500 ml of 0.2( M ) `H_(2) SO_(4)` are mixed with 250 ml of 0.1 ( M ) `Ba( OH)_(2)` , the normality of resulting solution is -

Options

A. 0.2
B. 0.1
C. 0.5
D. 0.25

Answer

A

Explanation

500ml of 0.2 ( M) `H_(2)SO_(4)`
`= 500 xx 0.2 xx 10^(-3) `moles of `H_(2)SO_(4)`
`= 0.5 xx 0.2 xx 2 ` equiv. of `H_(2)SO_(4)`
`= 0.2 `equiv . of `H_(2)SO_(4)` 250 ml of 0.1 ( M) `Ba(OH)_(2)`
`= 250 xx0.1 xx 10^(-3)` moles of `Ba(OH)_(2)`
`= 250 xx 0.1 xx 10^(_3) xx 2 `
`= 0.05` equiv. of `Ba(OH)_(2)`
`:.` Excess `H_(2)SO_(4) = 0.2 - 0.05 = 0.15` equiv.
Hence normality of `H_(2)SO_(4)` in resulting solution `= ( 0.15)/( 750) xx 1000 = 0.2 ( N)`