Question

A solution was prepared by dissolving 25.0 g of Ba(OH)₂ in water to make one litre of solution. How many millilitres of 0.200 M H₂SO₄ would be required to react with 25.0 ml of the Ba(OH)₂ solution?

Answer 1

Expressing both the substances in terms of normality

0.200 M H₂SO₄ = 0.200 x 20.400 N H₂SO₄

(Normality = Molarity x \(\frac{\text{Mol. wt.}}{\text{Eq. wt.}}\))

1.00 g eq. wt. Ba(OH)₂ = \(\frac{171.4 g}2 = 85.7 g\)

\(\therefore\) Normality of Ba(OH)₂ = \(\frac{\text{Strength in g/litre}}{\text{Eq. wt.}}\)

\(= \frac{25}{85.7}\)

\(= 0.292\)

Now applying normality formula

N₁V₁ (for H₂SO₄) = N₂V₂ [for Ba(OH)₂]

0.400 x V₁ = 0.292 x 25

\(\therefore\) V₁ = \(\frac{0.292 \times 25}{0.400}\) ml

= 18.3 ml of 0.200 M H₂SO₄