`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

When 0.1 N NaoH is used
`N_(1)V_(1) = N_(2)V_(2)`
(HCl) (NaOH)
` 0.2 xx V = 50 xx 0.1`
`V = (50 xx 0.1)/(0.2) = 25 cm^(3)`
When 0.5 N KOH is used,
`N_(1)V_(1) = N_(3)V_(3)`
(HCl) (KOH)
`0.2 N xx 25 = 0.5 N xx V_(3)`
`V_(3) = (0.2 xx 25)/(0.5) = 10 cm^(3)`