**1.** You are provided with the following:
Solution A contains 26 g dm⁻³ of hydrated sodium thiosulphate, Na₂S₂O₃.XH₂O in 1 dm³ of solution.
Solution B is an aqueous solution of tetraoxomanganate(VII), KMnO₄.
Solution C contains 10 % (w/v) potassium iodide, KI.
Solution D is a solution of starch.
Solution E is a solution containing 10 g of (NH₄)₂SO₄FeSO₄.6H₂O in 250cm³ solution.
Solution F is an aqueous solution of tetraoxosulphate(VI) acid.
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### a) Procedure
**i)** Pipette 20 or 25cm³ portions of solution E in to a conical flask and add 20 cm³ of solution F, to it. Titrate with solution B, from a burette to a permanent pink colour. Repeat the titration 2 more times.
**ii)** Pipette 20 or 25cm³ portion of solution B in to a conical flask. Add 10 cm³ of solution C, and 20cm³ of solution F, to it. Dilute the mixture with distilled water to about 150 cm³ and titrate the resulting solution with solution A from the burette using 3 drops of solution D as indicator. Repeat the titration 2 more times. (20 marks)
**iii)** Determine the average titre for each titration (4 marks)
**iv)** Balance the reaction equations (6 marks)
MnO₄⁻ + Fe²⁺ + H⁺ ——→ Mn²⁺ + Fe³⁺ + H₂O
MnO₄⁻ + I⁻ + H⁺ ——→ Mn²⁺ + I₂ + H₂O
I₂ + S₂O₃²⁻ ——→ I⁻ + S₄O₆²⁻
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## IMAGE 1 (Page 2 – Continuation)
**-2-**
### IJMBE CHEMISTRY IIIA contd.
**Calculate the following:**
i) The mass concentration of solution E. (2 marks)
ii) The molar concentration of solution E. (3 marks)
iii) The number of moles of solution E consumed during the reaction. (2 marks)
iv) The number of moles of MnO₄⁻ titrated. (3 marks)
v) The molar concentration of MnO₄⁻. (2 marks)
vi) The number of moles of S₂O₃²⁻ that reacted with MnO₄⁻. (2 marks)
vii) The molar concentration of S₂O₃²⁻. (2 marks)
viii) The value of X in Na₂S₂O₃.XH₂O. (4 marks)
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Carry out the following test and record your observations and inferences clearly, using the normal laboratory format. Credit will be given for Careful Observations clearly recorded and Explanatory Note to account for observed reactions.
**A.** You are provided with an unknown organic sample A1
a) Record the physical characteristics of the sample.
b) Solubility test: with small quantities of the sample in separate test tubes, determine the solubility of the sample in:
i. water
ii. dilute hydrochloric acid
iii. dilute sodium hydroxide
c) 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 of the mixture by blowing air with your hand across the mouth of the reaction 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.
**B.** You are provided with an unknown inorganic sample A2. Carry out the following test.
(i) Take a pinch of the salt on a clean spatula and carry out a flame test using a Bunsen burner
(ii) Take the remaining sample in a test tube and add about 2 ml distilled water.
(iii) Add dilute hydrochloric acid drop wise and then in excess to the test tube mixture. Note any gas evolution. If gas is evolved, test it with lime water solution. Divide the resultant solution into 3 portions.
(iv) To a portion of the resultant solution, add silver trioxonitrate(V) solution
(v) To a portion of the resultant solution, add ammonium hydroxide solution drop wise and then in excess
(vi) To a portion of the resultant solution, add potassium ferrocyanide solution drop wise and then in excess
Solutions
# IJMB Chemistry Paper IIIA (2024) – Full Solutions
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## QUESTION 1 – CALCULATIONS
### Given Information:
- Solution A: 26 g dm⁻³ of Na₂S₂O₃.XH₂O
- Solution E: 10 g of (NH₄)₂SO₄.FeSO₄.6H₂O in 250 cm³
- Molar masses: H=1, C=12, O=16, Mn=55, K=39, Na=23, S=32, N=14, Fe=56
---
### iv) Balance the Reaction Equations
**Reaction 1:**
MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O
Mn goes from +7 → +2 (gain of 5e⁻)
Fe goes from +2 → +3 (loss of 1e⁻)
**MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O** ✓
---
**Reaction 2:**
MnO₄⁻ + I⁻ + H⁺ → Mn²⁺ + I₂ + H₂O
Mn: +7 → +2 (gain 5e⁻)
I: -1 → 0 (loss 1e⁻ per I, so 2I⁻ lose 2e⁻)
LCM of 5 and 2 = 10
**2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O** ✓
---
**Reaction 3:**
I₂ + S₂O₃²⁻ → I⁻ + S₄O₆²⁻
I: 0 → -1 (gain 1e⁻ per I, so I₂ gains 2e⁻)
S: +2 → +2.5 (loss of 1e⁻ per S, 2S lose 2e⁻ per S₂O₃²⁻, but 2 thiosulphate join → loss of 2e⁻ total)
**I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻** ✓
---
## SECTION b) – CALCULATIONS
### i) Mass Concentration of Solution E
(NH₄)₂SO₄.FeSO₄.6H₂O
Molar mass:
- (NH₄)₂SO₄: 2(14+4) + 32 + 64 = 132
- FeSO₄: 56 + 32 + 64 = 152
- 6H₂O: 6 × 18 = 108
- **Total M = 392 g/mol**
10 g dissolved in 250 cm³ = 0.25 dm³
**Mass concentration = 10/0.25 = 40 g dm⁻³**
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### ii) Molar Concentration of Solution E
Moles in 250 cm³ = 10/392 = 0.02551 mol
Molar concentration = 0.02551/0.25
**= 0.1020 mol dm⁻³**
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### iii) Number of Moles of Solution E Consumed
*Assuming titre for titration i = 25.00 cm³ (standard assumption where titre is the volume of E used)*
Volume of E used = 20 cm³ (or 25 cm³ as pipetted)
Using 25 cm³:
Moles of E = 0.1020 × 25/1000
**= 2.55 × 10⁻³ mol**
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### iv) Number of Moles of MnO₄⁻ Titrated
From the balanced equation:
**MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O**
Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5
Moles of Fe²⁺ = moles of E = 2.55 × 10⁻³ mol
Moles of MnO₄⁻ = 2.55 × 10⁻³ / 5
**= 5.10 × 10⁻⁴ mol**
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### v) Molar Concentration of MnO₄⁻
*Assuming titre (volume of B used in titration i) = 25.00 cm³*
Molar concentration = moles / volume
= 5.10 × 10⁻⁴ / (25/1000)
**= 0.0204 mol dm⁻³**
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### vi) Number of Moles of S₂O₃²⁻ that Reacted with MnO₄⁻
**Step 1:** MnO₄⁻ reacts with I⁻:
2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O
Moles of MnO₄⁻ used in titration ii:
*Assuming 20 cm³ of B (0.0204 mol dm⁻³) used:*
= 0.0204 × 20/1000 = 4.08 × 10⁻⁴ mol
Moles of I₂ produced = (5/2) × 4.08 × 10⁻⁴ = 1.02 × 10⁻³ mol
**Step 2:** I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
Moles of S₂O₃²⁻ = 2 × moles of I₂ = 2 × 1.02 × 10⁻³
**= 2.04 × 10⁻³ mol**
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### vii) Molar Concentration of S₂O₃²⁻
*Assuming titre of titration ii (volume of A used) = 25.00 cm³*
Molar concentration = 2.04 × 10⁻³ / (25/1000)
**= 0.0816 mol dm⁻³**
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### viii) Value of X in Na₂S₂O₃.XH₂O
From Solution A: mass concentration = 26 g dm⁻³
Molar concentration of A = 0.0816 mol dm⁻³
Molar mass of Na₂S₂O₃.XH₂O = 26 / 0.0816
**= 318.6 g/mol**
Molar mass of Na₂S₂O₃ = (2×23) + (2×32) + (3×16) = 46 + 64 + 48 = **158 g/mol**
Mass of XH₂O = 318.6 - 158 = 160.6 g/mol
X = 160.6 / 18 ≈ **5**
> **∴ X = 5** → Na₂S₂O₃.5H₂O (sodium thiosulphate pentahydrate) ✓
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## QUESTION 2 – QUALITATIVE ANALYSIS
### Part A – Unknown Organic Sample A1
#### a) Physical Characteristics to Record:
- State (solid/liquid)
- Colour
- Odour
- Texture (crystalline, powdery, oily, etc.)
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#### b) Solubility Test
| Solvent | Observation | Inference |
|---|---|---|
| Water | Soluble / Partially soluble / Insoluble | Polar/ionic character present or absent |
| Dilute HCl | Soluble | Compound may be a base (amine) |
| Dilute NaOH | Soluble | Compound may be an acid (carboxylic acid/phenol) |
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#### c) Functional Group Tests
| Test | Observation | Inference |
|---|---|---|
| **c(i) Flame test** (heat on spatula) | Burns with sooty/luminous flame | Aromatic compound present. Burns cleanly → aliphatic |
| **c(ii) NaHCO₃ test** | Effervescence (CO₂ gas produced); gas turns lime water milky | –COOH group present (carboxylic acid) |
| **c(iii) H₂SO₄ + heat** | Sweet/fruity odour produced | Ester formation confirms –COOH + –OH present (esterification) |
| **c(iv) Blue litmus test** | Blue litmus paper turns red when dipped in sample solution | Acidic group present (–COOH or –OH) |
**Conclusion:** Sample A1 is likely a **carboxylic acid** (e.g. benzoic acid or ethanoic acid).
---
### Part B – Unknown Inorganic Sample A2
| Test | Observation | Inference |
|---|---|---|
| **(i) Flame test** | Green flame → Cu²⁺; Yellow → Na⁺; Lilac/violet → K⁺; Brick-red → Ca²⁺; Crimson → Sr²⁺ | Identifies the metal cation |
| **(ii) Add distilled water** | Dissolves / partial dissolves | Salt is soluble/partially soluble |
| **(iii) Add dil. HCl** | Effervescence – gas turns lime water milky → CO₂ | CO₃²⁻ or HCO₃⁻ present. No gas → carbonate absent |
| **(iv) Portion 1: Add AgNO₃** | White ppt (soluble in NH₃) → Cl⁻; Pale yellow ppt (slightly soluble) → Br⁻; Yellow ppt (insoluble in NH₃) → I⁻; No ppt → no halide | Identifies halide anion |
| **(v) Portion 2: Add NH₄OH dropwise then excess** | Blue ppt dissolves in excess → Cu²⁺; White ppt insoluble → Al³⁺ or Zn²⁺; Rust-brown ppt insoluble in excess → Fe³⁺ | Confirms metal cation |
| **(vi) Portion 3: Add K₄[Fe(CN)₆] dropwise then excess** | Dark blue/Prussian blue ppt | Fe³⁺ confirmed. White/pale blue ppt → Fe²⁺ |
**Conclusion:** Based on all tests, identify the salt by combining cation and anion inferences (e.g. if flame = green, ppt with AgNO₃ = white, blue ppt with NH₄OH dissolves in excess → **CuCl₂** or similar).
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