Cu+(aq) + 2 NH3(aq) → Cu(NH3)2+(aq) Kf = 6.3×1010
Cu+(aq) + Br‒(aq) Ksp = 6.3×10‒9
CuBr(s) + 2 NH3(aq) → Cu(NH3)2+(aq) + Br‒(aq) K = Kf × Ksp
K = 396.9
We can now plug in the equilibrium concentrations and solve for x, which is the molar solubility of CuBr in M or mol/L.
Copper(I) ions in aqueous solution react with NH 3(aq) according to
Cu+ (aq) + 2NH3 (aq) → Cu(NH3)2 + (aq) Kf = 6.3 x 1010
Calculate the solubility (in g.L-1) of CuBr(s) (Ksp = 6.3 x 10-9) in 0.49 M NH 3 (aq).
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