Question

Calculate the solubility (in g·L–1) of CuBr(s) (Ksp = 6.3× 10–9) in 0.90 M NH3(aq). Kf=6.3x10^10

Answer 1

1. Calculate K
Cu+2NH3 => Cu(NH3)2 Kf=6.3x10^10
CuBr <=> Cu+Br Ksp=6.3x10^-9
CuBr+2NH3 <=> Cu(NH3)2+Br K=?
When reactions are added, their equilibrium constants are multiplied.
K=(6.3x10^10)(6.3x10^-9)=396.9

2. Set up a concentration table
CuBr+2NH3 <=> Cu(NH3)2+Br
- 0.9 0 0
- -2x +x +x
- 0.9-2x x x
K=x^2/(0.9-2x)^2=396.9
x/(0.9-2x) = (396.9)^(1/2)
x=0.44 M

3. Convert the solubility of CuBr from mol/L to g/L
- CuBr is 143.45g/mol
0.44mol/L x 143.45g/mol = 63g/L