Calculate the pH of
(i) 1 M `H_(2)SO_(4) " "
(ii) 2 M H_(2)SO_(4) " "
(iii) 10^(-2)M H_(2)SO_(4)` solutions.
Given that the second ionization constant `(K_(a_(2)))` of `H_(2)SO_(4)` if `10^(-2)`.

Explanation

`H_(2)SO_(4)` ionizes in two steps as
`H_(2)SO_(4) rarr H^(+)+HSO_(4)^(-) ` (almost completely ionized )
`HSO_(4)^(-) hArr H^(+) + SO_(4)^(2-)` (partially ionized)
(i) In 1 M `H_(2)SO_(4), H^(+)` ions are obtained mainly from 1st ionization = 1 M and `H^(+)` ions obtained from 2nd step are negligible as compared to 1 M. Hence `[H^(+)] = 1 M , pH = - log 1 = 0 `.
(ii) In 2 M `H_(2)SO_(4)` , again `H^(+)` ions are obtained mainly from 1st ionization as explained above. Hence, `[H^(+)]=2 M, pH = - log 2 = - 0.303`.
Here note that there is nothing wrong with the negative value of pH but this offers no advantage over the actual value of `[H^(+)]`.
(iii) In `10^(-2) M H_(2)SO_(4),H^(+)` ions obtained from 2nd ionization cannot be neglected in comparison to `10^(-2) M H^(+)` ions obtained from 1 st ionization. Hence, we have
`{:(,H_(2)SO_(4),rarr,H^(+),+,HSO_(4)^(-),),("Initial",10^(-2)M,,,,,),("After 1 st ionization",~=0,,10^(-2)M,,10^(-2)M,),("Initial",HSO_(4)^(-),hArr,H^(+),+,SO_(4)^(-),),(,10^(-2)M,,,,,),("After ionization",10^(-2)-x,,10^(-2)+x,,x,):} (H^(+)=10^(-2) "from 1 st ionization + x from 2nd ionization")`
`:. K_(a_(2))=([H^(+)][SO_(4)^(2-)])/([HSO_(4)^(-)])=((10^(-2)+x)(x))/(10^(-2)-x)=10^(-2)` (Given)
`:. 10^(-2)x + x^(2)=10^(-4)-10^(-2)x or x^(2) + 2 xx 10^(-2)x-10^(-4) = 0 `
`:. x=(-b pmsqrt(b^(2)-4ac))/(2a) = (-2xx10^(-2)pmsqrt(4xx10^(-4)+4xx10^(-4)))/(2) = (-2xx10^(-2)+2.828xx10^(-2))/(2)`
`=(0.828xx10^(-2))/(2)` (ignoring the -ve value )
`=0.414xx10^(-2) M`
`:. [H^(+)]=10^(-2)+x=10^(-2)+0.414xx10^(-2) = 1.414xx10^(-2)`
`:. pH = - log [H^(+)]=-log(1.414xx10^(-2))=1.85`