Solubility Equilibria & Complex Ions
Know your solubility rules:
SOLUBILITY GUIDELINES
Soluble Compounds
NOT precipitates
Nitrates
Acetates
Chlorates
Chlorides
Bromides
Iodides
Sulfates
Exceptions
PRECIPITATES
None
None
None
Ag+1, Hg2+2, Pb+2
Ag+1, Hg2+2, Pb+2
Ag+1, Hg2+2, Pb+2
Ca+2, Sr+2, Ba+2, Hg2+2, Pb+2
Insoluble Compounds
PRECIPITATES
Sulfides
Carbonates
Phosphates
Hydroxides
Chromates
Exceptions
NOT Precipitates
NH4+1, Li+1, Na+1, K+1, Ca+2, Sr+2, Ba+2
NH4+1, Li+1, Na+1, K+1
NH4+1, Li+1, Na+1, K+1
Li+1, Na+1, K+1, Ca+2, Sr+2, Ba+2
NH4+1, Li+1, Na+1, K+1, Ca+2, Mg+2
17.4 Solubility Equilibria
The Solubility-Product Constant, Ksp
Consider a saturated solution of BaSO4 in contact with solid BaSO4.
We can write an equilibrium expression for the dissolving of the solid.
BaSO4(s)  Ba2+(aq) + SO42–(aq)
Since BaSO4(s) is a pure solid, the equilibrium expression depends only on the concentration of the ions.
Ksp =
Ksp is the equilibrium constant for the equilibrium between an ionic solid solute and its saturated aqueous solution.
Ksp is called the solubility-product constant
In general: the solubility product is equal to the product of the molar concentration of ions raised to powers corresponding to
their stoichiometric coefficients.
Al2(CO3)3  2Al3+ + 3CO32Ksp =
Solubility and Ksp
Solubility is the amount of substance that dissolves to form a saturated solution.
This can be expressed as grams of solid that will dissolve per liter of solution.
Molar solubility - the number of moles of solute that dissolve to form a liter of saturated solution.
Solubility can be used to find Ksp and Ksp can be used to find solubility (see problems)
Example 1:
a. If the molar solubility of CaF2 at 35oC is 1.24  10-3 mol/L, what is Ksp at this temperature?

CaF2
E .00124M actually dissolves
Ca+2
+
2 F-1
b.
It is found that 1.1  10-2 g of SrF2 dissolves per 100 mL of aqueous solution at 25oC. Calculate the solubility
product of SrF2.
SrF2

Sr+2
+
2 F-1
c.
The Ksp of Ba(IO3)2 at 25oC is 6.0  10-10. What is the molar solubility of Ba(IO3)2?
Ba(IO3)2

Ba+2
+
2 IO3-1
17.5 Factors That Affect Solubility
Factors that have a significant impact on solubility are:
- The presence of a common ion
- The pH of the solution
Common-Ion Effect
Solubility is decreased when a common ion is added.
This is an application of Le Châtelier’s principle:
Consider the solubility of CaF2:
CaF2(s)  Ca2+(aq) + 2F–(aq)
–
If more F is added (say by the addition of NaF), the equilibrium shifts left to offset the increase.
Therefore, more CaF2(s) is formed (precipitation occurs).
Example 2: Using Appendix D, calculate the molar solubility of AgBr in (a) pure water (b) 3.0  10-2 M AgNO3 solution
(c) 0.50 M NaBr solution
(a) AgBr  Ag+1 + Br-1
Ksp = 5.0  10-13
(b) AgBr

Ag+1
+
Br-1
Ksp = 5.0  10-13
(c) AgBr

Ag+1
+
Br-1
Ksp = 5.0  10-13
Notice the DECREASED solubility with the common ion in (b) and (c)
pH effects
Consider: Mg(OH)2(s)  Mg2+(aq) + 2 OH–(aq)
If OH– is removed, then the equilibrium shifts right and Mg(OH)2 dissolves.
OH– can be removed by adding a strong acid (lowering the pH):
OH–(aq) + H+(aq)  H2O(aq)
Another example:
CaF2(s)  Ca2+(aq) + 2 F–(aq)
–
If the F is removed, then the equilibrium shifts right and CaF2 dissolves.
F– can be removed by adding a strong acid (or lowering pH):
F–(aq) + H+(aq)  HF(aq)
Example 3: Calculate the molar solubility of Mn(OH)2 at (a) pH 7.0 (b) pH 9.5 (c) pH 11.8
the [OH-1] is set by the pH (or pOH)
(a)
pH = 7.0 so pOH = 7.0 so [OH-1] = 1.00  10-7
Mn(OH)2  Mn+2 + 2 OH-1
Ksp = 1.6  10-13
(b)
pH = 9.5 so pOH = 4.5 so [OH-1] = 3.16  10-5
Mn(OH)2  Mn+2 + 2 OH-1
Ksp = 1.6  10-13
(c)
pH = 11.8 so pOH = 2.2 so [OH-1] = 6.31  10-3
Mn(OH)2  Mn+2 + 2 OH-1
Ksp = 1.6  10-13
Common ion effect – increasing [OH-] decreases solubility
Example 4: Which of the following salts will be substantially more soluble in acidic solution than in pure water:
(a) ZnCO3 (b) ZnS (c) BiI3 (d) AgCN (e) Ba3(PO4)2
If the anion of the salt is the conjugate base of a weak acid, it will combine with H+1, reducing the concentration of the anion
and making the salt more soluble
ZnCO3  Zn+2 + CO3-2
the CO3 ion will react with the added H+
CO3-2 + H+  HCO3-1
Le Chatelier effect of removing CO3-2
-2
More soluble in acid:
17.6 Precipitation and Separation of Ions
Consider the following:
BaSO4(s)  Ba2+(aq) + SO42–(aq)
At any instant in time, Q = [Ba2+][ SO42– ]
If Q > Ksp, (too many ions) precipitation occurs until Q = Ksp.
If Q = Ksp equilibrium exists (saturated solution)
If Q < Ksp, (not enough ions) solid dissolves until Q = Ksp.
Selective Precipitation of Ions
Removal of one metal ion from a solution of two or more metal ions is called selective precipitation.
Ions can be separated from each other based on the solubilities of their salt compounds.
Example: If HCl is added to a solution containing Ag+ and Cu2+, the silver precipitates (as AgCl) while the Cu2+ remains in
solution
Generally, the less soluble ion is removed first!
Example 5: Will Ca(OH)2 precipitate if the pH of a 0.050 M solution of CaCl2 is adjusted to 8.0?
if Q > than Ksp then precipitation will occur
Example 6: A solution contains 0.00020 M Ag+1 and 0.0015 M Pb+2 .
Specify the [I-1] needed to begin precipitation for each cation.
If NaI is added, will AgI or PbI2 precipitate first?
the cation needing the lower [I-1] will precipitate first
AgI

Ag+1
+
I-1
Ksp = 8.3 x 10-17
PbI2

Pb+2
+
2 I-1
Ksp = 1.4 x 10-8
Complex Ions
Complex ion – a metal ion bonded to one or more Lewis bases. (We saw this with water in chapter 16)
It can happen with other Lewis bases (things that have lone pairs of electrons)
Rule of thumb: The number of Lewis bases (ligands) that a metal ion attracts is equal to double its charge. (Works about
75% of the time!)
Chapter 17 HW #2 (52, 54, 56, 58, 60, 62, 68, 72)