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20 Ml of KOH solution was titrated with ...

`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

A

`0.30 mol L^(-1)`

B

`0.15 mol L^(-1)`

C

`0.12 mol L^(-1)`

D

`0.075 mol L^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
a
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`
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