2022 IJMB chemistry paper 3

QUESTION 1

(a) Procedure:

(i) Pipette 20 or 25 cm³ of solution B into a conical flask. Add 2 or 3 drops of phenolphthalein indicator and titrate against solution A from a burette to a colourless end point.

(ii) Repeat the above procedure twice more using fresh portions of solution B. (16 marks)

(iii) Calculate the average titre. (2 marks)

(iv) Write the equation of the reaction. (2 marks)

(b) Calculate the following:

(i) Molar concentration of KOH in solution B (3 marks)

(ii) Number of moles of KOH titrated (3 marks)

(iii) Number of moles of HX consumed in the titration (2 marks)

(iv) Molar concentration of HX in solution A (3 marks)

(v) Molar mass of the weak acid (3 marks)

(vi) Relative molecular mass of X in the formula HX (2 marks)

(vii) pH of solution A (4 marks)

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(i) Distinguish between the end point and the equivalence point in acid-base volumetric analysis. (4 marks)

(ii) Explain why phenolphthalein and not methyl orange is used in the above titration. (3 marks)

(iii) Give one major source of error in the experiment (a) above. (1 mark)

(iv) Give two common features of all volumetric analyses. (2 marks)

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QUESTION 2

A. Sample A1 = Benzoic Acid

(a) Record the physical characteristics of the sample.

(b) Solubility test: with small quantities of the sample in separate test tubes, determine the solubility in:

• (i) Water

• (ii) Dilute hydrochloric acid

• (iii) Dilute sodium hydroxide

© Functional group test:

(i) Using a cleaned spatula, heat a small quantity of the sample in a Bunsen burner flame.

(ii) To a small quantity of the sample dissolved in ethanol, add small amount of NaHCO₃, warm the resulting mixture and note any gas evolution. If gas is evolved, test its reaction with lime water.

(iii) To a small quantity of the sample dissolved in ethanol, add 1 ml of H₂SO₄ and warm the resulting mixture. Perceive the odour by blowing air across the mouth of the test tube.

(iv) Dissolve a small amount of the sample in ethanol. Dip a moist red litmus paper into a dilute solution of NaOH or NH₃; the colour changes to blue. Then dip this blue litmus paper into the sample solution. (15 marks)

B. Sample A2 = Copper(II) trioxonitrate(V)

(a) Carry out two (2) confirmatory tests for the cation.

(b) Carry out two (2) distinguishing tests for the anion. (15 marks)

SOLUTIONS

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QUESTION 1(a)

(iii) Average Titre Calculation

A standard titration table (assumed typical values for this setup):

| Titration | Rough | 1st | 2nd |

|-----------|-------|-----|-----|

| Final (cm³) | 25.20 | 25.10 | 25.15 |

| Initial (cm³) | 0.00 | 0.00 | 0.00 |

| Titre (cm³) | 25.20 | 25.10 | 25.15 |

Average titre = (25.10 + 25.15) / 2 = 25.13 cm³

(Rough reading is discarded)

(iv) Equation of Reaction:

HX(aq) + KOH(aq) → KX(aq) + H₂O(l)

This is a 1:1 mole ratio neutralisation reaction.

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QUESTION 1(b)

(i) Molar Concentration of KOH in Solution B

Molar mass of KOH = 39 + 16 + 1 = 56 g/mol

$$C = \frac{mass}{M_r \times V(L)} = \frac{2.80}{56 \times 0.500} = \frac{2.80}{28} = \boxed{0.10 \text{ mol/dm}^3}$$

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(ii) Number of Moles of KOH Titrated

Volume of KOH used = 25.13 cm³ = 0.02513 dm³

$$n_{KOH} = C \times V = 0.10 \times 0.02513 = \boxed{2.513 \times 10^{-3} \text{ mol}}$$

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(iii) Number of Moles of HX Consumed

From the equation HX : KOH = 1 : 1

$$n_{HX} = n_{KOH} = \boxed{2.513 \times 10^{-3} \text{ mol}}$$

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(iv) Molar Concentration of HX in Solution A

Volume of HX used from burette = 25.13 cm³ = 0.02513 dm³

$$C_{HX} = \frac{n}{V} = \frac{2.513 \times 10^{-3}}{0.02513} = \boxed{0.10 \text{ mol/dm}^3}$$

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(v) Molar Mass of the Weak Acid HX

Total moles of HX in 250 cm³:

$$n = 0.10 \times 0.250 = 0.025 \text{ mol}$$

$$M_r(HX) = \frac{mass}{moles} = \frac{2.30}{0.025} = \boxed{92 \text{ g/mol}}$$

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(vi) Relative Molecular Mass of X

$$M_r(X) = M_r(HX) - M_r(H) = 92 - 1 = \boxed{91}$$

(Note: X with RMM 91 is consistent with the toluoate/tropylium group — HX here is consistent with a weak organic acid)

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(vii) pH of Solution A

Using Ka = 1.8 × 10⁻⁴ and C = 0.10 mol/dm³

For a weak acid: [H⁺] = √(Ka × C)

$$[H^+] = \sqrt{1.8 \times 10^{-4} \times 0.10} = \sqrt{1.8 \times 10^{-5}}$$

$$[H^+] = \sqrt{18 \times 10^{-6}} = 4.243 \times 10^{-3} \text{ mol/dm}^3$$

$$pH = -\log(4.243 \times 10^{-3}) = -\log(4.243) - \log(10^{-3})$$

$$pH = -0.6277 + 3 = \boxed{2.37}$$

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QUESTION 1©

(i) End Point vs Equivalence Point

| | End Point | Equivalence Point |

|—|---|—|

| Definition | The point where the indicator changes colour, signalling the titration should stop | The point where moles of acid exactly equal moles of base (stoichiometric completion) |

| Nature | Experimental/observable | Theoretical |

| Detection | By indicator colour change | By calculation or pH curve |

| Coincidence | May not be exact | The “true” neutralisation point |

The end point is ideally as close to the equivalence point as possible. Any difference between them constitutes titration error.

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(ii) Why Phenolphthalein and Not Methyl Orange

HX is a weak acid and KOH is a strong base. At the equivalence point, the solution is basic (pH > 7) due to hydrolysis of the salt KX. Phenolphthalein has a transition range of pH 8.3–10.0, which falls within the steep portion of the titration curve for a weak acid–strong base system. Methyl orange transitions at pH 3.1–4.4 (acidic range), which does not correspond to the equivalence point of this reaction and would give an inaccurate result.

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(iii) One Major Source of Error

Failure to rinse the burette with solution A (HX) before filling it, causing dilution of the acid and inaccurate titre readings.

(Other acceptable answers: parallax error in reading the meniscus; overshooting the end point; not swirling adequately)

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(iv) Two Common Features of All Volumetric Analyses

1. A standard solution (known concentration) must be used in the titration.

2. A sharp, detectable end point must be achievable — typically via an indicator or instrumental method.

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QUESTION 2A — Sample A1: Benzoic Acid (C₆H₅COOH)

(a) Physical Characteristics

• White crystalline solid

• Has a faint, characteristic aromatic odour

• Melting point ~122°C

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(b) Solubility Tests

| Solvent | Observation | Inference |

|—|---|—|

| Water | Sparingly soluble (slightly dissolves) | Weak/partial polar character |

| Dilute HCl | Insoluble | No reaction with acid; not a base |

| Dilute NaOH | Dissolves readily | Acidic compound; forms soluble sodium benzoate: C₆H₅COOH + NaOH → C₆H₅COONa + H₂O |

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© Functional Group Tests

(i) Flame Test:

| Observation | Inference |

|—|---|

| Burns with a sooty/luminous yellow flame | Aromatic compound with high carbon content (C:H ratio is high) |

(ii) NaHCO₃ Test:

| Observation | Inference |

|—|---|

| Effervescence (bubbles) produced; gas turns lime water milky | CO₂ gas evolved confirms –COOH (carboxylic acid) group present: C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂ |

(iii) H₂SO₄ + Ethanol (Esterification Test):

| Observation | Inference |

|—|---|

| A fruity/pleasant odour is perceived | Ester formed — confirms carboxylic acid group: C₆H₅COOH + C₂H₅OH ⇌ C₆H₅COOC₂H₅ + H₂O |

(iv) Litmus Test:

| Observation | Inference |

|—|---|

| Blue litmus paper turns red when dipped in the sample solution | Sample is acidic — confirms –COOH group |

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QUESTION 2B — Sample A2: Copper(II) Trioxonitrate(V) [Cu(NO₃)₂]

(a) Two Confirmatory Tests for the Cation (Cu²⁺)

Test 1 — Aqueous NaOH:

| Procedure | Observation | Inference |

|—|---|—|

| Add excess dilute NaOH to solution of sample | Blue gelatinous precipitate formed, insoluble in excess NaOH | Cu²⁺ confirmed: Cu²⁺ + 2OH⁻ → Cu(OH)₂↓ (blue) |

Test 2 — Aqueous Ammonia:

| Procedure | Observation | Inference |

|—|---|—|

| Add dilute NH₃ solution, then add in excess | Blue precipitate forms initially; dissolves in excess NH₃ to give a deep blue solution | Cu²⁺ confirmed: excess NH₃ forms tetraamminecopper(II) complex [Cu(NH₃)₄]²⁺ |

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(b) Two Distinguishing Tests for the Anion (NO₃⁻)

Test 1 — Brown Ring Test:

| Procedure | Observation | Inference |

|—|---|—|

| Add fresh FeSO₄ solution to sample; carefully add conc. H₂SO₄ down the side of the tube | A brown ring forms at the junction of the two liquids | NO₃⁻ confirmed: Fe²⁺ reduces NO₃⁻ to NO, which complexes with Fe²⁺ to form [Fe(NO)]²⁺ |

Test 2 — Dilute H₂SO₄ + Copper turnings:

| Procedure | Observation | Inference |

|—|---|—|

| Add dilute H₂SO₄ to sample, add copper turnings and warm | Colourless gas (turns brown in air) evolved; solution turns blue | NO₃⁻ confirmed: Cu + 4HNO₃(dilute) → Cu(NO₃)₂ + 2NO↑ + 2H₂O; NO oxidises in air to brown NO₂ |