A Level H2 Chemistry Practice Paper 3

TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key and Marking Scheme (Version 3)

Topic: Acids, Bases and Salts
Total Marks: 60

Section A: Structured Questions

1 (a) $pH = -\log_{10}[H^+]$ [1] (b) Assumption: $[H^+] = [A^-]$ and $[HA]_{eq} \approx [HA]_{initial}$ (since $K_a$ is small). [1] $K_a = \frac{[H^+]^2}{[HA]}$ $[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.74 \times 10^{-5} \times 0.100}$ [1] $[H^+] = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$ $pH = -\log(1.32 \times 10^{-3}) = 2.88$ [1] (c) (i) Since volumes and concentrations are equal, $[acid] = [salt]$. $pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right) = pK_a + \log(1) = pK_a$ [1] $pK_a = -\log(1.74 \times 10^{-5}) = 4.76$ [1] (ii) $CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$ [1] The added $H^+$ ions react with the conjugate base ($CH_3COO^-$) to form weak acid, minimizing the change in $[H^+]$. [1]

2 (a) $C_2H_5COOH + CH_3OH \rightleftharpoons C_2H_5COOCH_3 + H_2O$ [1 for reactants/products, 1 for equilibrium sign/conditions] (b) Let $x$ be the moles of ester formed. Initial: Acid=1.0, Alcohol=1.0, Ester=0, Water=0 Eq: Acid=$1.0-x$, Alcohol=$1.0-x$, Ester=$x$, Water=$x$ $K_c = \frac{[Ester][Water]}{[Acid][Alcohol]} = \frac{(x/V)(x/V)}{((1-x)/V)((1-x)/V)} = \frac{x^2}{(1-x)^2}$ [1] $\sqrt{4.0} = \frac{x}{1-x} \Rightarrow 2 = \frac{x}{1-x}$ [1] $2(1-x) = x \Rightarrow 2 - 2x = x \Rightarrow 3x = 2$ [1] $x = 0.67 \text{ mol}$ [1]

3 (a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1] (b) Let $s$ be solubility in $\text{mol dm}^{-3}$. $[Mg^{2+}] = s$, $[OH^-] = 2s$ $K_{sp} = (s)(2s)^2 = 4s^3$ [1] $1.8 \times 10^{-11} = 4s^3$ $s^3 = 4.5 \times 10^{-12}$ $s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$ [1] (c) $H^+$ ions from HCl react with $OH^-$ ions to form water: $H^+ + OH^- \rightarrow H_2O$. [1] This decreases $[OH^-]$, shifting the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ to the right (Le Chatelier), increasing solubility. [1]

4 (a) $[H^+] = 10^{-pH} = 10^{-2.88} = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$ [1] (b) $K_a = \frac{[H^+][A^-]}{[HA]}$ Assume $[H^+] = [A^-]$ and $[HA]_{eq} \approx 0.050$ $K_a = \frac{(1.32 \times 10^{-3})^2}{0.050}$ [2] $K_a = 3.48 \times 10^{-5} \text{ mol dm}^{-3}$ [1] (c) Phenolphthalein. [1] The equivalence point for weak acid-strong base titration is alkaline (pH 8-9). Phenolphthalein changes color in the range 8.3-10.0, which falls within the vertical section of the titration curve. [1]

5 (a) Pair 1: $NH_3$ (base) / $NH_4^+$ (acid) [1] Pair 2: $H_2O$ (acid) / $OH^-$ (base) [1] (b) A proton ($H^+$) acceptor. [1] (c) The methyl group in methylamine is electron-releasing (positive inductive effect, +I). [1] This increases the electron density on the nitrogen atom, making the lone pair more available for donation to a proton, thus making it a stronger base than ammonia. [1]

6 (a)

• Y-axis: pH (0-14), X-axis: Volume of acid added. [1]
• Start pH ~13, End pH ~1. [1]
• Vertical section centered at pH 7. [1] (b) Salt formed is NaCl, which is neutral. Neither $Na^+$ nor $Cl^-$ hydrolyzes. [1] $[H^+] = [OH^-] = 10^{-7} \text{ mol dm}^{-3}$. (c) Moles excess $H^+ = 0.10 \text{ cm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.01 \text{ mmol} = 1.0 \times 10^{-5} \text{ mol}$. [1] Total volume $\approx 50.0 \text{ cm}^3 = 0.050 \text{ dm}^3$. $[H^+] = \frac{1.0 \times 10^{-5}}{0.050} = 2.0 \times 10^{-4} \text{ mol dm}^{-3}$. [1] $pH = -\log(2.0 \times 10^{-4}) = 3.70$. [1]

7 (a) $Al^{3+}(aq) + 6H_2O(l) \rightarrow [Al(H_2O)_6]^{3+}(aq)$ [1] $[Al(H_2O)_6]^{3+}(aq) + H_2O(l) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H_3O^+(aq)$ [1] (b) Effervescence (bubbles of gas) and a white precipitate. [1] $Al^{3+}$ is acidic; $CO_3^{2-}$ is basic. They undergo mutual hydrolysis. $2Al^{3+} + 3CO_3^{2-} + 3H_2O \rightarrow 2Al(OH)_3(s) + 3CO_2(g)$. [2]

8 (a) $K_w = [H^+][OH^-]$ [1] (b) Since dissociation is endothermic, increasing T shifts equilibrium to the right. [1] $[H^+]$ and $[OH^-]$ both increase. [1] Since $pH = -\log[H^+]$, pH decreases (becomes < 7). Note: Water remains neutral as $[H^+]=[OH^-]$. [1]

9 (a) $K_a = \frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}$ [1] (b) $pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right)$ $pK_a = -\log(6.3 \times 10^{-5}) = 4.20$ $pH = 4.20 + \log\left(\frac{0.040}{0.020}\right) = 4.20 + \log(2) = 4.20 + 0.30 = 4.50$ [2] (c) The active preservative species is the undissociated benzoic acid molecule, which can penetrate bacterial cell membranes. [1] At low pH, equilibrium shifts towards the undissociated acid ($C_6H_5COOH$). At neutral pH, it exists mainly as the benzoate ion, which cannot penetrate cells effectively. [1]

10 (a)

1. $H_2SO_4 \rightarrow H^+ + HSO_4^-$ [1]
2. $HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$ [1] (b) Removing a proton from a neutral molecule ($H_2SO_4$) is easier than removing a positive proton from a negatively charged ion ($HSO_4^-$) due to electrostatic attraction. [2] (c) $[H^+]$ from 1st dissociation = 0.010 M. $[H^+]$ from 2nd dissociation $\approx$ 0.010 M (assuming strong/complete for simplicity in this context, though technically $K_{a2}$ is weak, usually A-Level questions specify "assume complete" or give $K_{a2}$. If complete: Total $[H^+] = 0.020$ M). $pH = -\log(0.020) = 1.70$. [2] (Note: If treating 2nd step as weak, calculation is more complex, but "assume complete" is standard for this mark allocation unless $K_{a2}$ is provided.)

Section B: Data-Based and Application Questions

11 (a) A < D < C < B (Increasing strength means lower pH for same conc, so order of strength: B < C < D < A. Question asks increasing strength: B, C, D, A). [1] Correction: Lowest pH is strongest. A(1.0) > D(1.9) > C(2.4) > B(2.9). Order of increasing strength: B, C, D, A. (b) Chlorine is electronegative and exerts a negative inductive effect (-I). [1] This withdraws electron density from the carboxyl group, weakening the O-H bond and stabilizing the conjugate base ($ClCH_2COO^-$) by dispersing the negative charge. [2] (c) $[H^+] = 10^{-2.4} = 3.98 \times 10^{-3} \text{ mol dm}^{-3}$. $K_a = \frac{[H^+]^2}{[HA]} = \frac{(3.98 \times 10^{-3})^2}{0.10}$ [2] $K_a = 1.58 \times 10^{-4} \text{ mol dm}^{-3}$. [1]

12 (a) $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$ [1] (b) (i) Moles $H^+ = 0.050 \times \frac{12.5}{1000} = 6.25 \times 10^{-4} \text{ mol}$. Ratio $H^+ : OH^-$ is 1:1. Moles $OH^- = 6.25 \times 10^{-4} \text{ mol}$. $[OH^-] = \frac{6.25 \times 10^{-4}}{0.025} = 0.025 \text{ mol dm}^{-3}$. [2] (ii) $[Ca^{2+}] = \frac{1}{2}[OH^-] = 0.0125 \text{ mol dm}^{-3}$. $K_{sp} = [Ca^{2+}][OH^-]^2 = (0.0125)(0.025)^2$ [2] $K_{sp} = 7.81 \times 10^{-6} \text{ mol}^3 \text{ dm}^{-9}$. [1] (c) Solubility decreases. [1] Adding $CaCl_2$ increases $[Ca^{2+}]$. By Le Chatelier’s principle, the equilibrium shifts to the left to remove excess $Ca^{2+}$, causing precipitation of $Ca(OH)_2$. [1]

13 (a) $^+H_3N-CH_2-COO^-$ [1] (b) The pH at which the amino acid exists primarily as a zwitterion and has no net electrical charge. [1] (c) (i) $pI = \frac{pK_{a1} + pK_{a2}}{2} = \frac{2.34 + 9.60}{2} = 5.97$. [1] (ii) Curve starts at low pH (~1). Two buffer regions (flat parts) centered at pH 2.34 and 9.60. Two equivalence points (vertical sections). [3]

14 (a) $K_{In} = \frac{[H^+][In^-]}{[HIn]}$ $\frac{[In^-]}{[HIn]} = \frac{K_{In}}{[H^+]}$ $\log\left(\frac{[In^-]}{[HIn]}\right) = \log K_{In} - \log[H^+] = -pK_{In} + pH$ $pH = pK_{In} + \log\left(\frac{[In^-]}{[HIn]}\right)$ [2] (b) (i) Red. (pH < pKa, acid form dominates). [1] (ii) The equivalence point for weak acid-strong base is ~pH 8-9. Methyl orange changes color at pH 3.1-4.4. The color change would occur long before the equivalence point, leading to a large titration error. [2]

15 (a) pH 4.0: $H_2CO_3$ [1] pH 8.0: $HCO_3^-$ [1] pH 12.0: $CO_3^{2-}$ [1] (b) pH = $pK_{a1} = 6.4$. [1] (c) $H^+(aq) + HCO_3^-(aq) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons H_2O(l) + CO_2(g)$. Added acid is removed by $HCO_3^-$. Added base is removed by $H_2CO_3$. This maintains blood pH around 7.4. [2]

Section C: Long Structured Questions

16 (a) $pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right)$ $5.0 = 4.76 + \log\left(\frac{[salt]}{0.10}\right)$ $0.24 = \log\left(\frac{[salt]}{0.10}\right)$ $\frac{[salt]}{0.10} = 10^{0.24} = 1.74$ $[salt] = 0.174 \text{ mol dm}^{-3}$. [2] For $1.0 \text{ dm}^3$, moles salt = 0.174 mol. Mass = $0.174 \times 82.0 = 14.27 \text{ g}$. [2] Procedure: Dissolve 14.3 g of sodium ethanoate in some $0.10 \text{ M}$ ethanoic acid, then make up to $1.0 \text{ dm}^3$ with the same acid. [1] (b) Initial moles: Acid = $0.10 \times 0.1 = 0.01$ mol. Salt = $0.174 \times 0.1 = 0.0174$ mol. Moles NaOH added = $0.010 \times 0.010 = 0.0001$ mol. Reaction: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$. New moles Acid = $0.01 - 0.0001 = 0.0099$ mol. New moles Salt = $0.0174 + 0.0001 = 0.0175$ mol. $pH = 4.76 + \log\left(\frac{0.0175}{0.0099}\right) = 4.76 + 0.25 = 5.01$. [4] (c) In water: $[OH^-] = \frac{0.0001}{0.11} \approx 9 \times 10^{-4}$. pOH ~3. pH ~11. Change from 7 to 11 is 4 units. [2] Buffer changed by only 0.01 pH units. Buffer is highly effective. [1]

17 (a) Precipitation of ions from a solution by careful addition of a precipitating agent, where salts precipitate in order of their solubility products. [2] (b) (i) $[Ag^+] = \frac{K_{sp}(AgCl)}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8} \text{ mol dm}^{-3}$. [2] (ii) $[Ag^+]^2 = \frac{K_{sp}(Ag_2CrO_4)}{[CrO_4^{2-}]} = \frac{1.1 \times 10^{-12}}{0.010} = 1.1 \times 10^{-10}$. $[Ag^+] = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \text{ mol dm}^{-3}$. [2] (iii) AgCl precipitates first because it requires a lower concentration of $Ag^+$ ($1.8 \times 10^{-8}$ vs $1.05 \times 10^{-5}$). [2] (c) $Ag^+$ reacts with $Cl^-$ first to form white AgCl. [1] Once all $Cl^-$ is consumed, excess $Ag^+$ reacts with $CrO_4^{2-}$ to form brick-red $Ag_2CrO_4$ precipitate. [1] The appearance of the red color indicates the endpoint. [1]

18 (a) Order of acidity: Ethanoic < Chloroethanoic < Dichloroethanoic. [1] Cl is electronegative (-I effect). [1] More Cl atoms withdraw more electron density, stabilizing the carboxylate anion more effectively. [1] This makes the O-H bond more polar and easier to break, increasing $K_a$. [1] (b) (i) Phenoxide ion ($C_6H_5O^-$) is stabilized by resonance delocalization of the negative charge into the benzene ring. Ethoxide ion has no such stabilization. [3] (ii) Ethanoate ion ($CH_3COO^-$) has resonance delocalization over two electronegative oxygen atoms, which is more effective than delocalization into the carbon ring of phenoxide. Also, O-H bond in carboxylic acids is more polar. [3]

19 (a) (i) Alkaline. $CH_3COO^-$ hydrolyzes: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$. [2] (ii) Acidic. $NH_4^+$ hydrolyzes: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$. [2] (iii) Neutral. Derived from strong acid and strong base; no hydrolysis. [1] (b) $K_a(NH_4^+) = \frac{K_w}{K_b(NH_3)} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$. [1] $[H^+] = \sqrt{K_a \times [Salt]} = \sqrt{5.56 \times 10^{-10} \times 0.10}$ [2] $[H^+] = 7.46 \times 10^{-6}$. $pH = -\log(7.46 \times 10^{-6}) = 5.13$. [1]

20 (a) Rinse electrode with distilled water. [1] Immerse in buffer solutions of known pH (e.g., 4.0 and 7.0) and adjust calibration settings. [1] (b) (i) $pK_a$ is the pH at the half-equivalence point (where volume of base added is half that required for equivalence). [2] (ii) The pH change at the equivalence point for very weak acids is gradual, not vertical. Visual indicators do not show a sharp color change, making detection difficult. A pH meter detects the inflection point accurately. [2] (c)

1. Parallax error in reading burette. Minimize by reading at eye level. [2]
2. Air bubbles in burette jet. Minimize by flushing jet before starting. [2] (Other valid errors: Wet conical flask, incorrect indicator choice, etc.)