Why excess of `NH_(4)OH` is used in precipitating the sulphides of group IV cations?

Explanation

The solubility product of group IV radicals is greater than that of group II ions. Therefore a higher concentration of `S^(2-)` ions is required to precipitate group IV ions as their sulphides now to get a higher concentration of `S^(2-)` ions `H_(2)S` gas is passed in presence of `NH_(4)OH` the `Ooverset(ɵ)(H)` ions from `NH_(4)OH` neutralise the `H^(o+)` ions from `H_(2)S` to form undissociated water molecules. the removal of `H^(o+)` ion causes more `H_(2)S` to dissociate or ionise, thus increasing the concentration of `S^(-)` ions so the product `[M^(2+)][S^(2-)]` exceeds the solubility products of the sulphides of Zn, Mn, Ni and Co and hence they are precipitated.
`(M=Zn,Mn,Co` or `Ni)`
`2NH_(4)OHhArr2NH_(4)^(o+)+2OH^(ɵ)`
`H_(2)ShArr2H^(o+)+S^(2-)`
`2H^(o+)+2OH^(ɵ)to2H_(2)O`