Question

An aqueous solution of a metal bromide `MBr_(2)` (0.05 M) is saturated with `H_(2)S`. What is the minimum pH at which MS will be precipitated ? `(K_(sp) "for MS" = 6.0 xx 10^(-21), "concentration of saturated " H_(2)S=0.1M, K_(1)=10^(-7) and K_(2)=1.3xx10^(-13) "for " H_(2)S)`.

Answer 1

`[M^(2+)]=[MBr_(2)]=0.05 M `
For the precipitation of MS, minimum `[S^(2-)]` can be calculated from
`[S^(2-)][M^(2+)]=K_(sp)` i.e., `[0.05][S^(2-)]=6.0xx10^(-21) or [S^(2-)]=1.2xx10^(-19)`
`S^(2-)` ions are obtained from following dissociations
`H_(2)S overset(K_(1)) hArr H^(+)+HS^(-)`
`HS^(-) overset(K_(2))hArr H^(+)+S^(2-)`
`K_(1) = ([H^(+)][HS^(-)])/([H_(2)S]), K_(2) = ([H^(+)][S^(2-)])/([HS^(-)])`
`:. K_(1)K_(2)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])`
`10^(-7)xx1.3xx10^(-13)=([H^(+)]^(2)[1.2xx10^(-19)])/((0.1)) or [H^(2)]^(2)=(1.3xx10^(-20)xx10^(-1))/(1.2xx10^(-19))=1.083xx10^(-2)`
or `[H^(+)]=1.041xx10^(-1)=0.1041 M`
`:. pH = - log (0.1041)=0.9826`