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50 cm^3 of 0.2N HCl is titrated against...

`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 :

A

`12 cm^3`

B

`10 cm^3`

C

`25cm^3`

D

`10.5 cm^3`

Text Solution

Verified by Experts

The correct Answer is:
B
`HCl+ NaOH to NaCl + H_2O`
` 50 cm^2 ` of 0.2 N `HCl = (0.2 )/(1000) xx 50`
= 0.010 g equiv .
`50 cm^3` of 0.1 N `NaOH = (0.1)/(1000) xx 50`
= 0.005 g equiv .
No. of g equiv. of HCl left = (0.10 - 0.005)
= 0.005 g equiv .
` therefore ` volume of KOH required
`= (1000)/(0.5) xx 0.005 = 10 cm^3`
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