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 :

0.30 m o l L^{- 1}

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0.15 m o l L^{- 1}

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0.12 m o l L^{- 1}

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0.075 m o l L^{- 1}

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Solution

The correct option is A $0.30 m o l L^{- 1}$
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.

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