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

Options

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

Answer

A

Explanation

We know,1 milliequivalent`=NxxV` (in mL)
Milliequivalents of HCl `=0.2xx50=10`
Milliequivalents of NaOH=`0.1xx50=5`
`therefore` HCl and NaOH neutralise each other with equal equivalents
M.eq. of HCl left=10-5=5
Volume of new solution =50+50=100`cm^(3)`
`N_(HCl)"left"=(5)/(100)=0.05` N (from above formula)
`100 cm^(3)` of 0.05 N titrated with 0.5 N KOH with volume,
`N_(HCl)V_(HCl)=NKOHV_(KOH)`
`V=(0.05xx100)/(0.5)=10cm(3)`