20 mL of KOH solution was titrated with 0.20 M H 2 SO 4 solution in a conductivity cell. The data obtained were plotted to give the graph shown below The concentration of the KOH solution was

Question

20 mL of KOH solution was titrated with 0.20 M H 2 SO 4 solution in a conductivity cell. The data obtained were plotted to give the graph shown below <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/7abacd0017b05c48564912c525c132ef-.png" /> The concentration of the KOH solution was (A) 0.30 mol L -1 (B) 0.15 mol L -1 (C) 0.12 mol L -1 (D) 0.075 mol L -1

Tardigrade Admin · Accepted Answer

Equivalent point =15 mL H 2 SO 4 (from graph) Using m E q of acid =m E q of base ⇒(2 × 0.2) × 15 =(1 × M KOH ) × 20 ⇒ M KOH =0.3 M

Q. $20 m L$ of $K O H$ 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 $K O H$ solution was

$0.30 m o l L^{- 1}$

$0.15 m o l L^{- 1}$

$0.12 m o l L^{- 1}$

$0.075 m o l L^{- 1}$

Equivalent point $= 15 m L H_{2} S O_{4}$ (from graph)
Using $m Eq$ of acid $= m Eq$ of base
$\Rightarrow (2 \times 0.2) \times 15 = (1 \times M_{K O H}) \times 20 \Rightarrow M_{K O H}$
$= 0.3 M$

Q. 20mL of KOH solution was titrated with 0.20MH2​SO4​ solution in a conductivity cell. The data obtained were plotted to give the graph shown below The concentration of the KOH solution was

0.30molL−1

0.15molL−1

0.12molL−1

0.075molL−1