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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)`. |
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Answer» `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` |
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