An aqueous solution of a metal bromide `MBr_2` (0.05M) is saturated with`H_2S`. What is the minimum pH at which MS will precipitate. `K_(sp)` for M.S`=6.0xx10^(-21)`, conc. Of saturated `H_2S`=0.1(M) and `K_1=10^(-7)` and `K_2=1.3xx10^(-13)` for `H_2S`

Explanation

`H_2S` is a diprotic acid In its aqueous solution, its ionisation involves the following two steps-
`H_2S(aq) `HS^-(aq) overall ionisation : `H_2S(aq) overall ionisation constant
`K=([H^+][S^(2-)])/([H_2S])=K-1xxK_2=1.3xx10^(-20)`
Given that `[H_2S]=0.1(M)`
`:.[H^+]^2[S^(2-)]=1.3xx10^(-20)xx0.1=1.3xx10^(-21)`
In a saturated solution of MS, the following equilibriumb
exists `MS(s) `:.K_(sp)=[M^(2+)][S^(2-)]`
Given that the concentration of `MBr_2` solution= 0.05(M).
So, in this solution, `[M^(2+]=0.05(M)`
Minimum concentration of `S^(2-)` necessary for the precipitation of MS,
`[S^(2-)]=(K_(sp))/([M^(2+)])=(6xx10^(-21))/(0.05)=1.2xx10^(-19)(M)`
Now. `[H^+]^2xx1.2xx10^(-19)=1.3xx10^(-21)`
`:.[H^+]=0.104(M)`
Therefore, the minimum concentration of `H^+` (aq) ions for the precipitation of MS = 0.104(M) and the minimum pH for precipitation `=-log_(10)[H^+]`
`=-log[0.104]=0.983`