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A Level H1 Chemistry Practice Paper 2

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TuitionGoWhere Practice Paper - Chemistry H1 A-Level

Answer Key and Marking Scheme

Subject: Chemistry H1
Topic: Acids, Bases, and Salts
Version: 2 of 5

Section A: Structured Questions

1.
(a) A weak acid is an acid that partially dissociates (or ionizes) in water. [1]
(b) CH3COOH(aq)⇌CH3COO−(aq)+H+(aq)CH_3COOH(aq) \rightleftharpoons CH_3COO^-(aq) + H^+(aq)
Note: Must use reversible arrow ⇌\rightleftharpoons and state symbols. [1]
(c) Ethanoic acid molecules can form hydrogen bonds between molecules due to the presence of the -OH group in the carboxyl group. Propane only has weak van der Waals forces (instantaneous dipole-induced dipole). Hydrogen bonds are stronger than van der Waals forces, requiring more energy to break. [2]

2.
(a)
[OH−]=0.100 mol dm−3[OH^-] = 0.100 \text{ mol dm}^{-3}
pOH=−log⁡(0.100)=1.0pOH = -\log(0.100) = 1.0
pH=14.0−1.0=13.0pH = 14.0 - 1.0 = 13.0 [2]
(b)

  • Start pH high (~13).
  • Gradual decrease, then steep drop around 25.0 cm³.
  • Equivalence point at pH 7 (strong acid + strong base).
  • End pH low (~1).
  • Label equivalence point at volume = 25.0 cm³. [3]
    (c)
    Indicator: Methyl orange OR Bromothymol blue.
    Colour change:
  • Methyl orange: Yellow to Orange/Red.
  • Bromothymol blue: Blue to Yellow.
    (Phenolphthalein is also acceptable but less ideal due to steep range, though technically works for strong/strong. Methyl orange is standard for strong acid into base if reversing, but here acid is added to base. Methyl orange changes 3.1-4.4. The vertical section is pH 10-4. So MO works. Phenolphthalein 8.3-10 also works. Accept either with correct colour change for Acid into Base).
    Correction: Titration is NaOH (in flask) + HCl (burette). Start Alkaline. End Acidic.
  • Phenolphthalein: Pink to Colourless.
  • Methyl Orange: Yellow to Orange. [2]

3.
(a) NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq) [1]
(b)
Pair 1: NH3NH_3 (base) and NH4+NH_4^+ (conjugate acid).
Pair 2: H2OH_2O (acid) and OH−OH^- (conjugate base). [2]
(c)
Kb=[NH4+][OH−][NH3]K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}
Assume [NH4+]=[OH−]=x[NH_4^+] = [OH^-] = x and [NH3]eq≈0.10[NH_3]_{eq} \approx 0.10.
1.8×10−5=x20.101.8 \times 10^{-5} = \frac{x^2}{0.10}
x2=1.8×10−6x^2 = 1.8 \times 10^{-6}
x=[OH−]=1.34×10−3 mol dm−3x = [OH^-] = 1.34 \times 10^{-3} \text{ mol dm}^{-3}
pOH=−log⁡(1.34×10−3)=2.87pOH = -\log(1.34 \times 10^{-3}) = 2.87
pH=14.0−2.87=11.13pH = 14.0 - 2.87 = 11.13 [3]

4.
(a) An amphoteric substance can act as both an acid and a base. [1]
(b)
(i) Al2O3(s)+6H+(aq)→2Al3+(aq)+3H2O(l)Al_2O_3(s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2O(l) [2]
(ii) Al2O3(s)+2OH−(aq)+3H2O(l)→2[Al(OH)4]−(aq)Al_2O_3(s) + 2OH^-(aq) + 3H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq)
Note: Accept 2AlO2−+H2O2AlO_2^- + H_2O depending on syllabus variant, but tetrahydroxoaluminate is preferred in modern A-Level. [2]

5.
(a)
pH=pKa+log⁡([salt][acid])pH = pK_a + \log \left( \frac{[salt]}{[acid]} \right)
pKa=−log⁡(1.8×10−4)=3.74pK_a = -\log(1.8 \times 10^{-4}) = 3.74
Since [salt]=[acid][salt] = [acid], log⁡(1)=0\log(1) = 0.
pH=3.74pH = 3.74 [2]
(b)
When H+H^+ is added, it reacts with the conjugate base (HCOO−HCOO^-) to form undissociated acid (HCOOHHCOOH):
H+(aq)+HCOO−(aq)→HCOOH(aq)H^+(aq) + HCOO^-(aq) \rightarrow HCOOH(aq)
This removes the added H+H^+, keeping the pH relatively constant. [2]

6.
(a) Ksp=[Mg2+][OH−]2K_{sp} = [Mg^{2+}][OH^-]^2 [1]
(b)
Let solubility be s mol dm−3s \text{ mol dm}^{-3}.
[Mg2+]=s[Mg^{2+}] = s, [OH−]=2s[OH^-] = 2s.
Ksp=(s)(2s)2=4s3K_{sp} = (s)(2s)^2 = 4s^3
5.6×10−12=4s35.6 \times 10^{-12} = 4s^3
s3=1.4×10−12s^3 = 1.4 \times 10^{-12}
s=1.4×10−123=1.12×10−4 mol dm−3s = \sqrt[3]{1.4 \times 10^{-12}} = 1.12 \times 10^{-4} \text{ mol dm}^{-3} [3]
(c)
Common ion effect. MgCl2MgCl_2 provides Mg2+Mg^{2+} ions.
According to Le Chatelier’s principle, increasing [Mg2+][Mg^{2+}] shifts the equilibrium position to the left (precipitate side), reducing the solubility of Mg(OH)2Mg(OH)_2. [2]

7.
(a) Ka=[H+][ClO−][HClO]K_a = \frac{[H^+][ClO^-]}{[HClO]} [1]
(b)
[H+]=10−pH=10−7.4=3.98×10−8 mol dm−3[H^+] = 10^{-pH} = 10^{-7.4} = 3.98 \times 10^{-8} \text{ mol dm}^{-3} [1]
(c)
pH=pKa+log⁡([ClO−][HClO])pH = pK_a + \log \left( \frac{[ClO^-]}{[HClO]} \right)
pKa=−log⁡(3.0×10−8)=7.52pK_a = -\log(3.0 \times 10^{-8}) = 7.52
Ratio is 1 (0.050/0.0500.050/0.050). log⁡(1)=0\log(1) = 0.
pH=7.52pH = 7.52 [2]

8.
Strength refers to the degree of dissociation/ionization of the acid in water (strong acids fully dissociate, weak acids partially dissociate).
Concentration refers to the amount of acid (moles) per unit volume of solution (mol dm−3\text{mol dm}^{-3}). [2]

9.
(a) CO2(g)+H2O(l)⇌H2CO3(aq)CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq) [1]
(b) HCO3−(aq)⇌H+(aq)+CO32−(aq)HCO_3^-(aq) \rightleftharpoons H^+(aq) + CO_3^{2-}(aq) [1]

10.
(a) At half-equivalence point, pH=pKapH = pK_a. [1]
(b)
pKa=4.8pK_a = 4.8
Ka=10−4.8=1.58×10−5 mol dm−3K_a = 10^{-4.8} = 1.58 \times 10^{-5} \text{ mol dm}^{-3} [2]

Section B: Data-Based and Application Questions

11.
(a) HCl is a strong acid and fully dissociates, producing a high [H+][H^+] (0.1 mol dm−30.1 \text{ mol dm}^{-3}). Ethanoic acid is a weak acid and only partially dissociates, producing a much lower [H+][H^+]. Lower [H+][H^+] means higher pH. [2]
(b)
[H+]=10−2.9=1.26×10−3 mol dm−3[H^+] = 10^{-2.9} = 1.26 \times 10^{-3} \text{ mol dm}^{-3}
% dissociation=[H+]eq[HA]initial×100\% \text{ dissociation} = \frac{[H^+]_{eq}}{[HA]_{initial}} \times 100
=1.26×10−30.10×100=1.26%= \frac{1.26 \times 10^{-3}}{0.10} \times 100 = 1.26 \% [2]
(c) NH3NH_3 is a weak base and partially dissociates to produce OH−OH^-. NaOHNaOH is a strong base and fully dissociates. Thus, [OH−][OH^-] in ammonia is lower than in NaOH, resulting in a lower pOH and therefore a lower pH (less alkaline). [2]

12.
(a) Acid production increases [H+][H^+]. H+H^+ reacts with OH−OH^- from the equilibrium to form water. This decreases [OH−][OH^-]. According to Le Chatelier’s principle, the equilibrium shifts to the right to restore [OH−][OH^-], causing more hydroxyapatite to dissolve. [3]
(b) Fluoroapatite is less soluble (has a lower KspK_{sp}) than hydroxyapatite. Therefore, the equilibrium concentration of ions is lower, making it harder for acid to shift the equilibrium sufficiently to cause significant dissolution/decay. [2]

13.
(a) [H+]=10−3.0=1.0×10−3 mol dm−3[H^+] = 10^{-3.0} = 1.0 \times 10^{-3} \text{ mol dm}^{-3} [1]
(b)
If strong, [H+][H^+] would be 0.010 mol dm−30.010 \text{ mol dm}^{-3} (pH 2.0).
Actual [H+][H^+] is 0.001 mol dm−30.001 \text{ mol dm}^{-3}.
Since [H+]<[HX]initial[H^+] < [HX]_{initial}, it is partially dissociated, so it is a weak acid. [2]
(c)
Ka=[H+][X−][HX]K_a = \frac{[H^+][X^-]}{[HX]}
Assume [H+]=[X−]=10−3[H^+] = [X^-] = 10^{-3}.
[HX]eq≈0.010[HX]_{eq} \approx 0.010 (since dissociation is small).
Ka=(10−3)20.010=10−610−2=1.0×10−4 mol dm−3K_a = \frac{(10^{-3})^2}{0.010} = \frac{10^{-6}}{10^{-2}} = 1.0 \times 10^{-4} \text{ mol dm}^{-3} [2]

14.
(a)
pH=pKa+log⁡([HCO3−][H2CO3])pH = pK_a + \log \left( \frac{[HCO_3^-]}{[H_2CO_3]} \right)
7.4=6.1+log⁡(ratio)7.4 = 6.1 + \log (\text{ratio})
1.3=log⁡(ratio)1.3 = \log (\text{ratio})
Ratio=101.3=19.95≈20\text{Ratio} = 10^{1.3} = 19.95 \approx 20 [3]
(b)
Lactic acid adds H+H^+. The HCO3−HCO_3^- (conjugate base) in the buffer reacts with the added H+H^+ to form H2CO3H_2CO_3. This removes the excess H+H^+, preventing a significant drop in blood pH. [2]

15.
(a) pH = pKa1pK_{a1} [1]
(b) pH = pKa2pK_{a2} [1]
(c) At pH 5.0 (between pKa1=3pK_{a1}=3 and pKa2=7pK_{a2}=7), the predominant species is HA−HA^-. [1]

Section C: Extended Response and Synthesis

16.
(a)
pH=pKa+log⁡([A−][HA])pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)
pKa=−log⁡(1.7×10−5)=4.77pK_a = -\log(1.7 \times 10^{-5}) = 4.77
5.0=4.77+log⁡(ratio)5.0 = 4.77 + \log (\text{ratio})
0.23=log⁡(ratio)0.23 = \log (\text{ratio})
Ratio=100.23=1.70\text{Ratio} = 10^{0.23} = 1.70 [3]
(b)
[HA]=0.10 mol dm−3[HA] = 0.10 \text{ mol dm}^{-3}.
[A−]0.10=1.70⇒[A−]=0.170 mol dm−3\frac{[A^-]}{0.10} = 1.70 \Rightarrow [A^-] = 0.170 \text{ mol dm}^{-3}.
Volume = 250 cm3=0.250 dm3250 \text{ cm}^3 = 0.250 \text{ dm}^3.
Moles of CH3COONa=0.170×0.250=0.0425 molCH_3COONa = 0.170 \times 0.250 = 0.0425 \text{ mol}.
Mass = moles ×Mr=0.0425×82.0=3.485 g\times M_r = 0.0425 \times 82.0 = 3.485 \text{ g}.
Answer: 3.49 g (or 3.5 g). [4]

17.
(a) Mg(OH)2(s)+2HCl(aq)→MgCl2(aq)+2H2O(l)Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l) [1]
(b) Mg(OH)2Mg(OH)_2 is sparingly soluble and a weak base. It neutralizes acid gradually without causing a sudden, dangerous spike in pH or damaging tissue, unlike NaOH which is a strong, corrosive base and fully soluble. [2]
(c)
Moles Mg(OH)2=0.5058.3=0.008576 molMg(OH)_2 = \frac{0.50}{58.3} = 0.008576 \text{ mol}.
From equation, 1 mol Mg(OH)2Mg(OH)_2 reacts with 2 mol HClHCl.
Moles HCl=2×0.008576=0.01715 molHCl = 2 \times 0.008576 = 0.01715 \text{ mol}.
Volume HCl=nc=0.017150.10=0.1715 dm3=171.5 cm3HCl = \frac{n}{c} = \frac{0.01715}{0.10} = 0.1715 \text{ dm}^3 = 171.5 \text{ cm}^3.
Answer: 172 cm³. [3]

18.
(a) Propanoic acid has lower conductivity than HCl. HCl fully dissociates into many ions (H+,Cl−H^+, Cl^-). Propanoic acid partially dissociates, resulting in fewer ions to carry charge. [2]
(b)
pH: Increases (becomes less acidic) because [H+][H^+] decreases upon dilution.
% Dissociation: Increases. According to Ostwald’s dilution law (or Le Chatelier), diluting adds water. The equilibrium HA⇌H++A−HA \rightleftharpoons H^+ + A^- shifts to the right (more particles) to counteract the decrease in concentration, leading to a higher fraction of dissociated molecules. [4]

19.
(a) NH4Cl(s)→NH4+(aq)+Cl−(aq)NH_4Cl(s) \rightarrow NH_4^+(aq) + Cl^-(aq) [1]
(b) NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)
(Or NH4+⇌NH3+H+NH_4^+ \rightleftharpoons NH_3 + H^+) [1]
(c) The ammonium ion acts as a weak acid, donating protons to water to form H3O+H_3O^+ (or H+H^+). The chloride ion is the conjugate base of a strong acid and does not hydrolyze. The production of H3O+H_3O^+ makes the solution acidic. [2]

20.
(a) Acidic. [1]
(b) The salt formed is ammonium chloride (NH4ClNH_4Cl). The NH4+NH_4^+ ion is a weak acid (conjugate of weak base NH3NH_3) and undergoes hydrolysis to produce H+H^+ ions. The Cl−Cl^- ion is neutral. Thus, the solution contains excess H+H^+. [2]
(c) The equivalence point for a Weak Base + Strong Acid titration is at pH < 7 (approx pH 5-6). Phenolphthalein changes colour at pH 8.3–10.0, which is far from the equivalence point. It would change colour before the equivalence point is reached, leading to a large titration error. [2]