In an excess of `NH_(3(aq.)),Cu^(2+)` ion form a deep blue complex ion `[Cu(NH_(3))_(4)]^(2+)` having formation constant `K_(f)=5.6xx10^(11)`. Calculate the concentration of `Cu^(2+)` in a solution prepared by adding `5.0xx10^(-3)` mole of `CuSO_(4)` to `0.50` litre of `0.40M NH_(3)`.

Explanation

`{:(,Cu^(2+)+,4NH_(3(l))hArr,[Cu(NH_(3))_(4)]^(2+),),("Initial mole", 0.005,0.5xx0.4,0):}`
`=0.2`
`K_(f)=5.6xx10^(11)`
`K_(f)` is larger and thus all the `Cu^(2+)` will give `[Cu(NH_(3))_(4)]^(2+)`. Let `Cu^(2+)` left is a, then
`[Cu(NH_(3))_(4)]^(2+)= 0.005-a=0.005 "mole"`
`=(0.005)/(0.5)M` , `(0.005gtgt a)`
`[Cu^(2+)]=(a)/(0.5)M`
`[NH_(3)]=0.2-4xx0.005+2a`
`0.2-0.2+2a(0.18gtgt2a)`
`=0.18 "mole"= (0.18)/(0.5)M`
`:. K_(f)= ([Cu(NH_(3))_(4)]^(2+))/([Cu^(2+)][NH_(3)]^(4))`
`5.6xx10^(11)=((0.005)/(0.5))/((a)/(0.5)xx[(0.18)/(0.5)]^(4))`
`:. a= (0.005xx(0.5)^(4))/((0.18)^(4)xx5.6xx10^(11))= 5.32xx10^(-13)`