An aqueous solution of aniline of concentration `0.24` M is prepared. What concentration of sodium hydroxide is needed in this solution so that anilinium ion concentration remains at `1xx10^(-8)M` ? `(K_(a)for C_(6)H_(5)NH_(3)^(+)=2.4xx10^(-5)M)`

Explanation

`C_(6) H_(5) NH_(2) + H_(2) O rarr C_(6) H_(5) NH_(3)^(+) + OH^(-)`
`K_(b) = ([ C_(6) H_(5) NH_(3)^(+) ] [OH^(-) ] ) /([C_(6)H_(5)NH_(2) ] )`
Also `K_(b)` for `C_(6) H_(5) NH_(2) = ( K_(w))/(K_(a)" for " C_(6) H_(5) NH_(3)^(+)) = ( 10^(-14))/( 2.4 xx 10^(-5))`
Since dissociation of `C_(6) H_(5) NH_(2)` occurs in presence of NaOH and thus dissociation of `C_(6)H_(5)NH_(2)` will suppress.
`[C_(6)H_(5) NH_(2) ] = 0.24 , [ C_(6) H_(5) NH_(3) ^(+) ]= 10^(-8), [OH^(-) ] = 1 `
`( 10^(-14))/(2.4 xx 10^(-5)) = ( 10^(-8)xx[OH^(-)])/( 0.24 ) rArr[ OH^(-) ] = ( 0.24 xx 10^(-14))/( 2.4 xx 10^(-5) xx 10^(-8)) = 0.01 `
`[NaOH ] = 0.01 M `