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 titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is:

The correct option is D

16 ml

In the complete neutralization of acid and base, the Milli-equivalent of acid and base must be equivalent.

Milli-equivalent of HCl = ${\mathrm{N}}_1\times {\mathrm{V}}_1=0.1\times 100=10$

Milli-equivalent of NaOH = ${\mathrm{N}}_2\times {\mathrm{V}}_2=0.2\times 30=6$

Let the titration be completed by adding a ‘V₃’ volume of 0.25 N KOH solution.

${\mathrm{N}}_1\times {\mathrm{V}}_1={\mathrm{N}}_2\times {\mathrm{V}}_2+{\mathrm{N}}_3\times {\mathrm{V}}_3$

$0.1\times 100=0.2\times 30+0.25\times {\mathrm{V}}_3$

$10=6+0.25\times {\mathrm{V}}_3$

$4=0.25\times {\mathrm{V}}_3$

$\frac{4}{0.25}={\mathrm{V}}_3$

$16={\mathrm{V}}_3$

The titration is completed by adding a ‘16ml’ volume of 0.25 N KOH solution.

Hence, option (d) is correct.