A Level H1 Chemistry Practice Paper 3

TuitionGoWhere Practice Paper - Chemistry H1 A-Level

Answer Key and Marking Scheme

Version: 3 of 5
Subject: Chemistry H1
Topic: Acids, Bases and Salts

Section A: Structured Questions

1
(a) A weak acid is an acid that partially dissociates (or ionizes) in water. [1]
(b) $CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)$ [1]
Note: Must use reversible arrow $\rightleftharpoons$ and state symbols.
(c)
For a weak acid: $[H^+] \approx \sqrt{K_a \times [HA]}$
$[H^+] = \sqrt{1.7 \times 10^{-5} \times 0.10}$
$[H^+] = \sqrt{1.7 \times 10^{-6}} = 1.30 \times 10^{-3} \text{ mol dm}^{-3}$ [1]
$pH = -\log(1.30 \times 10^{-3})$ [1]
$pH = 2.88$ (or 2.9) [1]
(d) HCl is a strong acid and fully dissociates, producing a higher concentration of $H^+$ ions ($0.10 \text{ mol dm}^{-3}$) compared to ethanoic acid which only partially dissociates. [1]

2
(a) Since volumes and concentrations of acid and salt are equal, $[acid] = [salt]$.
$pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right)$
$pH = -\log(1.7 \times 10^{-5}) + \log(1)$
$pH = 4.77 + 0$
$pH = 4.77$ [2]
1 mark for $pK_a$ calculation, 1 mark for final pH.
(b) The added $H^+$ ions from HCl react with the ethanoate ions ($CH_3COO^-$) in the buffer to form undissociated ethanoic acid ($CH_3COOH$). [1]
This removes most of the added $H^+$, so the equilibrium position shifts to the left, keeping the $[H^+]$ (and thus pH) relatively constant. [1]

3
(a) Chloroethanoic acid is a stronger acid than ethanoic acid (lower pH indicates higher $[H^+]$). [1]
The chlorine atom is highly electronegative and exerts an electron-withdrawing inductive effect (-I effect). [1]
This withdraws electron density from the carboxylate group/O-H bond, weakening the O-H bond and stabilizing the resulting carboxylate anion ($CH_2ClCOO^-$) by dispersing the negative charge. This facilitates greater dissociation. [1]
(b) Larger. [1]

4
(a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1]
(b) Let solubility be $s \text{ mol dm}^{-3}$.
Then $[Mg^{2+}] = s$ and $[OH^-] = 2s$.
$K_{sp} = (s)(2s)^2 = 4s^3$ [1]
$1.2 \times 10^{-11} = 4s^3$
$s^3 = 3.0 \times 10^{-12}$
$s = \sqrt[3]{3.0 \times 10^{-12}}$
$s = 1.44 \times 10^{-4} \text{ mol dm}^{-3}$ [2]
1 mark for substitution, 1 mark for correct answer.
(c) In acidic solution, $H^+$ ions react with $OH^-$ ions to form water ($H^+ + OH^- \rightarrow H_2O$). [1]
This decreases $[OH^-]$, causing the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ to shift to the right (Le Chatelier’s Principle), dissolving more solid. [1]

5
(a) Moles of $Na_2CO_3 = C \times V = 0.050 \times \frac{25.0}{1000} = 1.25 \times 10^{-3} \text{ mol}$. [1]
(b) From equation, mole ratio $Na_2CO_3 : H_2SO_4$ is 1:1.
Moles of $H_2SO_4 = 1.25 \times 10^{-3} \text{ mol}$. [1]
Concentration of $H_2SO_4 = \frac{n}{V} = \frac{1.25 \times 10^{-3}}{20.0/1000} = \frac{1.25 \times 10^{-3}}{0.020} = 0.0625 \text{ mol dm}^{-3}$. [1]
(c) The salt formed is sodium sulfate (from strong acid and strong base component of carbonate reaction effectively going to completion with strong acid), but technically titration of carbonate with strong acid has two endpoints. The final endpoint (bicarbonate to carbonic acid/CO2) occurs in acidic pH range (approx pH 4). [1]
Methyl orange changes color in the acidic range (3.1-4.4), matching the equivalence point. Phenolphthalein changes in basic range, which would correspond to the first endpoint (carbonate to bicarbonate) or miss the final completion, leading to error. [1]
Accept: Equivalence point is acidic due to formation of weak acid $H_2CO_3$/CO2 saturated solution.

6
(a) An amphoteric substance can act as both an acid and a base. [1]
(b)
(i) $Al_2O_3(s) + 6HCl(aq) \rightarrow 2AlCl_3(aq) + 3H_2O(l)$ [1]
(ii) $Al_2O_3(s) + 2NaOH(aq) + 3H_2O(l) \rightarrow 2Na[Al(OH)_4](aq)$ (or $2NaAlO_2 + H_2O$) [1]
Note: Formation of tetrahydroxoaluminate is preferred in modern syllabi.

7
(a) Acid: $H_2CO_3$ [0.5]
Conjugate Base: $HCO_3^-$ [0.5]
(b) The added $H^+$ (from lactic acid) reacts with the hydrogencarbonate ions ($HCO_3^-$) in the buffer. [1]
Equation: $H^+ + HCO_3^- \rightarrow H_2CO_3 \rightarrow H_2O + CO_2$.
This removes the excess $H^+$, preventing a large drop in pH. [1]

8
(a) Ammonium chloride ($NH_4Cl$). [1]
(b) The ammonium ion ($NH_4^+$) is the conjugate acid of a weak base ($NH_3$). It undergoes hydrolysis in water:
$NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)$ [1]
This produces $H_3O^+$ (or $H^+$) ions, making the solution acidic (pH < 7). [1]
Note: $Cl^-$ is the conjugate base of a strong acid and does not hydrolyze significantly.

Section B: Data-Based and Application Questions

9
(a) The $pK_a$ is approximately equal to the pH at the half-equivalence point. [1]
From the description, the equivalence point is at 25 cm³. Half-equivalence is at 12.5 cm³. At this point, $[HA] = [A^-]$, so $pH = pK_a$. Reading from a standard curve of this type, the pH at half-volume is typically around 4.8 (for ethanoic-like acids). Accept any value between 4-5 if justified as half-equivalence. [1]
(b) $K_a = 10^{-pK_a}$. If $pK_a = 4.8$, $K_a = 1.58 \times 10^{-5} \text{ mol dm}^{-3}$. [1]
(c) Sketch:

• Starts at lower pH (approx 1.0 for 0.1 M HCl). [1]
• Vertical section is larger and centered at pH 7. [1]
• Equivalence point at 25 cm³. [1]

10
(a)
(i) Precipitation occurs when $[Ag^+][X^-] > K_{sp}$.
For AgCl: $[Ag^+] = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8} \text{ mol dm}^{-3}$.
For AgI: $[Ag^+] = \frac{8.3 \times 10^{-17}}{0.010} = 8.3 \times 10^{-15} \text{ mol dm}^{-3}$.
Since a lower $[Ag^+]$ is required to precipitate AgI, AgI precipitates first. [3]
1 mark for each calculation, 1 mark for conclusion.
(ii) The second salt is AgCl. It starts precipitating when $[Ag^+]$ reaches $1.8 \times 10^{-8} \text{ mol dm}^{-3}$. [2]
Note: The question asks for concentration to start precipitation of the SECOND salt.

(b) AgCl dissolves because $Ag^+$ forms a soluble complex ion with ammonia: $[Ag(NH_3)_2]^+$. [1]
This reduces free $[Ag^+]$, shifting the solubility equilibrium of AgCl to the right.
For AgI, the $K_{sp}$ is so small that the concentration of $Ag^+$ is too low to form the complex significantly / the equilibrium constant for complex formation is not sufficient to overcome the very low solubility product. [1]

11
(a) The carboxylate anion ($RCOO^-$) has negative charge delocalized over two oxygen atoms via resonance. [1]
The phenoxide ion has charge delocalized into the ring, but less effectively onto electronegative atoms compared to the two oxygens in carboxylate. Also, the carbonyl group withdraws electrons inductively. Thus, the carboxylate ion is more stable, making the acid stronger. [1]
(b)
Moles of aspirin = $\frac{0.50}{180.0} = 2.78 \times 10^{-3} \text{ mol}$. [1]
Mole ratio Aspirin : NaOH is 1:1.
Moles NaOH = $2.78 \times 10^{-3} \text{ mol}$.
Volume NaOH = $\frac{n}{C} = \frac{2.78 \times 10^{-3}}{0.10} = 0.0278 \text{ dm}^3 = 27.8 \text{ cm}^3$. [2]
(c) The volume of NaOH required would increase. [1]
(Hydrolysis of the ester group produces a phenol group and ethanoic acid, both of which can react with NaOH, consuming more base.)

12
(a) $K_w$ is the ionic product of water, defined as $[H^+][OH^-]$. [1]
(b) In pure water, $[H^+] = [OH^-]$.
$[H^+]^2 = 1.0 \times 10^{-14} \Rightarrow [H^+] = 1.0 \times 10^{-7}$.
$pH = -\log(10^{-7}) = 7.0$. [1]
(c) Since dissociation is endothermic, increasing temperature shifts equilibrium to the right (more dissociation). [1]
$[H^+]$ increases. Since $pH = -\log[H^+]$, the pH decreases (becomes less than 7). [1]
Note: Water remains neutral because $[H^+] = [OH^-]$, but neutral pH is not always 7.