1.

A 100 ml solution of 0.1N-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 0.25N-KOH solution. The volume of KOH required for completing the titration is

Answer»

16 ml
32 ml
35 ml
70 ml

Solution :In the neutralization of acid and base `NxxV` of both must be EQUIVALENT
`NxxV` of HCl=0.1`xx100=10`
`NxxV` of NaOH=0.2`xx30=6`
as to OBTAIN `10NxxV` of base
`4NxxV` of base is required
`NxxV` of KOH=0.25`xx16=4`
`N_(1)V_(1)=underset(NaOH)(NxxV)+underset(KOH)(NxxV)`
`0.1xx100=0.2xx30+0.25xxV`
`10=6+0.25V`
`V=(400)/(0.25)impliesV=16ml`


Discussion

No Comment Found

Related InterviewSolutions