What is `[Cd^(2+)]` in `1.0L` of solution prepared by dissolving `0.001mol Cd(NO_(3))_(2)` and `1.5 mmol. NH_(3) ? K_(d)` for the dissociation of `Cd(NH_(3))_(4)^(2+)` into `Cd^(2+)` and `4NH_(3)` is `1.8 xx 10^(-7)`. Neglect the amount of `Cd` in complexes containing fewer than `4` ammonia molecules.

Explanation

`{:(,Cd^(2+)+,4NH_(3)hArr,Cd(NH_(3))_(4)^(2+),),("Initial",rArr0.001,1.5,0,),("Used up",rArr0.001-x,4(0.001-x),0.001-x,),("Eq.",rArr~~x,~~1.5,~~0.001,):}`
`Cd(NH_(3))_(4)^(2+) hArr Cd^(2+) + 4NH_(3)`.
`K_(d) = ([Cd^(2+)][NH_(3)]^(4))/([Cd(NH_(3))_(4)^(2+)])`
`1.8 xx 10^(-7) = ((x)(1.5)^(4))/(0.001)`
`:.x = 3.6 xx 10^(-16) = [Cd^(2+)]`