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Standard X
Chemistry
Electrolysis and Electrolytes
Question

20 mL of KOH solution was titrated with 0.20 M H2SO4 solution in a conductivity cell. The data obtained were plotted to give the graph shown below.
The concentration of the KOH solution was :
259483_7c4e7de87daa4622984cce0d24f44051.png
  1. 0.30molL−1
  2. 0.15molL−1
  3. 0.12molL−1
  4. 0.075molL−1

A
0.12molL−1
B
0.15molL−1
C
0.075molL−1
D
0.30molL−1
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Solution
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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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20 mL of KOH solution was titrated with 0.20 M H2SO4 solution in a conductivity cell. The data obtained were plotted to give the graph shown below.
The concentration of the KOH solution was :
259483_7c4e7de87daa4622984cce0d24f44051.png
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