Question

Copper(I) ions in aqueous solution react with NH3(aq) according to Cu+(aq) + 2NH3 (aq) --> Cu(NH3)2 Kf = 6.3*10^10

Calculate the solubility (in g

Answer 1

First we calculate equilibrium constant K
Cu + 2NH₃ => Cu(NH₃)₂ ............K_f = 6.3 x 10^10
CuBr <=> Cu + Br.....................K_sp = 6.3 x 10^-9
CuBr + 2NH₃ <=> Cu(NH₃)₂ + Br.........K=?
When reactions are added, their equilibrium constants are multiplied.
K = (6.3 x 10^10) (6.3 x 10^-9)
K = 396.9

Now, Set up a concentration table
..............CuBr + 2NH₃ <=> Cu(NH₃)₂ + Br
Initial..................0.61...........0..............0
Change...............-2x...........+x.............+x
Equi................0.61-2x..........x..............x

K = x^2 / (0.61 - 2x)^2 = 396.9
x / (0.61 - 2x) = (396.9)^(1/2)
x / (0.61 - 2x) = 19.92
x = 12.153 - 39.84x
40.84x = 12.153
x = 0.3 M

Now, convert the solubility of CuBr from mol/L to g/L
CuBr is 143.45g/mol
0.3 mol/L x 143.45g/mol = 42.68 g/L
Hence Solubility = 42.68 g/L