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

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. $0.075\, mol \,L ^{-1}$

Answer 1

Answer: (A)

Equivalent point $=15\, mL H _2 SO _4$ (from graph)
Using $m E q$ of acid $=m E q$ of base
$\Rightarrow(2 \times 0.2) \times 15 =\left(1 \times M_{ KOH }\right) \times 20 \Rightarrow M_{ KOH }$
$=0.3 \,M$