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

Given, `MBr_(2)(s) hArr MBr_(2)(aq) rarr M^(2+) +2Br^(Θ)`
`MBr_(2) + H_(2)S rarr MS + 2HBr`
`K_(sp)` of `mS = [M^(2+)] [S^(2-)]`
`6 xx 10^(-21) = (0.05) [S^(2-)] :. [S^(2-)] =1.2 xx 10^(-19)M`
Thus, `MS` will be precipitated if `H_(2)S` provides `1.2 xx 10^(-19) M` ions of `S^(2-)`
Now for `H_(2)S`
`H_(2)S hArr 2H^(o+) + S^(2-)`
`K_(1) xx K_(2) = ([H^(o+)]^(2)[S^(2-)])/([H_(2)S])`
`10^(-7) xx1.3 xx 10^(-13) = ([H^(o+)][1.2 xx 10^(-19)])/([0.1)]`
`[H^(o+)] = 1.04 xx 10^(-1)` and `pH = 0.9826`