Question

20 mL of KOH solution was titrated with 0.20 M $H_{2}S{O}_4$ solution in a conductivity cell. The data obtained were plotted to give the graph shown below.
The concentration of the KOH solution was :

• A. $0.30mol{L}^{-1}$
• B. $0.15mol{L}^{-1}$
• C. $0.12mol{L}^{-1}$
• D. $0.075mol{L}^{-1}$

Answer 1

The correct option is A $0.30mol{L}^{-1}$

From the graph given, the equivalent point is reached at 15mL of H₂SO₄.

Milli-equivalents of an acid or base = n-factor x Molarity x volume (in mL)

n-factor for H₂SO₄ = 2 and for KOH = 1

We know that, at equivalent point,

m.eq. of acid = m.eq. of base

⇒ 2 x 0.20 x 15 = 1 x M x 20

⇒ M = 0.3 mol/L

Hence, option A is correct.