Question

A $100ml$ solution of $0.1N-HCl$ was titrated with $0.2N-NaOH$ solution .The titration was discontinued after adding$30ml$ 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. $32ml$
• B. $16ml$
• C. $35ml$
• D. $70ml$

Answer 1

The correct option is B $16ml$

In the neutralization of acid and base $N\times V$ of both must be equivalent $N\times H$ of $HCl=0.1\times 100=10N\times V$ of $NaOH=0.2\times 30=6$ as to obtain $10N\times V$ of base $4N\times V$of base is required $N\times V$ of $KOH=0.25\times 16=4N1V=N_{NaOH}\times V+N_{KOH}\times V$ $0.1\times 100=0.2\times 30+0.25V10=6+0.25v$
$V=\frac{400}{0.25}$
$V=16ml$