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

Volume of HCl neutralised by NaOH (Caustic soda) = $V_1$

 $\begin{array}{l}{\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2;0.1\times {\mathrm{V}}_1=0.2\times 30;{\mathrm{V}}_1=60\mathrm{ml}(\mathrm{here} 2 \mathrm{stands} \mathrm{for} \mathrm{NaOH})\\ {\mathrm{V}}_{\mathrm{total}}\left(\mathrm{HCl}\right)=100\mathrm{ml}\\ {\mathrm{V}}_1=60\mathrm{ml}\\ \mathrm{HCl} \mathrm{remains} \mathrm{unreacted}=100-60=40\mathrm{ml}\\ 40\mathrm{ml}0.1\mathrm{N}\mathrm{HCl}\mathrm{is}\mathrm{now}\mathrm{neutralised}\mathrm{by}\mathrm{KOH}\left(0.25\mathrm{N}\right)\\ {\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2\left(\mathrm{KOH}\right)\\ 0.1\times 40=0.25\times {\mathrm{V}}_2;{\mathrm{V}}_2=16\mathrm{ml}.\end{array}$