Question

A solution contains 50g/dm3 of impure sodium hydroxide. 25cm3 of it got neutralised by 14cm3 of sulfuric acid. Sulfuric acid = 1mol/dm3
H2SO4+2NaOH = Na2SO4+2H2O
-No. Of moles of sulfuric acid present in 14cm3 of the solution.
-Concentration of sodium hydroxide
-Percentage purity of sodium hydroxide.

Answer 1

Answer on Question #34706, Chemistry, Other

**Task:**

A solution contains 50 g/dm³ of impure sodium hydroxide. 25 cm³ of it got neutralised by 14 cm³ of sulfuric acid. Sulfuric acid = 1 mol/dm³

H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O

-No. Of moles of sulfuric acid present in 14 cm³ of the solution.

-Concentration of sodium hydroxide

-Percentage purity of sodium hydroxide

**Answer:**

$$C = \frac{v}{V} \quad v = CV$$ $$v(H_2SO_4) = 1 \cdot \frac{14}{1000} = 0.014 \text{ mol}$$ $$v(NaOH) = 2 \cdot v(H_2SO_4)$$ $$v(NaOH) = 2 \cdot 0.014 = 0.028 \text{ mol}$$ $$C(NaOH) = \frac{v(NaOH)}{V(NaOH)} = \frac{0.028}{0.025} = 1.12 \text{ mol/l}$$ $$v = \frac{m}{M} \quad M(NaOH) = 40 \text{ g/mol}$$ $$C(NaOH) = 1.12 \cdot 40 = 44.8 \text{ g/l}$$ $$\%NaOH = \frac{44.8}{50} \cdot 100 = 89.6\%$$

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