Question

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

• A. $16 \, ml$
• B. $32 \, ml$
• C. $35 \, ml$
• D. $70 \, ml$

Answer 1

Answer: (A)

In the neutralization of acid and base $N\times V$ of both must be equivalent

$N\times V \, of \, HCl \, = \, 0.1\times 100 \, = \, 10$

$N\times V \, of \, NaOH= \, 0.2\times 30=6$

$N_{1}V_{1}\left(ml\right)=\underset{NaOH}{N \times V}\left(ml\right)+\underset{KOH}{N \times V \left(ml\right)}$

$0.1\times 100 \, = \, 0.2\times 30+0.25\times V$

$10 \, = \, 6+0.25 \, V$

$V=\frac{400}{25} \, \, V \, = \, 16 \, ml$