1. Solubility and Complex-Ion Equilibria
Chapter 18
Solubility and Complex-Ion Equilibria
John A. Schreifels
Chemistry 212

2. Overview Solubility Equilibria Complex-Ion Equilibria
Solubility Product Constant
Solubility and Common Ion Effect
Precipitation Calculations
Effect of pH on Solubility
Complex-Ion Equilibria
Complex Ion Formation
Complex Ions and Solubility
Application of Solubility Equilibria
Qualitative analysis of metal ions
John A. Schreifels
Chemistry 212

3. Solubility Equilibria
Solubility of a solid treated as with other equilibria. Solution is saturated. No more solid will dissolve since dynamic equilibrium.
AgCl(s)  Ag+(aq) + Cl(aq) Ksp = [Ag+][Cl]
Solid not included in the equilibrium expression.
MyXz(s)  yM+p(aq)+zXq(aq) Ksp=[M+p]y[Xq]z
where Ksp = solubility product.
E.g. determine the equilibrium expression for each: PbCl2, Ag2SO4,Al(OH)3.
Ksp can be determined if the solubility is known.
E.g. Determine Ksp for silver chromate (Ag2CrO4) if its solubility in water is g/L at 25C.
Determine molar solubility.
Determine Ksp.
E.g. 2 Determine Ksp of CaF2 if its solubility is 2.20x104M.
John A. Schreifels
Chemistry 212

4. PbCl2(s)  Pb2+(aq) + 2Cl(aq)
SOLUBILITY FROM Ksp
Can be determined by using stoichiometry to express all quantities in terms of one variable- solubility, x. i.e. for the reaction. Use equilibrium table to write concentration of each in terms of the compound dissolving
E.g. determine the solubility of PbCl2 if its Ksp = 1.2x105
PbCl2(s)  Pb2+(aq) + 2Cl(aq)
E.g.1 Determine solubility of AgCl if its Ksp = 1.8x1010M2.
E.g.2 Determine solubility of Ag2CO3 if its Ksp = 8.1x1012M3.
E.g.3 Determine solubility of Fe(OH)3 if its Ksp = 4x1038M4
John A. Schreifels
Chemistry 212

5. Factors that Affect Solubility
The common–Ion effect (Remember LeChatelier’s Principle)
E.g. Determine solubility of PbCl2 (Ksp = 1.2x105)in 0.100M NaCl.
Write equilibrium table in terms of x and [Cl]
a common–ion reduces the solubility of the compound.
Assume that [Cl]NaCl >>x
Solve for x.
E.g. determine the solubility of CaF2 in a solution of CaCl2. Ksp = 3.9x1011.
John A. Schreifels
Chemistry 212

6. Precipitation of Ionic Compounds
Starting with two solutions, Qsp used to predict precipitation and even the extent of it.
Precipitation = reverse of dissolution
Precipitation occurs when Qsp > Ksp until Qsp = Ksp
If Qsp < Ksp, precipitation won’t occur.
E.g. determine if precipitation occurs after mixing mL 3.00x103 M BaCl2 and mL 3.00x103 M Na2CO3.
Solution:
CBaCl2 = 1.50x103 M; CNa2CO3 = 1.50x103 M
Qsp = 1.50x103 M1.50x103 M = 2.25x106
Qsp >1.1x1010.= Ksp precipitation.
E.g. 2 determine equilibrium concentration of each after precipitation occurs.
assume complete precipitation occurs;
set up equilibrium table; and solve for equilibrium concentration of barium and carbonate ion concentrations.
John A. Schreifels
Chemistry 212

7. Precipitation of Ionic Compounds
Eg. 3 determine the fraction of Ba2+ that has precipitated.
Solution:
Use the amount remaining in solution (results of E.g. 2) divided by starting concentration to determine the fraction of barium that is left in solution.
Subtract from above.
E.g.4 determine the Br concentration when AgCl starts to precipitate if the initial concentration of bromide and chloride are M. Ksp(AgBr) = 5.0x1013; Ksp(AgCl) = 1.8x1010.
John A. Schreifels
Chemistry 212

8. Factors that Affect Solubility-pH
pH of the Solution: LeChatelier’s Principle again.
E.g. determine the solubility of CaF2 at a pH of Ksp = 3.9x1011. Ka(HF) = 6.6x104.
Strategy:
Determine the ratio of [F] and [HF] from the pH and Ka.
Write an expression for solubility in terms of Ka and pH and
Substitute into solubility equation to determine the solubility.
Solution:
Ksp = 3.9x1011 = x[F]2 (pH changes the amount of free Fluoride.)
Let x = solubility. Then 2x = [F] + [HF]
From equilibrium equation:
2x = [F](1+1/0.066) = 16.15*[F] or
[F] = 2*x/16.15 = 0.124*x
3.9x1011 = x(0.0124*x)2
x = 1.36x103 M vs. 2.13x104 M (normal solubility)
John A. Schreifels
Chemistry 212

9. Separation of Ions By Selective Precipitation
Metal ions with very different Ksp can be separated.
Divalent metal ions are often separated using solubility variations for the metal sulfides.
Solution is saturated with H2S at M; pH adjusted to keep one component soluble and the other insoluble.
H2S is diprotic acid; the overall reaction to get to sulfide is:
Combine with solubility equilibrium reaction to get the overall equilibrium expression and constant.
E.g. determine the solubility of M Zn2+ in M H2S at pH = 1. Ksp = 1.10x1021.
John A. Schreifels
Chemistry 212

10. Ag+(aq) + 2NH3(aq)  Ag(NH3)2(aq) Kf = 1.7x107
Complex Ions
Formation of Complex Ions (Coordination Complexation ) = an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond.
Ag+(aq) + 2NH3(aq)  Ag(NH3)2(aq) Kf = 1.7x107
Large equilibrium constant indicates that “free” metal is completely converted to the complex.
Eg. What is the concentration of the silver amine complex above in a solution that is originally M Ag+ and 1.00 M NH3?
E.g. determine the [Ag+] (free silver concentration) in M AgNO3 that is also 1.00 M NaCN.
John A. Schreifels
Chemistry 212

11. Factors that Affect Solubility: Complexation
Free metal ion concentration in solution is reduced when complexing agent added to it;
Free metal ion concentration needed in solubility expression.
E.g. determine if precipitation will occur in a solution containing M AgNO3 and M Nal in 1.00 M NaCN. Recall Kf = 5.6x1018
Agl(s)Ag+ + l Ksp = 8.5.x1017
Strategy:
Determine the free metal concentration in the solution.
Use free metal concentration with iodide concentration to get Qsp
If Qsp < Ksp, no precipitation
If Qsp > Ksp, precipitation
If Qsp = Ksp, precipitation is starting.
John A. Schreifels
Chemistry 212

12. Solubility with Complexing Agent
E.g. Determine the solubility of AgI in 1.00 M NaCN. Recall Kf = 5.6x1018
Agl(s)Ag+ + l Ksp = 8.5.x1017
Strategy:
Combine to equilibria equations to find a single equation describing the equilibrium.
the presence of a complexing agent increases the solubility
Setup equilibrium table and solve.
John A. Schreifels
Chemistry 212