Study Companion

For more NEET and AIIMS mcq on Chemical arithmetic and volumetric analysis Visit
Q110) A 100 ml solution of 0.1N HCl was titrated with 0.2N NaOH solution. The titration was discontinued after adding 30ml of NaOH solution. The remaining titration was completed by adding 0.25N KOH solution. The volume of KOH required for completing the titration is ... [ DCE 1999]


Step I
miliequivalent of HCl=0.1×100=10
miliequivalent of NaOH=0.2 × 30=6
miliequivalent of HCl remains=10 -6=4
Thus 4 milimole of KOH required
4=0.25 × V
V=16 mlAnswer: (c)