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)`.

Explanation

The minimum `[OH^(-)]` at which there will be no precipitation of `Mg(OH)_(2)` canbe obtained by
`K_(SP)=[Mg^(2+)][OH^(-)]^(2)`
`9.0xx10^(-12)=[0.05][OH^(-)]^(2)`
`:. [OH^(-)]=1.34xx10^(-5)M`
thus, a solution having `[OH^(-)]=1.34xx10^(-5)M` will not show precipitation of `Mg(OH)_(2)` in `0.05M Mg^(2+)` solution. These hydroxyl ions are to be derived by a buffer of `NH_(4)CI` and `NH_(4)OH` , i.e.,
`NH_(4)OHhArrNH_(4)^(+)+OH^(-)`
`NH_(4)CIrarr NH_(4)+CI^(-)`
For `NH_(4)OH`,
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])`
`:. [NH_(4)^(+)]=0.067M`
or `[NH_(4)CI]=0.067M`