Question

A `100 ml` solution of `0.1 N HCl` was titrated with `0.2`? `N NaOH` solution. The titration. The remaining titration war completed by adding `0.25 N KOH` solution. The volume of `KOH` required for completing the titration is
A. (a)` 70 ml`
B. (b)` 32 ml`
C. (c )` 35 ml`
D. (d)` 16 ml`

Answer 1

Correct Answer - D
Volume m of HCl neutralised by `NaOH`
`=("Caustic soda")=V_(1)`
`N_(1)V_(1)=N_(2)V_(2), 0.1xxV_(1)=0.2xx60, V_(1)=60 ml`
V total (HCl)=100 ml
`V_(1)=60 ml`
`=40 ml`
`40 ml 0.1 N HCl` is now netralised by `KOH (0.25N) rarr (HCl)`
`N_(1)V_(1)=N_(2)V_(2) (KOH)`
`0.1xx40=0.25xxV_(2), V_(2)=16 ml`.