Question

25 ml of a solution of hydrochloric acid containing 7.3 g of the acid per litre neutralized 30 ml of caustic soda solution; 20 ml of this alkali solution neutralized 24 ml of a solution of sulphuric acid. Calculate the normality and strength (g/litre) of the sulphuric acid.

Answer 1

Strength of HCl = 7.3 g/litre

Eq. wt. of HCl = 36.5

\(\therefore\) Normality of HCl = \(\frac{7.3}{36.5} = \frac 15 \) N

30 ml of NaOH solution = 25 ml of N/5 HCl solution

\(\therefore\) Normality of NaOH solution = \(\frac{25}{5 \times 30} = \frac 16 \) N

Again, 24 ml of H₂SO₄ solution = 20 ml of \(\frac N6\) NaOH solution

\(\therefore\) Normality of H₂SO₄ solution = \(\frac{20}{6 \times 24} = \frac5{36}\) N = 0.139 N

\(\therefore\) Strength of H₂SO₄ solution = 0.139 x 49 = 6.8 g/litre

(\(\because\) Eq. wt. of H₂SO₄ = 49)