Question

In a titration experiment, 13.3 mL of an aqueous H2SO4 solution was titrated with 0.6 M NaOH solution. The equivalence point in the titration was reached when 13.8 mL of the NaOH solution was added. What is the molarity of the H2SO4 solution?

Answer 1

Answer on Question #74792, Chemistry / General Chemistry

In a titration experiment, 13.3 mL of an aqueous H₂SO₄ solution was titrated with 0.6 M NaOH solution. The equivalence point in the titration was reached when 13.8 mL of the NaOH solution was added. What is the molarity of the H₂SO₄ solution.

Solution:

$$2 \mathrm{NaOH} + \mathrm{H_2SO_4} \rightleftharpoons \mathrm{Na_2SO_4} + 2 \mathrm{H_2O}$$ $$mols \left(NaOH\right) = M \times V$$ $$mols \left(NaOH\right) = \frac{0.6 M}{1 L} \times 0.0138 L = 0.00828 mol$$

Look at the coefficients in the balanced equation

$$mols \left(H_2SO_4\right) = \frac{1}{2} \times mols \left(NaOH\right)$$ $$mols \left(H_2SO_4\right) = \frac{1}{2} \times 0.00414 mol$$ $$M \left(H_2SO_4\right) = \frac{mols \left(H_2SO_4\right)}{V \left(H_2SO_4\right)}$$ $$M \left(H_2SO_4\right) = \frac{0.00414 mol}{0.0133 L} = 0.3 M$$

Answer: 0.3 M (H₂SO₄)