A 6.60\text{ g} sample of impure ammonium tetraoxosulfate(VI), (\text{NH}_4)_2\text{SO}_4,… — JAMB UTME

A $6.60\text{ g}$ sample of impure ammonium tetraoxosulfate(VI), $(\text{NH}_4)_2\text{SO}_4$ , is heated with excess sodium hydroxide solution. The evolved ammonia gas, $\text{NH}_3$ , is absorbed completely in $100.0\text{ cm}^3$ of $0.50\text{ mol dm}^{-3}$ tetraoxosulfate(VI) acid solution, $\text{H}_2\text{SO}_4$ . The unreacted acid requires $40.0\text{ cm}^3$ of $0.50\text{ mol dm}^{-3}$ sodium hydroxide solution for complete neutralization. What is the percentage purity of the ammonium tetraoxosulfate(VI) sample? $[\text{N} = 14, \text{H} = 1, \text{S} = 32, \text{O} = 16]$

Answer: 80 %

Answer

80%

The correct answer is 80.0%. Through back-titration analysis, 0.020 mol of NaOH neutralizes 0.010 mol of unreacted excess H₂SO₄ out of the initial 0.050 mol, leaving 0.040 mol of H₂SO₄ to react with 0.080 mol of evolved NH₃ gas. Since 1 mole of pure ammonium tetraoxosulfate(VI) produces 2 moles of NH₃ gas, the sample contained 0.040 mol of pure (NH₄)₂SO₄. Multiplying by its molar mass (132 g/mol) yields 5.28 g of pure compound. Dividing 5.28 g by the total sample mass of 6.60 g and multiplying by 100 gives exactly 80.0%.

Step-by-Step Solution

Calculate the initial moles of H₂SO₄ acid solution used for absorbing ammonia.

0.050 mol H₂SO₄

Total acid available = Volume (in dm³) × Concentration (in mol dm⁻³).

Calculate the unreacted moles of H₂SO₄ from the titration with NaOH.

0.010 mol excess H₂SO₄

1 mole of H₂SO₄ reacts with 2 moles of NaOH, so excess H₂SO₄ = 0.5 × moles of NaOH used.

Calculate moles of H₂SO₄ neutralized by evolved NH₃ gas.

0.040 mol H₂SO₄ reacted

Reacted acid = Initial total acid - Excess unreacted acid.

Calculate the moles of NH₃ evolved from the sample.

0.080 mol NH₃

2 moles of NH₃ react with 1 mole of H₂SO₄.

Determine the mass of pure (NH₄)₂SO₄ present in the original sample.

5.28 g of pure (NH₄)₂SO₄

1 mole of (NH₄)₂SO₄ yields 2 moles of NH₃. Mass = moles (0.040 mol) × molar mass (132 g/mol).

Compute the percentage purity of the sample.

80%

Percentage Purity = (Mass of pure substance / Total mass of impure sample) × 100%.

Key Concept

Back-titration quantitative analysis for determining percentage purity

Estimated Time:3m 0s

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