`K_(1)` and `K_(2)` for dissociation of `H_(2)A` are `4xx10^(-3)` and `1xx10^(-5)`. Calculate concentration of `A^(2-)` ion in `0.1M H_(2)A` solution. Also report `[H^(+)]` and pH.

Explanation

`H_(2)A hArr H^(+)+HA^(-)`
`K_(1)= ([H^(+)][HA^(-)])/([H_(2)A])= 4xx10^(-3)`
`.: [H^(+)]=Calpha, [HA^(-)]=Calpha, [H_(2)A]=C(1-alpha)`
or `4xx10^(-3)= (Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/((1-alpha)) , (C=0.1M)`
or `4xx10^(-3)= (0.1xxalpha^(2))/((1-alpha)) , [(1-alpha)` should not be neglected]
`:. alpha=0.18`
`:. [H^(+)]=Calpha=0.1xx0.18=0.018M`
`:. pH = 1.7447`
`:. [HA^(-)]=C alpha=0.1xx0.18)=0.082M`
Now `HA^(-)` further dissociated to `H^(+)` and `A^(2-)`,
`C_(1)=[HA^(-)]=0.018M`
`{:(HA^(-)hArr, H^(+)+,A^(2-)), (1,0,0),((1-alpha_(1)),alpha_(1),alpha_(1)):}`
`:. K_(a_(2))=1xx10^(-5)= ([H^(+)][A^(2-)])/([HA^(-)])`
`.: [H^(+)]` already in solution `=0.018` and thus, dissociation of `HA^(-)` further suppresses due to common ion effect and `(1-alpha)~~1`
`:. 1xx10^(-5)= (0.018xxC_(1)alpha_(1))/(C_(1)(1-alpha_(1)))= 0.018xxalpha_(1)`
`:. alpha_(1)= (1xx10^(-5))/(0.018)= 5.55xx10^(-4)`
`= 10^(-5)M`
`:. [HA^(-)]=C_(1)(1-alpha_(1))=C_(1)= 0.018M`