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` .

Options

A. `0.982`
B. `0.0983`
C. `1.96`
D. `2.96`

Answer

A

Explanation

given
`MBr_(2) (g) hArr MBr_(2)(aq) rarr M^(+2) + 2Br^(-)`
`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 `1.2xx 10^(-19) M` ions of `S^(-2)`
now for `H_(2)S H_(2)S hArr 2H^(+) + S^(-2)`
`k_(a_(1)) xx k_(a_(2)) = ([H^(+)]^(2)[S^(-2)])/([H_(2)S])`
`10^(-7) xx 1.3 xx 10^(-13) = ([H^(+)]^(2)[1.2 xx 10^(-19)])/([0.1])`
`[H^(+)] = 1.04 xx 10^(-1)`
`:. p^(H) = 0.9826`