50 cm^(2)of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after is completed by adding 0.5 NKOH. The volume of KOH required for completing the titration is

50 cm^(2)of 0.2 N HCl is titrated against 0.1 N NaOH solution. The tit

1.	50 cm^(2)of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after is completed by adding 0.5 NKOH. The volume of KOH required for completing the titration is
Answer» `12 cm^(3)` `10 cm^(3)` `25 cm^(3)` `10.5cm^(3)` Solution :`50 cm^(3) ` of 0.2 N HCL `=50xx0.2` millieq =10 millieq `50 cm^(3)` of 0.1 N NaOH = 5 millieq Remaining HCl = 10 - 5 millieq = 5 millieq It is neutralized by `V cm^(3)` of 0.5 N KOH i.e. `0.5xxV` millieq = 5 millieq or `V=10 cm^(3)`.