Secondary 4 Pure Chemistry Preliminary Examination Paper 5

TuitionGoWhere Practice Paper - Pure Chemistry Secondary 4

ANSWER KEY & MARKING SCHEME
PRELIMINARY EXAMINATION 2024
Version 5 of 5

Subject: Pure Chemistry
Level: Secondary 4
Total Marks: 60

SECTION A: Structured Questions (40 Marks)

1.
(a) Observations:

• Effervescence / Bubbles of gas produced. [1]
• Green solid (copper(II) carbonate) dissolves / disappears to form a blue solution. [1]
  (Note: "Solid disappears" alone is insufficient; must mention colour change or effervescence)

(b) Equation:
$CuCO_3(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + H_2O(l) + CO_2(g)$
[1] for correct formulae, [1] for balancing and state symbols.

(c) Reason for excess:
To ensure all the sulfuric acid is reacted / neutralised. [1]

(d) Method:

1. Filter the mixture to remove excess copper(II) carbonate. [1]
2. Heat the filtrate to evaporate some water / until saturated (crystallisation point). [1]
3. Allow the solution to cool to form crystals, then filter and dry between filter papers. [1]

2.
(a) Ionic Equation:
$Mg(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H_2(g)$
[1]

(b) Graph:

• Curve A (HCl) starts steeper than Curve B (Ethanoic). [1]
• Both curves level off at the same final volume of gas (since Mg is limiting or acids are in excess/equivalent moles, but typically Mg is limiting in these comparisons unless stated otherwise; assuming excess acid, same moles of H+ available? No, weak acid has fewer H+ initially but total moles are same. If Mg is limiting, final volume is same. If Acid is limiting, final volume is same because total moles of acid are same). Standard assumption: Mg is limiting or both acids have same total moles of replaceable H.
• Curve A reaches plateau earlier than Curve B. [1]
• Labels: Curve A = HCl, Curve B = Ethanoic Acid. [1]

(c) Explanation:

• Hydrochloric acid is a strong acid and fully ionised, giving a higher concentration of $H^+$ ions. [1]
• Ethanoic acid is a weak acid and partially ionised, giving a lower concentration of $H^+$ ions.
• Higher concentration of $H^+$ leads to a higher frequency of effective collisions between $H^+$ and Mg atoms. [1]

3.
(a) Calculation:

• $M_r$ of $(NH_4)_2SO_4 = 2(14 + 4(1)) + 32 + 4(16) = 2(18) + 32 + 64 = 36 + 32 + 64 = 132$. [1]
• Mass of N = $2 \times 14 = 28$.
• % N = $(28 / 132) \times 100 = 21.2\%$. [1]

(b)
(i) Acid: Sulfuric acid ($H_2SO_4$). [0.5]
Alkali: Aqueous ammonia ($NH_3(aq)$) or Ammonium hydroxide. [0.5]
(Note: Must be ammonia/ammonium hydroxide, not NaOH, to get ammonium salt)

(ii) Reason for indicator:
To detect the end-point / neutralisation point. [1]

(iii) Reason for repeat without indicator:
To prevent the indicator from contaminating the salt / to obtain pure salt crystals. [1]

4.
(a) Cation: Zinc ion ($Zn^{2+}$). [1]
(Note: Al³⁺ also gives white ppt soluble in excess NaOH, but Al(OH)₃ is insoluble in excess ammonia. Zn(OH)₂ is soluble in excess ammonia. Pb²⁺ is insoluble in excess ammonia. So it must be Zn²⁺.)

(b) Anion: Sulfate ion ($SO_4^{2-}$). [1]

(c) Ionic Equation:
$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$
[1]

(d)
(i) Gas: Ammonia ($NH_3$). [1]
(ii) Confirmation: Confirms the presence of ammonium ions ($NH_4^+$).
*(Wait, Test 1 and 2 identified Zn²⁺. Test 4 identifies NH₄⁺. This implies Substance X contains BOTH Zn²⁺ and NH₄⁺? Or is X a mixture? The question says "Substance X". If X is a single compound, it could be a double salt or the question implies X is a mixture. However, standard QA questions usually test one cation. Let's re-read Test 1 & 2. White ppt soluble in excess NaOH AND excess NH₃ is characteristic of Zn²⁺. Test 4 produces NH₃ gas with NaOH. This confirms NH₄⁺. So X contains Zn²⁺ and NH₄⁺? Or did I misinterpret Test 1/2?
Actually, if X is $(NH_4)_2Zn(SO_4)_2$ or similar, it fits. Or simply, the question asks what Test 4 confirms. It confirms Ammonium ions.
Correction: If Test 1/2 proved Zn²⁺, and Test 4 proves NH₄⁺, then X contains both. But usually, these questions target one cation. Let's look at Test 1 again. "White precipitate... dissolves in excess NaOH". Could be Al³⁺, Zn²⁺, Pb²⁺. Test 2: "Dissolves in excess ammonia". Only Zn²⁺ and Cu²⁺ (blue) do this. Al³⁺ and Pb²⁺ do not. So Cation is Zn²⁺.
Test 4: Heating with NaOH produces NH₃. This means NH₄⁺ is present.
So Substance X contains Zn²⁺ and NH₄⁺?
Alternative interpretation: Maybe Test 1/2 was on a different sample? No, "solution of X".
Standard Exam Logic: Often, students miss that Zn²⁺ does NOT produce NH₃ gas. NH₄⁺ does. So X is likely a mixture or a complex salt.
However, for the purpose of the mark scheme:
(ii) It confirms the presence of ammonium ions ($NH_4^+$). [1]

5.
(a) Definition:
An oxide that reacts with both acids and bases to form salt and water. [1]

(b)
(i) $ZnO(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2O(l)$ [2]
(ii) $ZnO(s) + 2NaOH(aq) \rightarrow Na_2ZnO_2(aq) + H_2O(l)$
(Or $ZnO + 2OH^- \rightarrow [Zn(OH)_4]^{2-}$)
[2] for correct formulae and balancing.

6.
(a)
(i) Calcium hydroxide (slaked lime) / Calcium carbonate (limestone). [1]
(ii) Calcium hydroxide/carbonate is less corrosive / cheaper / safer to handle than sodium hydroxide. [1]

(b)
Gas: Sulfur dioxide ($SO_2$) OR Nitrogen oxides ($NO_x$). [1]
Source: Burning of fossil fuels / Vehicle exhausts. [1]

7.
(a) Observation: White precipitate. [1]

(b) Reason:
To remove carbonate ions (or other interfering ions) that might also form a white precipitate with barium ions (e.g., $BaCO_3$). [1]

(c) Student Correct?
No. [1]
Silver sulfate is slightly soluble / does not form a precipitate under standard dilute conditions, OR Silver nitrate tests for halides. While $Ag_2SO_4$ can precipitate if concentrated, the standard test for sulfate is Barium. More importantly, AgCl is white, AgI is yellow. Sulfate does not give a characteristic white ppt with AgNO₃ in the same definitive way as Chloride.
Better Answer: No, because silver sulfate is sparingly soluble and may not precipitate immediately, whereas barium sulfate is insoluble. Also, silver nitrate is the specific test for halides. [1]

8.
(a) Reason:
Potassium is a very reactive metal (Group 1). The reaction with acid would be violent / explosive / dangerous. [1]

(b) Method:

1. Perform titration using pipette (alkali) and burette (acid) with indicator to find exact volume needed. [1]
2. Repeat titration without indicator using the exact volumes determined. [1]
3. Evaporate the solution to crystallisation point (saturation). [1]
4. Cool, filter, wash with cold distilled water, and dry. [1]

SECTION B: Free-Response Questions (20 Marks)

9.
(a) Equation:
$2FeSO_4 \cdot 7H_2O(s) \rightarrow Fe_2O_3(s) + SO_2(g) + SO_3(g) + 7H_2O(g)$
*(Note: Balancing Fe: 2 on left, 2 on right. S: 2 on left, 1+1 on right. H: 14 on left, 14 on right. O: $2(4+7)=22$ on left. Right: $3 + 2 + 3 + 7 = 15$? Wait.
Let's balance properly.
$2(FeSO_4 \cdot 7H_2O) \rightarrow Fe_2O_3 + SO_2 + SO_3 + 14H_2O$?
Left O: $2 \times (4 + 7) = 22$.
Right O: $3 (Fe_2O_3) + 2 (SO_2) + 3 (SO_3) + 14 (H_2O) = 22$.
Yes.
Equation: $2FeSO_4 \cdot 7H_2O(s) \rightarrow Fe_2O_3(s) + SO_2(g) + SO_3(g) + 14H_2O(g)$
[2]

(b)
(i) U-tube: Colourless liquid / water droplets condense. [1]
(ii) KMnO₄: Purple solution turns colourless / decolourises. [1]
Explanation: Sulfur dioxide is a reducing agent and reduces the manganate(VII) ions. [1]

(c)
(i) Reaction Type: Oxidation. [1]
(ii) Ionic Equation:
$Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-$
(Or full equation with $O_2$ and $H^+$, but ionic half-equation is often accepted for "change")
Better: $4Fe^{2+} + O_2 + 4H^+ \rightarrow 4Fe^{3+} + 2H_2O$.
However, simple oxidation of ion: $Fe^{2+} \rightarrow Fe^{3+} + e^-$. [1]

10.
Tests:

1. Add dilute nitric acid:

   • Sodium Carbonate: Effervescence / bubbles of gas ($CO_2$) produced. Gas turns limewater milky. [2]
   • Sodium Chloride & Iodide: No observable change.

2. Add aqueous silver nitrate to the remaining two solutions (after acidifying with nitric acid):

   • Sodium Chloride: White precipitate ($AgCl$) formed. [2]
   • Sodium Iodide: Yellow precipitate ($AgI$) formed. [2]

(Alternative: Use Lead(II) nitrate. Chloride = White ppt, Iodide = Yellow ppt. Carbonate = White ppt but dissolves in acid with effervescence.)
Marking:

• Correct reagent for Carbonate (Acid) + Observation. [2]
• Correct reagent for Halides (AgNO₃ or Pb(NO₃)₂) + Observations distinguishing Cl and I. [4]

11.
(a) Solutions:

1. Lead(II) nitrate solution. [1]
2. Potassium iodide solution (or Sodium iodide). [1]

(b) Ionic Equation:
$Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s)$
[1]

(c) Method:

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

(d) Reason:
Lead(II) iodide is insoluble, so it would coat the unreacted lead(II) oxide, preventing further reaction. [1]

12.
(a) Difference:

• Strong acid ionises / dissociates completely in water. [1]
• Weak acid ionises / dissociates partially in water. [1]

(b)
(i) Lower pH: Hydrochloric acid. [1]

(ii) Explanation:

• Neutralisation depends on the total number of moles of acid available, not just the initial $H^+$ concentration. [1]
• As $H^+$ ions are consumed by the alkali, the equilibrium of the weak acid shifts to the right (Le Chatelier's principle), causing more ethanoic acid to ionise until all acid molecules have reacted. Thus, the total moles of $H^+$ neutralised are the same for both acids. [1]

(End of Marking Scheme)