`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. `10 cm^(3)`
B. `12 cm^(3)`
C. `16.2 cm^(3)`
D. `21.0 cm^(3)`

Answer

A

Explanation

Equation of HCl `=(0.2 xx 50)/1000 = 5.0 xx 10^(-3)`
`5.0 xx 10^(-3)` equivalent of NaOH will neutralise `5.0 xx 10^(-3)` equivalent of HCl
Equivalent of HCl left `=(10-5) xx 10^(-3) = 5 xx 10^(-3)`
Volume of solution `=50 + 50 = 100 cm^(3)`
Normality `=(5 xx 10^(-3))/100 xx 1000 = 0.05` N
`100 cm^(3)` of 0.05 N HCl is titrated against 0.5 N KOH.
`(N_(1) xx V_(1))/(HCl) = (N_(2) xx V_(2))/(KOH)`
`therefore V_(2) = (0.05 xx 100)/0.5 = 10 cm^(3)`.