First, determine the chemical reactions
Solubility: ZnCO3(s) → ← Zn2+(aq) + CO32–(aq)
Complex Ion: Zn2+(aq) + 4 NH3(aq) → ← [Zn(NH3)4]2+(aq)
Net: ZnCO3(s) + 4 NH3(aq) → ← [Zn(NH3)4]2+(aq) + CO32–(aq)
Kc = [[Zn(NH3)4]2+]e[CO32–]e [NH3]e4
Kc = Kf×Ksp = (4.1×108)×(1.4×10–11) = 5.7×10–3
Using the law of multiple equilibria
Initial1.000
Change– 4x+x+x
Equilibrium1.0 – 4xxx
5.7×10–3 = (x)(x)/(1.0 – 4x)4
Taking the square root of both sides gives:
0.0755 = x/(1.0 – 4x)2
Doing some algebra gives:
(16x2 –8x + 1)×0.0755 = x
1.208x2 – 1.604x + 0.0755 = 0
Solving the quadratic equation gives
x = 4.9×10–2 or x = 1.3
The molar solubility of ZnCO3 in 1.0 NH3 is 4.9×10–2 M.
(x ≠ 1.3 because that would give [NH3]e = 1.0 – 4(1.3) = –4.2 M, which is chemically unreasonable.)
The solubility increased by a factor of 104 in the presence of the ammonia!