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

Explanation

Give, `MBr_(2(g))hArr MBr_(2(aq.))rarrM^(2+)+2Br^(-)`
`MBr_(2)+H_(2)SrarrMS+2HBr`
`K_(SP)of MS=[M^(2+)][S^(2-)]`
`6xx10^(-21)=[0.05][S^(2-)]`
`:. [S^(2-)]=1.2xx10^(-19)M`
Thus, MS will be precipitated if `H_(2)S` provided `1.2xx10^(-19)M` ions of `S^(2-)`
Now for `H_(2)S H_(2)ShArr2H^(+)+S^(2-)`
`K_(1)xxK_(2)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])`
`10^(-7)xx1.3xx10^(-13)= ([H^(+)]^(2)[1.2xx10^(-19)])/([0.1])`
`:. [H^(+)]=1.04xx10^(-1)`
and `pH=0.9826`