How much `NH_(3)`should be added to a solution of `0.01M Cu(NO_(3))_(2)` to reduce `[Cu^(2+)` to `10^(-13)`. Neglect the amount of copper in complexes containing fewer than `4` ammonia molecules per copper atom. Given `K_(d)` for `Cu(NH_(3))_(4)^(2+) = 1.0 xx 10^(-12)`

Explanation

Since the sum of the concentrations of copper in complex and in the free ionic state must be equal to `0.001 mol L^(-1)` and since in the amount of the free ion is very small, the concentration of the complex is taken to be `0.001M`.
Let `x = [NH_(3)]`. Then
`Cu(NH_(3))_(4)^(2+) hArr Cu^(2+) + 4NH_(3)`
`K_(d) = ([Cu^(2+)][NH_(3)]^(4))/([CU(NH_(3))_(4)^(2+)])`
`10^(-12) = ((10^(-13))(x^(4)))/(0.001) rArr x^(4) = 1.0 xx 10^(-2)` or `x = 0.32`
`:. [NH_(3)]` equilibrium `= 0.32M`
the amount of `NH_(3)` used up in forming `0.001M` of complex is `0.004M`, an amount negligible compared with teh amount remaining at equilibrium. Hence, the amount of `NH_(3)` to be added is `0.32M`.