A Level H1 Chemistry Practice Paper 1

TuitionGoWhere Exam Practice (AI) - Chemistry H1 A-Level

Answer Key & Marking Scheme
Paper: Practice Paper 1 (Version 1 of 5)
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]
(Note: Do not accept "dilute". Must mention equilibrium/partial dissociation.)

(b) $\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{CH}_3\text{COO}^-\text{(aq)} + \text{H}^+\text{(aq)}$ [1]
(1 mark for correct species, 1 mark for reversible arrow and state symbols. If arrow is single, max 0.)

(c) Ethanoic acid molecules can form hydrogen bonds between the carboxyl groups (dimerization). [1]
Ethanal has permanent dipole-dipole forces but cannot form hydrogen bonds between molecules (no H attached to O/N/F). [1]
Hydrogen bonds are stronger than permanent dipole-dipole forces, requiring more energy to break.

2.
(a) Moles of $\text{NaOH} = c \times V = 0.050 \times \frac{22.4}{1000}$ [1]
$= 1.12 \times 10^{-3} \text{ mol}$

(b) From equation, ratio $\text{Acid} : \text{Base} = 1 : 1$.
Moles of acid = $1.12 \times 10^{-3} \text{ mol}$ [1]
Concentration = $\frac{n}{V} = \frac{1.12 \times 10^{-3}}{25.0/1000}$ [1]
$= 0.0448 \text{ mol dm}^{-3}$

(c) The salt formed is sodium benzoate ($\text{C}_6\text{H}_5\text{COO}^-\text{Na}^+$). [1]
The benzoate ion ($\text{C}_6\text{H}_5\text{COO}^-$) is the conjugate base of a weak acid and undergoes hydrolysis:
$\text{C}_6\text{H}_5\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{COOH} + \text{OH}^-$ [1]
This produces $\text{OH}^-$ ions, making the solution alkaline (pH > 7).

3.
(a) $K_a = \frac{[\text{HCO}_3^-][\text{H}^+]}{[\text{H}_2\text{CO}_3]}$ [1]
(Square brackets required. Water omitted.)

(b) Assumption: $[\text{H}^+] = [\text{HCO}_3^-]$ and dissociation is small so $[\text{H}_2\text{CO}_3]_{eq} \approx [\text{H}_2\text{CO}_3]_{initial}$. [1]
$K_a = \frac{[\text{H}^+]^2}{[\text{H}_2\text{CO}_3]}$
$[\text{H}^+] = \sqrt{K_a \times [\text{H}_2\text{CO}_3]} = \sqrt{4.3 \times 10^{-7} \times 0.010}$ [1]
$[\text{H}^+] = \sqrt{4.3 \times 10^{-9}} = 6.56 \times 10^{-5} \text{ mol dm}^{-3}$
$\text{pH} = -\log(6.56 \times 10^{-5}) = 4.18$ [1]

(c) $\text{SO}_2$ dissolves in water to form sulfurous acid ($\text{H}_2\text{SO}_3$). [1]
Sulfurous acid is a stronger acid than carbonic acid (or dissociates more), producing a higher concentration of $\text{H}^+$ ions, thus lowering the pH further. [1]

4.
(a) Amphoteric substances can react as both an acid and a base. [1]

(b) (i) $\text{Al}_2\text{O}_3 + 6\text{H}^+ \rightarrow 2\text{Al}^{3+} + 3\text{H}_2\text{O}$ [1]
(ii) $\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-$ [1]
(Accept $\text{AlO}_2^-$ if balanced correctly, but tetrahydroxoaluminate is preferred in modern syllabi.)

5.
(a) The buffer contains high concentrations of $\text{CH}_3\text{COOH}$ and $\text{CH}_3\text{COO}^-$. [1]
When $\text{H}^+$ is added, it reacts with the conjugate base $\text{CH}_3\text{COO}^-$:
$\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$ [1]
This removes most of the added $\text{H}^+$, keeping the pH relatively constant. [1]

(b) $\text{pH} = \text{p}K_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$
$\text{p}K_a = -\log(1.7 \times 10^{-5}) = 4.77$ [1]
$\text{pH} = 4.77 + \log\left(\frac{0.20}{0.10}\right) = 4.77 + \log(2) = 4.77 + 0.30 = 5.07$ [1]

6.
(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]
$1.8 \times 10^{-11} = 4s^3$
$s^3 = 4.5 \times 10^{-12}$
$s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$ [1]

(c) Common ion effect. [1]
Adding $\text{NaOH}$ increases $[\text{OH}^-]$. To maintain constant $K_{sp}$, the equilibrium $\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)}$ shifts to the left, precipitating more solid and decreasing solubility. [1]

7.
(a) $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.3 \times 10^{-5} \times 0.050}$ [1]
$[\text{H}^+] = \sqrt{6.5 \times 10^{-7}} = 8.06 \times 10^{-4} \text{ mol dm}^{-3}$
$\text{pH} = -\log(8.06 \times 10^{-4}) = 3.09$ [1]
(Assumption: dissociation is small. Check: $8.06 \times 10^{-4} / 0.050 \approx 1.6\% < 5\%$, valid.) [1 for correct working/answer]

(b) Curve starts at pH ~3.1. [1]
Buffer region: gradual rise, pH = pKa (~4.9) at half-equivalence. Vertical section at equivalence point (pH ~8-9). [1]
Equivalence point volume = $25.0 \text{ cm}^3$. Curve levels off at high pH (~12-13). [1]
(Labels required for marks.)

8.
(a) $\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$ [1]

(b) $\text{NH}_4\text{Cl}$ dissociates into $\text{NH}_4^+$ and $\text{Cl}^-$. [1]
$\text{NH}_4^+$ is a weak acid (conjugate of 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 of a strong acid and does not hydrolyze.
Net result: increase in $[\text{H}^+]$, so solution is acidic. [1]

9.
(a) $\text{H}_2\text{SO}_4 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_4 + 2\text{H}_2\text{O}$ [1]

(b) Calcium hydroxide (lime) is cheaper/more abundant than sodium hydroxide. [1]
It is less corrosive/safer to handle than concentrated NaOH. [1]

10.
(a) HCl is a strong acid and fully dissociates, giving high $[\text{H}^+]$. [1]
Ethanoic acid is weak and partially dissociates, giving lower $[\text{H}^+]$ for the same concentration. [1]

(b) Chlorine is electronegative and exerts an electron-withdrawing inductive effect. [1]
This pulls electron density away from the $\text{O-H}$ bond in the carboxyl group, weakening it. [1]
It also stabilizes the resulting carboxylate anion ($\text{CH}_2\text{ClCOO}^-$) by dispersing the negative charge, favoring dissociation. [1]

Section B: Data-Based & Application Questions

11.
(a) $[\text{H}^+] = 10^{-\text{pH}} = 10^{-7.4} = 3.98 \times 10^{-8} \text{ mol dm}^{-3}$ [1]

(b) $\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$
$\text{p}K_a = -\log(1.4 \times 10^{-4}) = 3.85$ [1]
$3.85 = 3.85 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$
$\log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 0$
Ratio = $1$ (or $1:1$) [1]

12.
(a) $\text{MgO}$: Reacts slowly with water to form $\text{Mg(OH)}_2$. pH is alkaline (~9-10). [1]
$\text{SiO}_2$: Insoluble in water, no reaction. pH remains neutral (~7). [1]

(b) $\text{Mg}$ is a metal with low electronegativity; $\text{Mg-O}$ bond is ionic. Oxide ion ($\text{O}^{2-}$) accepts protons (basic). [1]
$\text{Si}$ is a non-metal with higher electronegativity; $\text{Si-O}$ bonds are covalent/giant covalent. [1]
$\text{SiO}_2$ reacts with bases (acidic oxide) but not acids. The high oxidation state and covalent nature make it acidic. [1]

13.
(a) The concentration of $\text{HCl}$ decreases as it is consumed. [1]
Fewer collisions per unit time between $\text{H}^+$ ions and $\text{CaCO}_3$ surface, reducing rate. [1]

(b) Ethanoic acid is a weak acid and has a lower $[\text{H}^+]$ than $\text{HCl}$ of the same concentration. [1]
Rate depends on $[\text{H}^+]$; lower concentration leads to fewer effective collisions and slower initial rate. [1]

14.
(a) $\text{p}K_{In}$ is the pH at which the indicator is at its mid-point color change ($[\text{HIn}] = [\text{In}^-]$). [1]

(b) Strong acid-strong base titrations have a vertical pH change spanning pH 3-10. Bromothymol blue (range 6-8) changes color within this vertical section. [1]
Weak acid-strong base titrations have an equivalence point in the alkaline range (pH ~8-9). The vertical section is shorter and higher. [1]
Bromothymol blue would change color before the equivalence point is reached (or the color change would be gradual/not sharp), leading to error. [1]

15.
(a) Bacteria ferment sugar to produce lactic acid (or $\text{H}^+$ ions). [1]
$\text{H}^+$ reacts with $\text{OH}^-$ in the equilibrium, removing $\text{OH}^-$. [1]
Equilibrium shifts right to restore $\text{OH}^-$, causing dissolution of hydroxyapatite (demineralization). [1]

(b) Fluoroapatite has a lower $K_{sp}$ (is less soluble) than hydroxyapatite. [1]
This means the equilibrium lies further to the left (solid form), making it more resistant to acid attack/dissolution. [1]

16.
(a) Moles $\text{H}^+ = 0.10 \times \frac{50}{1000} = 0.0050 \text{ mol}$ [1]

(b) Moles $\text{OH}^- = 0.08 \times \frac{50}{1000} = 0.0040 \text{ mol}$ [1]

(c) $\text{H}^+$ is in excess.
Excess moles $\text{H}^+ = 0.0050 - 0.0040 = 0.0010 \text{ mol}$ [1]
Total volume = $100 \text{ cm}^3 = 0.100 \text{ dm}^3$
$[\text{H}^+] = \frac{0.0010}{0.100} = 0.010 \text{ mol dm}^{-3}$ [1]
$\text{pH} = -\log(0.010) = 2.0$ [1]

17.
(a) $\text{HOOCCH}_2\text{CH}_2\text{COOH} + 2\text{NaOH} \rightarrow \text{NaOOCCH}_2\text{CH}_2\text{COONa} + 2\text{H}_2\text{O}$ [1]

(b) Removing the first $\text{H}^+$ leaves a negative charge on the molecule. [1]
It is more difficult to remove a positive $\text{H}^+$ ion from a negatively charged species due to electrostatic attraction. [1]

18.
(a) $\text{NaCl}$. [1]
$\text{Na}^+$ (from strong base) and $\text{Cl}^-$ (from strong acid) do not hydrolyze. Solution is neutral. [1]

(b) $\text{CH}_3\text{COONa}$. [1]
$\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$. Production of $\text{OH}^-$ makes it alkaline. [1]

19.
(a) $[\text{H}^+] = 10^{-2.5} = 3.16 \times 10^{-3} \text{ mol dm}^{-3}$ [1]

(b) $K_a = \frac{[\text{H}^+]^2}{[\text{HA}]} = \frac{(3.16 \times 10^{-3})^2}{0.1}$ [1]
$K_a = \frac{1.0 \times 10^{-5}}{0.1} = 1.0 \times 10^{-4} \text{ mol dm}^{-3}$ [1]

(c) No. [1]
For a weak acid, dilution increases the degree of dissociation ($\alpha$). The $[\text{H}^+]$ does not drop by a factor of 10 exactly; it drops by less than 10. Thus pH increases by less than 1 unit. [1]

20.
(a) $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$ [1]

(b) The reaction is highly exothermic. [1]
It produces a mist of sulfuric acid aerosol which is difficult to condense and corrosive to equipment. Absorption in conc. $\text{H}_2\text{SO}_4$ forms oleum, which is then safely diluted. [1]