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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 . |
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Answer» Solution :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))))` |
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