

(b) $[\text{H}^+] = 10^{-2.87} = 1.35 \times 10^{-3} \, \text{mol dm}^{-3}$. $K_a = \frac{[\text{H}^+]^2}{[\text{HB}]} = \frac{(1.35 \times 10^{-3})^2}{0.100} = 1.82 \times 10^{-5} \, \text{mol dm}^{-3}$. [2]

(c) Chloroethanoic acid (HC) has an electronegative chlorine atom. This exerts an electron-withdrawing inductive effect, stabilizing the conjugate base (carboxylate ion) by dispersing the negative charge. This makes the O-H bond more polar and easier to break, increasing acidity compared to ethanoic acid (HB). [2]

Section B: Data Interpretation & Practical Analysis

6 (a) Graph: Curve starts at origin, steep initial gradient, curves over to horizontal asymptote. [2]

(b) Curve B: Initial gradient is half of A (lower concentration). Final volume is half of A (fewer moles of HCl). [2]

(c) Initial rate is lower for B because concentration of $\text{H}^+$ is lower, leading to fewer effective collisions per second. Final volume is lower for B because the number of moles of limiting reagent (HCl) is halved. [2]

7 (a) $\text{Al}^{3+}$ (Aluminium). [1] (b) $\text{Br}^-$ (Bromide). [1] (c) $\text{Al}^{3+}(aq) + 3\text{OH}^-(aq) \rightarrow \text{Al(OH)}_3(s)$. [1] (d) $[\text{Ag}(\text{NH}_3)_2]^+$. [1]

8 (a) $[\text{H}^+] = 10^{-4.2} = 6.31 \times 10^{-5} \, \text{mol dm}^{-3}$. [1]

(b) Sulfur dioxide ($\text{SO}_2$) and Nitrogen oxides ($\text{NO}_x$). Equation: $\text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3$ or $2\text{SO}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4$. [3]

(c) $\text{CaCO}_3(s) + 2\text{H}^+(aq) \rightarrow \text{Ca}^{2+}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$. [1]

9 (a) $K_{pc} = \frac{[\text{Solute}]_{\text{hexane}}}{[\text{Solute}]_{\text{water}}} = \frac{0.04}{0.80} = 0.05$. [2]

(b) Dimerization in hexane reduces the concentration of monomeric species in the organic layer relative to what would be expected if it remained monomeric. This lowers the numerator in the $K_{pc}$ expression (if defined as organic/aqueous monomer), making the apparent $K_{pc}$ lower than for a non-associating solute. Alternatively, the equilibrium involves 2\text{HA}_{\text{water}} \rightleftharpoons (\text{HA})_2_{\text{hexane}}. [2]

10 (a) $[\text{Ag}^+] = \frac{K_{sp}}{[\text{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8} \, \text{mol dm}^{-3}$. [2]

(b) $[\text{Ag}^+]^2 = \frac{K_{sp}}{[\text{CrO}_4^{2-}]} = \frac{3.0 \times 10^{-12}}{0.010} = 3.0 \times 10^{-10}$. $[\text{Ag}^+] = \sqrt{3.0 \times 10^{-10}} = 1.73 \times 10^{-5} \, \text{mol dm}^{-3}$. [2]

(c) $\text{AgCl}$ precipitates first because it requires a lower $[\text{Ag}^+]$ ($1.8 \times 10^{-8}$ vs $1.73 \times 10^{-5}$). [1]

Section C: Long Structured Questions

11 (a) Thermal stability increases down the group. As the cation size increases, charge density decreases. The polarizing power of the cation on the carbonate ion decreases, causing less distortion of the C-O bond. Thus, more heat energy is required to break the bond and release $\text{CO}_2$. [3]

(b) $\text{Mg}^{2+}$ is smaller than $\text{Ba}^{2+}$. Therefore, the lattice energy of $\text{MgO}$ is more exothermic (more negative) than $\text{BaO}$ due to stronger electrostatic attraction between ions. Higher lattice energy requires more energy to overcome, resulting in a higher melting point. [3]

(c) (i) $\text{Be}^{2+}$ is very small with high charge density. It has high polarizing power, distorting the electron cloud of $\text{Cl}^-$, introducing covalent character. $\text{Mg}^{2+}$ is larger with lower charge density, forming predominantly ionic bonds. [2] (ii) $[\text{Be}(\text{H}_2\text{O})_4]^{2+} + \text{H}_2\text{O} \rightleftharpoons [\text{Be}(\text{H}_2\text{O})_3(\text{OH})]^+ + \text{H}_3\text{O}^+$. [2]

12 (a) $\text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^-$. [1]

(b) (i) $[\text{H}^+] = 10^{-2.17} = 6.76 \times 10^{-3} \, \text{mol dm}^{-3}$. $K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \approx \frac{(6.76 \times 10^{-3})^2}{0.100} = 4.57 \times 10^{-4} \, \text{mol dm}^{-3}$. [3] (ii) % dissociation = $\frac{6.76 \times 10^{-3}}{0.100} \times 100 = 6.76\%$. [2]

(c) (i) Alkaline. [1] (ii) $\text{NO}_2^-$ is the conjugate base of a weak acid. It hydrolyzes: $\text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^-$. The production of $\text{OH}^-$ ions makes the solution alkaline. [2]

13 (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) (i) $\text{pI} = \frac{2.34 + 9.60}{2} = 5.97$. [1] (ii) The pH at which the amino acid exists primarily as a zwitterion and has no net electrical charge. It does not migrate in an electric field. [2]

(c) $\text{H}_2\text{NCH}_2\text{COO}^-$ and $^+\text{H}_3\text{NCH}_2\text{COO}^-$. (The species involved in the equilibrium around pKa 9.60). [2]

14 (a) $[\text{Fe}(\text{H}_2\text{O})_6]^{3+} + \text{H}_2\text{O} \rightleftharpoons [\text{Fe}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}_3\text{O}^+$. [1]

(b) The high charge density of $\text{Fe}^{3+}$ polarizes the O-H bonds in the coordinated water molecules, weakening them and facilitating the release of $\text{H}^+$ ions. [2]

(c) (i) Iron(III) hydroxide, $\text{Fe(OH)}_3$. [1] (ii) Carbon dioxide, $\text{CO}_2$. [1] (iii) $\text{Fe}^{3+}$ is strongly acidic. The carbonate ion ($\text{CO}_3^{2-}$) is basic. The acid-base reaction between hydrated $\text{Fe}^{3+}$ and $\text{CO}_3^{2-}$ is preferred over precipitation of the carbonate, leading to hydrolysis and $\text{CO}_2$ evolution. [2]

15 (a) Carboxylic acid (-COOH). [1]

(b) $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ (Butanoic acid). [1]

(c) $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{CH}_2\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$. [2]

(d) C could be hydroxybutanone or an ester like methyl propanoate? No, reacts with Tollens'. Must be an aldehyde. Isomer $\text{C}_4\text{H}_8\text{O}_2$ with aldehyde group? Hydroxy-aldehyde or keto-aldehyde? Example: 3-hydroxybutanal or 2-hydroxybutanal. Or an ester? Esters don't react with Tollens'. Wait, $\text{C}_4\text{H}_8\text{O}_2$ degree of unsaturation = 1. If it reacts with Tollens', it has an aldehyde group. Possible structure: $\text{HO-CH}_2\text{-CH}_2\text{-CH}_2\text{-CHO}$ (4-hydroxybutanal) or similar. Functional groups: Hydroxyl (-OH) and Aldehyde (-CHO). [2]

Section D: Advanced Application

16 (a) $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$. [1]

(b) $\text{AgCl}(s) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+(aq) + \text{Cl}^-(aq)$. [1]

(c) Ammonia reacts with $\text{Ag}^+$ ions to form the stable complex ion $[\text{Ag}(\text{NH}_3)_2]^+$. This decreases the concentration of free $\text{Ag}^+$ ions in solution. According to Le Chatelier's principle, the equilibrium $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ shifts to the right to restore $[\text{Ag}^+]$, causing more $\text{AgCl}$ to dissolve. [3]

17 (a) $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$. $K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$. $1.8 \times 10^{-5} = \frac{x^2}{0.100} \Rightarrow x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}$. $\text{pOH} = -\log(1.34 \times 10^{-3}) = 2.87$. $\text{pH} = 14 - 2.87 = 11.13$. [3]

(b) At equivalence, moles $\text{NH}_3$ = moles $\text{HCl} = 0.0025 \, \text{mol}$. Volume = $50.0 \, \text{cm}^3$. $[\text{NH}_4^+] = \frac{0.0025}{0.050} = 0.050 \, \text{mol dm}^{-3}$. $\text{NH}_4^+$ is a weak acid. $K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$. $[\text{H}^+] = \sqrt{K_a \times [\text{NH}_4^+]} = \sqrt{5.56 \times 10^{-10} \times 0.050} = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}$. $\text{pH} = -\log(5.27 \times 10^{-6}) = 5.28$. [4]

(c) Methyl orange. The equivalence point is at pH 5.28 (acidic). Methyl orange changes colour in the range 3.1–4.4, which is close to the steep part of the titration curve for weak base-strong acid. Phenolphthalein changes too early (alkaline range). Bromothymol blue changes around neutral, which is not the equivalence point. Correction: Actually, for weak base-strong acid, the pH drop is steep around pH 4-6. Methyl orange (3.1-4.4) is often cited, but Bromocresol green (3.8-5.4) is better. Between the options, Methyl Orange is the standard choice for weak base/strong acid titrations in A-Level contexts as the endpoint is acidic. [2]

18 (a) Pure water contains very few ions ($[\text{H}^+] = [\text{OH}^-] = 10^{-7} \, \text{mol dm}^{-3}$). Electrical conductivity depends on the concentration of mobile ions, which is negligible in pure water. [2]

(b) (i) $[\text{H}^+] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \, \text{mol dm}^{-3}$. [1] (ii) Neutral. $[\text{H}^+] = [\text{OH}^-]$. [2]

(c) (i) $[\text{H}^+] = \sqrt{5.48 \times 10^{-14}} = 2.34 \times 10^{-7}$. $\text{pH} = -\log(2.34 \times 10^{-7}) = 6.63$. [2] (ii) The dissociation of water is endothermic. Increasing temperature shifts equilibrium to the right, increasing both $[\text{H}^+]$ and $[\text{OH}^-]$ equally. Since they remain equal, the water is neutral, but the higher $[\text{H}^+]$ results in a lower pH value. [2]

19 (a) Add excess aqueous ammonia to the mixture. $\text{AgCl}$ will dissolve to form a colourless solution, while $\text{NaCl}$ remains as a solid (or dissolves if water is present, but AgCl separation is key). Filter the mixture. The residue is NaCl (if solid mixture was dry and minimal water used) or filtrate contains Ag complex. Better description: Add dilute ammonia. AgCl dissolves. NaCl is soluble in water anyway. Standard Separation: If both are solids, add water. Both dissolve? No, AgCl is insoluble in water. Wait, NaCl is soluble in water. AgCl is insoluble. Method: Add water. NaCl dissolves. AgCl remains solid. Filter. Residue is AgCl. Filtrate contains NaCl. Question asks to use aqueous ammonia. If we use ammonia: AgCl dissolves. NaCl dissolves. Both in solution. Then add $\text{HNO}_3$ to precipitate AgCl again? Let's stick to the prompt's implication of using ammonia's specific property. Add aqueous ammonia. AgCl dissolves ($[\text{Ag}(\text{NH}_3)_2]^+$). NaCl dissolves. To separate: Add $\text{HNO}_3$ to the filtrate. AgCl reprecipitates. Filter. Alternative interpretation: Maybe the mixture is AgCl and another insoluble chloride? No, NaCl is soluble. Let's assume the question implies separating AgCl from a mixture where solubility in ammonia is the distinguishing factor vs another insoluble salt? But it says NaCl. Actually, NaCl is soluble in water. AgCl is not. If the question insists on ammonia: 1. Add water/ammonia. AgCl dissolves in ammonia. NaCl dissolves in water. 2. This doesn't separate them easily if both go into solution. Correction: Perhaps the question implies AgCl and AgI? No, it says NaCl. Let's assume standard "Qualitative Analysis" logic: Add water. NaCl dissolves. AgCl does not. Filter. But the question asks to use ammonia. Maybe: Add ammonia. AgCl dissolves. Filter? No, NaCl is also in solution. Let's look at Test 4 in Q7. AgCl dissolves in conc ammonia. If the mixture is solid AgCl and solid NaCl: Add water. NaCl dissolves. AgCl stays. Filter. Why ammonia? Maybe the question meant AgCl and AgBr? Given the text "solid sodium chloride and solid silver chloride": Method: Add water. Stir. Filter. Residue is AgCl. Filtrate is NaCl(aq). If forced to use ammonia: Add ammonia. Both dissolve (NaCl is soluble in water part of aq ammonia). This fails to separate. Likely intended answer for "Chemical method using ammonia": This might be a trick or poorly phrased. However, AgCl is soluble in dilute ammonia. NaCl is soluble in water. If we add limited ammonia? No. Let's provide the standard separation for AgCl/NaCl which is water, but note the ammonia property. Actually, if the question implies separating Ag+ from Na+ in solution: Add HCl -> AgCl ppt. Let's assume the question meant AgCl and AgI or similar. However, answering strictly: (a) Add dilute nitric acid and silver nitrate? No. Let's assume the question allows water as the solvent for ammonia. If I add aqueous ammonia, AgCl dissolves. NaCl dissolves. I cannot separate them by filtration. I must re-precipitate AgCl. 1. Add excess aqueous ammonia. AgCl dissolves. NaCl dissolves. 2. Add dilute $\text{HNO}_3$ to acidify. AgCl reprecipitates. 3. Filter. Residue is AgCl. Filtrate contains NaCl and ammonium nitrate. This is a valid chemical method. [3]

(b) $\text{AgCl}(s) + 2\text{NH}_3(aq) \rightarrow [\text{Ag}(\text{NH}_3)_2]^+(aq) + \text{Cl}^-(aq)$ (Dissolution) AND $[\text{Ag}(\text{NH}_3)_2]^+(aq) + \text{Cl}^-(aq) + 2\text{H}^+(aq) \rightarrow \text{AgCl}(s) + 2\text{NH}_4^+(aq)$ (Reprecipitation). The question asks for the reaction during separation. The dissolution is the key step distinguishing it if it were mixed with something insoluble in ammonia. But here both dissolve. Let's provide the dissolution equation as the primary "ammonia reaction". [1]

(c) Acidify the solution containing the complex with dilute nitric acid. AgCl will precipitate out. Filter and wash. [2]

20 (a) $K_a = \frac{x^2}{0.010}$. $3.0 \times 10^{-4} = \frac{x^2}{0.010} \Rightarrow x^2 = 3.0 \times 10^{-6} \Rightarrow x = 1.73 \times 10^{-3}$. $\text{pH} = -\log(1.73 \times 10^{-3}) = 2.76$. [3]

(b) The sodium salt (sodium acetylsalicylate) is an ionic compound. Ionic compounds generally have higher solubility in polar solvents like water due to ion-dipole interactions, compared to the covalent acid form which relies on weaker hydrogen bonding and has a non-polar benzene ring. [2]

(c) $\text{p}K_a = -\log(3.0 \times 10^{-4}) = 3.52$. At pH 7, $\text{pH} > \text{p}K_a$. Therefore, the deprotonated form (anion) predominates. [2]