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A 100 ml solution of 0.1N-HCl was titrat...

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

A

16 mL

B

32 mL

C

35 mL

D

70 mL

Text Solution

Verified by Experts

The correct Answer is:
A
In the neutralization of acid and base N × V of both must be equivalent
`N xx V` of HCl ` = 0.1 xx 100 = 10`
`N xx V` of NaOH `= 0.2 xx 30 = 6`
as to obtain 10 N `xx` V of base
`4 N xx V` of base is required
`N_(1)V_(1) = underset("NaOH")(N xx V) + underset("KOH")(N xx V)`
`0.1 xx 100 = 0.2 xx 30 + 0.25 V`
` 10 = +6 0.25 V`
`V = (400)/(0.25) rArr V = 16mL`
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