Drew asked in Science & MathematicsChemistry · 2 months ago

54.0mL of sulfuric acid is neutralized with 14.7mL of 1.09 M sodium hydroxide. Determine the concentration of the sulfuric acid solution.?

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  • 2 months ago
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    H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(ℓ)

    Mole ratio  H₂SO₄ : NaOH = 1 : 2

    Millimoles of NaOH = (1.09 mmol/mL) × (14.7 mL) = 16.0 mmol

    Millimoles of H₂SO₄ = (16.0 mmol) × (1/2) = 8.00 mmol

    Concentration of H₂SO₄ = (8.00 mmol) / (54.0 mL) = 0.148 M

    ====

    OR:

    (1.09 mol NaOH / 1000 mL NaOH solution) × (14.7 mL NaOH solution) × (1 mol H₂SO₄ / 2 mol NaOH) / (54.0/1000 L H₂SO₄ solution)

    = 0.148 M

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