`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 graphe shown below.
The concentration of the `KOH` solution was

Options

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

a

Explanation

Equivalent point `=15mL H_(2)SO_(4)(` from graph`)`
Using `mEq` of acid `=mEq` of base
`implies 2xx0.2)xx15=(1xxM_(KOH))xx20impliesM_(KOH)=0.3M`