20 cm³ of 0.09 mol dm³ solution of tetraoxosulphate (VI) acid requires 30cm³ of sodium hydroxide solution for complete neutralization. The molar…
20 cm³ of 0.09 mol dm³ solution of tetraoxosulphate (VI) acid requires 30cm³ of sodium hydroxide solution for complete neutralization. The molar concentration of sodium hydroxide is [H = 1; S = 32; 0 = 16; Na = 23]
Tetraoxosulphate (VI) acid is H₂SO₄. When it reacts with sodium hydroxide (NaOH), the balanced equation is: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. This means 1 mole of H₂SO₄ reacts with 2 moles of NaOH. The mole ratio of acid to base is 1:2.
We use the dilution/neutralization formula: CₐVₐ / CᵦVᵦ = nₐ / nᵦ, where Cₐ and Cᵦ are concentrations, Vₐ and Vᵦ are volumes, and nₐ and nᵦ are the mole ratios from the equation.
Substituting the values: (0.09 × 20) / (Cᵦ × 30) = 1/2. So 1.8 / (30 × Cᵦ) = 0.5. Cross-multiplying: 1.8 = 0.5 × 30 × Cᵦ = 15 × Cᵦ. Therefore Cᵦ = 1.8 / 15 = 0.12 mol dm⁻³.
The molar concentration of the sodium hydroxide solution is 0.12 mol dm⁻³.
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