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) = V1 N1V1 = N2V2;   0.1  ×  V1=0.2  × 30;  V1=60  ml(here 2 stands for NaOH)Vtotal  HCl = 100 mlV1=60  ml HCl remains unreacted=100-60=40  ml40  ml  0.1  N  HCl  is now  neutralised  by  KOH  0.25  N N1V1 = N2V2KOH0.1 × 40=0.25 × V2 ;  V2=16  ml.