1. Ch 16 Lecture 3 Solubility Equilibria
Solubility Basics
The Solubility Equilibrium
Dissolution of an ionic compound is an equilibrium process
CaF2 (s) Ca2+ (aq) F- (aq)
Ksp = Solubility Product = [Ca2+][F-]2
Remember, neither solids nor pure liquids (water) effect the equilibrium constant
Dissolving and Reforming change proportionately to the amount of solid
Solvent water is at such a high concentration as not to be effected
The Solubility Product is an equilibrium constant, so it has only one value at a given temperature
Solubility = the equilibrium position for a given set of conditions
There are many different conditions that all must obey Ksp
Common ions effect the solubility much as they effect pH

2. Ksp values of some slightly soluble ionic solids
Most NO3- salts are soluble
Most alkali metal and NH4+ salts are soluble
Most Cl-, Br-, and I- salts are soluble (except: Ag+, Pb2+, and Hg22+)
Most SO42- salts are soluble (except Ba2+, Pb2+, Hg22+, and Ca2+)
Most OH- salts are insoluble (except NaOH, KOH)
Most S2-, CO32-, CrO42-, and PO43- salts are insoluble
Note: These values may
differ from the ones in
your text. Use the values
from your text for all
homework problems

3. Example: What is the Ksp of a CuBr solution with a solubility of 2
Example: What is the Ksp of a CuBr solution with a solubility of 2.0 x 10-4 M?
CuBr (s) Cu+ (aq) + Br- (aq)
Ksp = [Cu+][Br-] = [2.0 x 10-4][2.0 x 10-4] = 4 x 10-8
Example: Ksp = ? for Bi2S3 with solubility of 1.0 x M
Example: Find the solubility of Cu(IO3)2 (Ksp = 1.4 x 10-7)
Cu (IO3)2 (s) Cu2+ (aq) IO3- (aq)
Ksp = [Cu2+][IO3-]2 = 1.4 x 10-7
(x)(2x)2 = 4x3 = 1.4 x x = 3.3 x 10-3 M
Relative Solubilities
If the salts being compared produce the same number of ions, we can compare solubilities by comparing Ksp values
AgI Ksp = 1.5 x 10-16
CuI Ksp = 5.0 x 10-12
CaSO4 Ksp = 6.1 x 10-5
Ksp = [Xn+]1[Yn-]1 for all of these, so we can directly compare them
Solubility of CaSO4 > solubility of CuI > solubility of AgI

4. If the salts being compared produce different numbers of ions, we must calculate the actual solubility values; we can’t use Ksp values to compare.
a) CuS (8.5 x 10-45) > Ag2S (1.6 x 10-49) > Bi2S3 (1.1 x 10-73) by Ksp alone
2 ions ions ions
b) Bi2S3 (1.0 x 10-15) > Ag2S (3.4 x 10-17) > CuS (9.2 x 10-23) in solubility
The Common Ion Effect
Common Ion = any ion in the solid we are trying to dissolve that is present in solution from another source.
What is the solubility of Ag2CrO4 (Ksp = 9 x 10-12) in 0.1 M AgNO3?
Ag2CrO4 (s) Ag+ (aq) CrO42- (aq)
Init
Equil x x
Ksp = 9 x = [ x]2[x] ~~ (0.1)2(x)
x = 9 x / 0.01 = 9 x = solubility
5% rule: 9 x / 0.1 < 5%, so the approximation is ok
Solubility in pure water is 1.3 x 10-4 M. Why does the solubility decrease?

5. Example: Find solubility of CaF2 (Ksp = 4.0 x 10-11) in 0.025 M NaF
pH and Solubility
Many salts contain hydroxide anion: Mg(OH) Mg OH-
High pH means large OH- common ion concentration
[OH-] would shift the equilibrium to the left
[H+] would shift the equilibrium to the right by using up OH- ions
Any Basic Anion will be effected by pH
OH-, S2-, CO32-, C2O42-, CrO42-, and PO43- are all basic anions
H+ will increase the solubility of their salts by removing the anions
Ag3PO4 (s) Ag+ (aq) + PO43- (aq)
PO H HPO42-
Acidic pH has no effect on non-basic anions or on most cations: Cl-, Br-, NO3-
AgCl (s) Ag+ (aq) + Cl- (aq)
H+ doesn’t react with either ion

6. Precipitation and Qualitative Analysis
We can use the solubility product to predict precipitation
If Q > Ksp, precipitation occurs until Ksp is reached
If Q < Ksp, no precipitation will occur
Example: 750 ml M Ce(NO3)3 is added to 300 ml 0.02 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 x 10-10) precipitate?
Ksp = [Ce3+][IO3-]3 We need to know concentrations.
Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x > Ksp Precipitate
We can also calculate the equilibrium concentrations after precipitation.
Examine the stoichiometry of the precipitation reaction allowed to go to completion
Becomes a Ksp problem with a common ion (ion in excess)

7. Calculate the equilibrium concentrations after precipitation when 100 ml of 0.05 M Pb(NO3)2 is added to 200 ml 0.10 M NaI. Ksp for PbI2 = 1.4 x 10–8.
PbI2 (s) Pb2+ (aq) I- (aq) Ksp = [Pb2+][I-]2
[Pb2+] = 1.67 x 10-2, [I-] = 6.67 x 10-2, Q = 7.43 x 10-5 > Ksp precipitate
Stoichiometry: Pb I PbI2
Initial 5mmol 20 mmol ----
Completion mmol ----
Equilibrium: some PbI2 redissolves, with I- common ion present
[I-] common ion = 10mmol / 300ml = M
PbI2 (s) Pb2+ (aq) I- (aq) Ksp = [Pb2+][I-]2
Initial M
Equilibrium x x
Ksp = 1.4 x 10–8 = [Pb2 +][I-]2 = (x)( x)2 ~~(x)(0.033)2
x = 1.3 x 10-5 M = [Pb2 +], [I-] = M
Example: ml 0.01 M Mg(NO3) ml 0.1 M NaF. Find [Mg2+] and [F-] at equilibrium. Ksp for MgF2 = 6.4 x 10-9

8. Qualitative Analysis
Selective Precipitation = addition of an ion that causes only one of a mixture of ions to precipitate
Add NaCl to Ag+ (aq) + Ba2+ (aq) AgCl(s) Na+ (aq) + Ba2+ (aq)
Example: Which ion will precipitate first? I- is added to a solution of M Cu+ and M Pb2+? Ksp CuI = 5.3 x Ksp PbI2 = 1.4 x 10-8.
Use Ksp CuI to find [I-] that will just start precipitation
Use Ksp PbI2 to find [I-] that will just start precipitation
Whichever has the lowest [I-] will precipitate first
Sulfide Ion (S2-) is particularly useful for selective cation precipitation
Metal sulfides have very different solubilities
MnS Ksp = 2.3 x 10-13
FeS Ksp = 3.7 x 10-19
Fe2+ would precipitate first from an equal mixture of Fe2+ and Mn2+
[S2-] can be controlled by pH
H2S H HS- Ka1 = 1 x 10-7
HS H S2- Ka2 = 1 x 10-19

9. S2- is quite basic. At low pH, there will be very little S2-
At high pH, there is much more S2-
We can selectively precipitate metal ions by adding S2- in acidic solution, and then slowly adding base.
CuS Ksp = 8.5 x 10-45
HgS Ksp = 1.6 x 10-54
MnS Ksp = 2.3 x 10-13
NiS Ksp = 3.0 x 10-21
Hg, then Cu, then Ni, then Mn
would precipitate as we raise pH
Solution of
Mn2+, Ni2+, Cu2+, Hg2+
Mn2+, Ni2+
Precip. Of
MnS, NiS
Precipitate of
CuS, HgS
Add H2S, pH = 2
Add OH- to pH = 8

10. Qualitative Analysis = scheme to separate and identify mixtures of cations by precipitation
Group I Insoluble Chlorides:
Add HCl. AgCl, PbCl2, Hg2Cl2 precipitate
Group II Sulfides Insoluble in Acid Solution:
Add H2S. Low [S2-] due to [H+]. HgS, CdS, Bi2S3, CuS, and SnS2 precip.
Group III Sulfides Insoluble in Basic Solution:
Add OH-. CoS, ZnS, MnS, NiS, FeS, Cr(OH)3, and Al(OH)3 precipitate.
Group IV Insoluble Carbonates:
Add CO32-. BaCO3, CaCO3, and MgCO3 precipitate.
Group V Alkali Metal and Ammonium
These ions (Na+, K+, NH4+) are soluble with common ions
f) Further tests would tell us which specific ions in each group are present

12. Complex Ion Equilibria
A complex ion = Lewis Acid—Lewis Base complex of a metal ion
Ligand = generic name for the Lewis Base bonded to a metal ion
H2O, NH3, Cl-, CN- all have lone pairs to donate; can be ligands
Coordination Number (CN) = the number of ligands bonded to a metal ion
CN = 6 is common: Co(H2O)62+, Ni(NH3)62+
CN = 4 CoCl42-, Cu(NH3)42+
CN = 2 Ag(NH3)2+
Coordination number depends on size and properties of the metal ion
Formation (or Stability) Constants
Ligand addition to the metal ion is stepwise
Ag NH Ag(NH3) K1 = 2.1 x 103
Ag(NH3) NH Ag(NH3) K2 = 8.2 x 103
Ag NH Ag(NH3) Kf = 1.7 x 107
Usually, [Ligand] >> [Mn+] to force the equilibria to the right
What are the equilibrium conditions of 100 ml 2.0 M NH3 added to 100 ml of M AgNO3?
K1 x K2 = Kf is large, favoring complete reaction
[NH3] is much larger than [Ag+], so assume constant

13. Do stoichiometry of the reaction:
Ag NH3 Ag(NH3)2+
Initial x 10-4 M 1 M
Equilib M 5 x 10-4 M
iv. Then use equilibrium expressions to find concentrations

14. Example: Find [Ag+], [Ag(S2O3)-], [Ag(S2O3)23-] for 150 ml 0
Example: Find [Ag+], [Ag(S2O3)-], [Ag(S2O3)23-] for 150 ml M AgNO ml 5.0 M Na2S2O3. K1 = 7.4 x 108, K2 = 3.9 x 104.
Complex Ions and Solubility
How do we dissolve AgCl(s)? Ksp = 1.6 x 10-10
Ag NH Ag(NH3) K1 = 2.1 x 103
Ag(NH3) NH Ag(NH3) K2 = 8.2 x 103
Adding NH3 to a suspension of AgCl, forces more of it to dissolve. The NH3, removes Ag+ from solution, forcing the equilibrium to the right.
[Ag+]T = [Ag+] [Ag(NH3)+] [Ag(NH3)2+]
The overall reaction is the sum of three individual reactions:
AgCl Ag Cl Ksp = 1.6 x 10-10
Ag NH Ag(NH3) K1 = 2.1 x 103
Ag(NH3) NH Ag(NH3) K2 = 8.2 x 103
AgCl NH Ag(NH3) Cl- K = ???
K = Ksp x K1 x K2 = (1.6 x 10-10)(2.1 x 103)(8.2 x 103) = 2.8 x 10-3

15. Example: Find the solubility of AgCl(s) in 10 M NH3.
AgCl NH Ag(NH3) Cl-
Initial
Equil – 2x x x
Solve using the expression for K as in a normal equilibrium problem.
Strategies for Dissolving Insoluble Solids
If the anion is basic, add acid
If the cation will form a complex, add ligand
Heat often increases solubility (temp. effects all equil. constants)
Example: HgS Hg S Ksp = 1 x 10-54
S2- is a basic anion, so add HCl
S H HS-
Hg2+ will form a chloride complex
Hg Cl HgCl42-

16. Precipitation of AgCl, Hg2Cl2, PbCl2
The qualitative analysis of the Group I cations illustrates complex ion equilibria
Solution of
Ag+, Hg22+, Pb2+
Add cold HCl
Precipitation of AgCl, Hg2Cl2, PbCl2
Heat
Precipitate of
AgCl, Hg2Cl2
Solution of Pb2+
Add CrO42-
Add NH3
Precipitate of
PbCrO4
Solution of
Ag(NH3)2+, Cl-
Precipitate of
Hg, HgNH2Cl
Add H+
Precip. of AgCl