It is given that `0.001 mol` each of `Cd^(2+)` and `Fe^(2+)` ions are contained in `1.0L` of `0.02M HC1` solution. This solutions is now saturated with `H_(2)S` gas at `25^(@)C`.
a. Determine whether or not each of these ions will be precipitated as sulphide?
b. How much `Cd^(2+)` ions remains in the solution at equilibrium?
`K_(1)(H_(2)S) = 1.0 xx 10^(-7), K_(2) (H_(2)S) = 1.0 xx 10^(-14)`: ltbRgt `K_(sp) (CdS) = 8 xx 10^(-27): K_(sp) (FeS) = 3.7 xx 10^(-19)`.

Explanation

Note that two salts are of same type (i.e. `AB` type)and `K_(sp)` of `CdS` is lower than that of `FeS`. This means `CdS` will precipitation first if at all any precipitation takes place. Calculate the minimum concentration of sulphide ion required to initiate the precipitation of each of the of the metal sulphide.
`[S^(2-)]_("min for Cds") = (K_(sp) CdS)/([Cd^(2+)]) = (8 xx 10^(-27))/(0.001) = 8 xx10^(-24)M`
`[S^(2-)]_("min for Cds") = (K_(sp) FeS)/([Fe^(2+)]) = (3.7 xx10^(-19))/(0.001) = 3.7 xx 10^(-16)M`
Now calculate the sulphide ion concentration in the saturated solution using:
`K_(a) = K_(1)xx K_(2) = ([H^(o+)]^(2)[S^(2-)])/([H_(2)S]) = 10^(-21)`
`rArr [S^(2-)] = (10^(-21)xx0.1)/((0.02)^(2)) = 2.5 xx 10^(-19)M`
Thus, only `CdS` will get precipiated
To calculate the remaining concentration of `Cd^(2+)` ion in the solution, first assume that whole of `Cd^(2+)` has been precipitated as `Cd^(2+) +H_(2)S rarr CdS +2H^(o+)`
Thus `[H^(o+)] "new" = 0.02 + 0.001 xx2 = 0.022M`
`rArr [S^(2-)]_("At new equilibrium") = (10^(-21)xx0.1)/((0.022)^(2)) = 2.5 xx 10^(-19)M`
`rArr [Cd^(2+)] "Left" = (8 xx 10^(27))/(2.07xx10^(-19)) = 3.86 xx 10^(-8)M`
Note : `FeS` will not precipitate at all.