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O Level Chemistry Practice Paper 2

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TuitionGoWhere Practice Paper - Chemistry O-Level

Answer Key & Marking Scheme
Paper: Practice Paper 2 (Version 2 of 5)
Topic: Acids, Bases and Salts

Section A: Structured Questions

1.
(a) A weak acid is an acid that only partially dissociates (or ionises) in water. [1]
(b)
(i) Hydrochloric acid will have a lower pH (around 1) than ethanoic acid (around 3) because it has a higher concentration of H+H^+ ions. [1]
(ii) Magnesium reacts more vigorously (faster rate of effervescence) with hydrochloric acid than with ethanoic acid. [1]
The concentration of H+H^+ ions is higher in hydrochloric acid, leading to more frequent successful collisions. [1]

2.
(a) CuCO3(s)+H2SO4(aq)→CuSO4(aq)+H2O(l)+CO2(g)CuCO_3(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + H_2O(l) + CO_2(g)
[1 for correct formulae, 1 for balancing and state symbols]
(b) To ensure all the sulfuric acid reacts / to ensure the salt formed is not contaminated with acid. [1]
(c)

  1. Filter the mixture to remove excess copper(II) carbonate. [1]
  2. Heat/Evaporate the filtrate to the point of crystallisation (or until saturated). [1]
  3. Allow to cool for crystals to form, then filter and dry between filter papers/in a desiccator. [1]

3.
(a) Iron (Fe). [1]
(b) High pressure increases the yield because there are fewer moles of gas on the product side (2 moles vs 4 moles), shifting equilibrium to the right. [1]
High pressure also increases the rate of reaction by increasing the frequency of collisions. [1]
(c) NH3+H2O⇌NH4++OH−NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- [1]

4.
(a) Curve starts at high pH (13), decreases gradually, then drops sharply through pH 7, and levels off at low pH. Equivalence point marked at pH 7. [2]
(b) Indicator: Methyl Orange [1]
Colour change: Yellow to Red (or Orange) [1]
(Note: Phenolphthalein is also acceptable: Pink to Colourless)

5.
(a) An amphoteric oxide is one that reacts with both acids and bases to form a salt and water. [1]
(b)
(i) ZnO+2HCl→ZnCl2+H2OZnO + 2HCl \rightarrow ZnCl_2 + H_2O [1]
(ii) ZnO+2NaOH→Na2ZnO2+H2OZnO + 2NaOH \rightarrow Na_2ZnO_2 + H_2O (Sodium zincate) [1]

6.
(a) White precipitate formed. [1]
(b) Ba2+(aq)+SO42−(aq)→BaSO4(s)Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) [2]
(c) To remove carbonate ions (or other interfering ions) which also form white precipitates with barium ions. [1]

7.
(a) Use: To neutralise acidic soil. [1]
Explanation: Calcium hydroxide is a base/alkali and reacts with acids in the soil to raise the pH. [1]
(b)
(i) Ca(OH)2+2NH4Cl→CaCl2+2H2O+2NH3Ca(OH)_2 + 2NH_4Cl \rightarrow CaCl_2 + 2H_2O + 2NH_3 [2]
(ii) Test: Damp red litmus paper. [1]
Observation: Turns blue. [1]

8.
(a) Iron(II) / Fe2+Fe^{2+} [1]
(b) Chloride / Cl−Cl^- [1]
(c) Iron(II) chloride [1]
(d) Fe2+(aq)+2OH−(aq)→Fe(OH)2(s)Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s) [1]

9.
(a) Moles of Mg = 0.12/24=0.0050.12 / 24 = 0.005 mol [1]
(b) From equation, 1 mol Mg produces 1 mol H2H_2.
Moles of H2H_2 = 0.005 mol.
Volume = 0.005×24=0.120.005 \times 24 = 0.12 dm³ (or 120 cm³). [2]

10.
(a) Titration. [1]
(b) Copper(II) oxide is insoluble in water, so the endpoint cannot be detected using an indicator in the same way (or excess solid can be filtered off, making titration unnecessary/unsuitable for precise neutralisation detection without indicator change visibility). Accept: Titration is for soluble base + acid; CuO is insoluble. [1]

Section B: Free Response Questions

11.
(a) Diprotic means one molecule of the acid can release two protons (H+H^+ ions). [1]
(b)
(i) Na2CO3+H2SO4→Na2SO4+H2O+CO2Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O + CO_2 [2]
(ii)
Moles of H2SO4=0.50×(25.0/1000)=0.0125H_2SO_4 = 0.50 \times (25.0/1000) = 0.0125 mol. [1]
Ratio Na2CO3:H2SO4Na_2CO_3 : H_2SO_4 is 1:1.
Moles of Na2CO3=0.0125Na_2CO_3 = 0.0125 mol. [1]
Mass = 0.0125×106=1.3250.0125 \times 106 = 1.325 g. [1]

12.
(a) Reacting iron with hydrochloric acid produces Iron(II) chloride (FeCl2FeCl_2), not Iron(III) chloride (FeCl3FeCl_3). [1]
Direct combination with chlorine ensures the iron is oxidised to the +3 state. [1]
(b) Hygroscopic means the substance absorbs water vapour from the air. [1]
It must be stored in a dry/airtight container to prevent it from dissolving in the absorbed water (deliquescence) or becoming impure. [1]
(c)
(i) Iron(III) hydroxide / Fe(OH)3Fe(OH)_3 [1]
(ii) Fe3+(aq)+3OH−(aq)→Fe(OH)3(s)Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s) [1]

13.
Step 1: Add dilute nitric acid to each solid.

  • The solid that effervesces (produces gas turning limewater milky) is sodium carbonate. [2]
  • The other two show no visible change.

Step 2: To the remaining two solutions (or solids dissolved in water), add dilute nitric acid followed by aqueous silver nitrate.

  • The one forming a white precipitate is sodium chloride. [2]
  • The one with no precipitate (or remaining clear) is sodium sulfate. (Note: Silver sulfate is slightly soluble, but usually no ppt forms in dilute conditions compared to AgCl. Alternatively, use Barium Nitrate for sulfate confirmation).

Alternative for Step 2: Add dilute nitric acid followed by barium nitrate to the remaining two.

  • The one forming a white precipitate is sodium sulfate. [2]
  • The one with no change is sodium chloride.

(Award marks for logical sequence, correct reagents, and distinct observations for all three). [6]

14.
(a)
(i) Brown / Reddish-brown. [1]
(ii) Cu+4HNO3→Cu(NO3)2+2NO2+2H2OCu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O [2]
(b) Magnesium is above hydrogen in the reactivity series, so it can displace hydrogen from dilute acids. [1]
Copper is below hydrogen in the reactivity series, so it cannot displace hydrogen from dilute acids. [1]

15.
(a) Lead(II) nitrate solution AND Potassium iodide solution (or Sodium iodide). [2]
(b) Pb2+(aq)+2I−(aq)→PbI2(s)Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s) [1]
(c)

  1. Filter the mixture to collect the precipitate. [1]
  2. Wash the residue with distilled water to remove soluble impurities. [1]
  3. Dry the residue in an oven or between filter papers. [1]

End of Marking Scheme