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

Options

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

Answer

B

Explanation

`{:("Acid",=,"Base"),(N_(1)V_(1),=,N_(2)V_(2)),(0.2 xx V_(1),=,0.1 xx 50):}`
`:. V_(1)=(0.1 xx 50)/(0.2 ) =25cm^(3)`
Thus , `25cm^(3)` is the volume of acid neutralized by `50 cm^(3)` of 0.1 N NaOH
`:. ` volume of 0.2 N HCl left behind
`=50-25=25 cm^(3)`
This is neutralized by 0.5 KOH
Acid= Base
`N_(1)V_(1) = N_(2)V_(2)`
`0.2 xx 25 =0.5 xx V_(2) :. V_(2)=(0.2 xx 25)/(0.5 ) =10cm^(3)`
`:. `Volume of 0.5 N KOH required `=10cm^(3)`