`"50 cm"^(3)` of `0.2" N HCl"` is titrated against `0.1" N NaOH"` The remaining titration adding `"50 cm"^(3)` of NaOH. The remaining titration is completed by adding 0.5 N KOH The volume of KOH required for completing the titration is

Options

A. A) `"12 cm"^(3)`
B. B) `"10 cm"^(3)`
C. C) `"25 cm"^(3)`
D. D) `10.5" cm"^(3)`

Answer

B

Explanation

Milli - equivqlents of `HCl=50xx0.2=10` milli-eq.
Milli-equivalents of `NaOH=50xx0.1=5` milli - eq.
Remaining milli - equivalents of acid = 5
Required volume of `KOH=0.5xxV_(1)=5`
`V_(1)=10cm^(3)`