A piece of `Al` wieghing `2.7 g` is titrated with `75.0 mL` of `H_(2) SO_(4)` (specific gravity `1.8 mL^(-1)` and 24.7% `H_(2) SO_(4)` by weight). After the metal is completely dissolved, the solution is diluted to `400 mL`. Calculate the molarity of free `H_(2) SO_(4)` solution.

Explanation

Normality of the given `H_(2)SO_(4) = 5.95 N `
` :. ` m.e of 75 mL of `H_(2)SO_(4) = 5.95 xx75`
` = 446.25 `
Equivalent of Al ` = (2.7)/9 = 0.3 " " …(Eqn .4)`
`{ " eq .wt. of Al " = ("al .wt")/("valency") = 27/3 = 9 }`
` :. " m.e of Al " = 0.3 xx 1000 = 300 " " ...(Eqn .3)`
Since 300 m.e of Al will react with 300 m.e of `H_(2)SO_(4)`
m.e of free `H_(2)SO_(4)` = total m.e of `H_(2)SO_(4)-300`
`= 446.25 - 300`
`= 146.25`
Now the free `H_(2)SO_(4)` is diluted to 400 mL and we knw that the m.e of `H_(2)SO_(4)` does not change on dilution .
` :. ` normality of the diluted free `H_(2)SO_(4)` in the resulting solution
` = ( " m.e of free" H_(2)SO_(4))/(" volume (mL)")`
` = (146.25)/400 = 0.366` N .
` :. " molarity " = (0.366)/2 = 0.183 M " " ...(Eqn . 6i)` ( basicity of `H_(2)SO_(4) = 2`)