Question

A solution has `0.05M Mg^(2+)` and `0.05M NH_(3)`. Calculate the concentration of `NH_(4)CI` required to prevent the formation of `Mg(Oh)_(2)` in solution. `K_(SP)` for `Mg(OH)_(2)=9.0xx10^(-12)` and ionisation constant of `NH_(3)` is `1.8xx10^(-5)`.

Answer 1

The minimum `[overset(Theta)OH]` at which there will be no precipitation of `Mg(OH)_(2)` can be obtained by
`K_(sp) = [Mg^(2+)] [overset(Theta)OH]^(2)`
`9.0 xx 10^(-12) = [0.05] [overset(Theta)OH]^(2)`
`:. [overset(Theta)OH] = 1.34 xx 10^(-5)M`
Thus, a solution having `[overset(Theta)OH] = 1.34 xx 10^(-5)` will not show precipitation of `Mg(OH)_(2)` in `0.05 M Mg^(2+)` solution. These hydroxy`1` ions are to be derived by a buffer of `NH_(4)CI` and `NH_(4)OH`, i.e.,
`NH_(4)OH hArr NH_(4)^(o+) + overset(Theta)OH`
`NH_(4)CI rarr NH_(4) + CI^(Theta)`
For `NH_(4)OH, K_(b) = ([NH_(4)^(o+)][overset(Theta)OH])/([NH_(4)OH])`
In presence of `NH_(4)CI`, all the `[NH_(4)^(o+)]` are provided by `NH_(4)CI` since common ion effect decreases dissociation of `NH_(4)OH`.
`:. 1.8 xx 10^(-5) = ([NH_(4)^(o+)][1.34 xx 10^(-5)])/([0.05])`
`:. [NH_(4)^(o+)] = 0.067 M or [NH_(4)CI] = 0.067M`