Calculate solubility of `MnS` in a buffer solution of given `pH.K_(sp)` of `MnS` and `K_(a_(1))&K_(a_(2))` for `H_(2)S` are given .

Explanation

Let the new solubility of `MnS=x`
`[Mn^(2+)]=x`=Initial concentration of `S^(2-)` ions but free `S^(2-)` ions will be less because some of the `S^(2-)` ions will react with `H^(+)` from buffer to form `HS^(-)` and `H_(2)S`
`[Mn^(2+)]=x=underset("Free")[S^(2-)]+[Hs^(-)]+[H_(2)S]....(1)`
Let us calculate the value of `[HS^(-)]`&`[H_(2)S]` in terms of free `[s^(2-)]` ion.For that consider:
`HS^(-)hArr H^(+)+S^(2-)"",,,,,,""H_(2)ShArr H^(+)+HS^(-)`
`K_(2)=([H^(+)][S^(2-)])/([HS^(-)])"",,,,""K_(1)=([H^(+)][HS^(-)])/([H_(2)S])`
`[HS^(-)]=([H^(+)][S^(2-)])/(K_(2))....(2)`
and `[H_(2)S]=([H^(+)][S^(-)])/(K_(1))=([H^(+)][S^(2-)])/(K_(1)K_(2)]....(3)`
Put (2) &(3) in (1)
`x=[S^(2-)](1+([H^(+)])/(K_(2))+([H^(+)]^(2))/(K_(1)K_(2)))`
`x=(K_(sp))/([Mn^(2+)])(1+([H^(+)])/(K_(2))+(H^(+)]^(2)/(K_(1)K_(2)))`
`x=sqrt(K_(sp)(1+([H^(+)])/(K_(2))+([H^(+)]^(2))/(K_(1)K_(2))))`