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

Step-by-step solution

In the neutralization of acid and base N × V of both must be equivalent N × V   of   HCl   =   0 .1 × 100   =   10 N × V   of   NaOH =   0 .2 × 30 = 6 N 1 V 1 ml = N × V NaOH ml + N × V ml KOH 0 .1 × 100   =   0 .2 × 30 + 0 .25 × V 10   =   6 + 0 .25   V V = 400 25 =   16   ml