A popular lecture demonstration involves the sequential precipitation and dissolution of several sli — Equilibrium / Solubility Chemistry Question
Problem Context
A popular lecture demonstration involves the sequential precipitation and dissolution of several slightly soluble silver compounds beginning with a [Ag+] = 0.0050 M. Use the information below to answer the following questions about this demonstration.
[Ksp values; AgCl 1.8×10-10, AgBr 5×10-13, AgI 8.3×10-17, Ag2SO4 1.4×10-5]
What must the [SO4 2-] be in order to start precipitation in a solution in which [Ag+] = 0.005 M?
Model Answer
Ksp = [Ag+]2[SO4 2-] so [SO4 2-] = Ksp / [Ag+]2 = 1.4×10-5 / (0.005)2 = 0.56 M
State the order in which the halide ions should be added to a concentration of 0.10 M so that each precipitate will form from the [Ag+] in equilibrium with the previous precipitate. Support your answer with appropriate calculations.
Model Answer
Cl– then Br– then I–. For Cl, 1.8×10-10 = [Ag+](0.10), so [Ag+] = 1.8×10-9, for 5×10-13 = [Ag+](0.10), so [Ag+] = 5×10-12 and for 8.3×10-17 = [Ag+](0.10), so [Ag+] = 8.3×10-16
The [Ag+] in equilibrium with AgCl in 0.1 M Cl– is sufficient to cause precipitation of AgBr in 0.1 M Br– . In turn, the [Ag+] in equilibrium with AgBr in 0.1 M Br– will cause precipitation of AgI in 0.1 M I– . However, the reverse order of addition of the anions would not lead to this behavior.
As a way of making this demonstration more striking, one of the silver halides in this series is dissolved by adding aqueous ammonia before precipitating the next silver halide. Which silver halide(s) dissolve in 0.60 M NH3? Support your answer with calculations. [Kf Ag(NH3)2 + 1.7×107]
Model Answer
We need the net equilibrium constant for the combined reaction in each case.
AgCl(s) → Ag+(aq) + Cl-(aq) Ksp = 1.8×10-10
Ag+(aq) + 2NH3(aq) → Ag(NH3)2+ Kf = 1.7×107
So K = Ksp × Kf = 1.8×10-10 × 1.7×107 = 3.06×10-3
and AgBr(s) → Ag+(aq) + Br-(aq) Ksp = 5×10-13
Ag+(aq) + 2NH3(aq) → Ag(NH3)2+ Kf = 1.7×107
So K = Ksp × Kf = 5×10-13 × 1.7×107 = 8.5×10-6
and AgI(s) → Ag+(aq) + I-(aq) Ksp = 8.3×10-17
Ag+(aq) + 2NH3(aq) → Ag(NH3)2+ Kf = 1.7×107
So K = Ksp × Kf = 8.3×10-17 × 1.7×107 = 1.41×10-9
Now calculate the reaction quotient, Q, for the given case (using chloride as the example),
AgCl(s) + 2NH3(aq) → Ag(NH3)2+ + Cl-(aq)
So, Q = [Cl-][Ag(NH3)2+] / [NH3]2 = (0.10)(0.005) / (0.60)2 = 1.39×10-3
This value is smaller than only one of the calculated values for the net equilibrium constant (the case of chloride) so the silver chloride is the only one that will dissolve in 0.60 M NH3.