For buffer solution,
`NH_(4)CI=0.25M`
and `NH_(4)OH=0.05M`
`pOH= -log K_(b)+log(["Salt"])/(["Base"])`
`:. pOH= -log 1.8xx10^(-5)+log ([0.25])/(0.05)]`
`:. [OH^(-)]=3.6xx10^(-6)M`
Now `Al(OH)_(3)` and `Mg(OH)_(2)` are stirred vigorously in it.
`:. [Al^(3+)][OH^(-)]^(3)= K_(SP)of Al(OH)_(3)`
`[Al^(3+)][3.6xx10^(-6)]^(3)=6xx10^(-32)`
`:. [Al^(3+)]=1.28xx10^(-15)M`
Also `[Mg^(2+)][OH^(-)]^(2)= K_(SP)of Mg(OH)_(2)`
`[Mg^(2+)][3.6xx10^(-6)]^(2)=8.9xx10^(-12)`
`[Mg^(2+)]=0.686M`