A 100 ml solution of 0.1 N – HCl was titrated with 0.2 N – NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25 N – KOH solution. The volume of KOH required for completing the titration is

Options

A. 16 ml
B. 32 ml
C. 35 ml
D. 70 ml

Answer

A

Explanation

In the neutralization of acid and base `NxxV` of both must be equivalent
`NxxV` of `HCl=0.1xx100=10`
`NxxV` of `NaOH=0.2xx30=6`
`N_(1)V_(1)(ml)=underset(NaOH)(NxxV)(ml)+Nxxunderset(KOH)(V)(ml)`
`0.1xx100=0.2xx30+0.25xxV`
10=6+0.25V
`V=(400)/(25)" "V=16ml`