1. Solubility and Equilibrium

2. Ionic Equilibria in Aqueous Systems
19.1 Equilibria of Acid-Base Buffers
19.2 Acid-Base Titration Curves
19.3 Equilibria of Slightly Soluble Ionic Compounds
19.4 Equilibria Involving Complex Ions

3. Equilibria of Slightly Soluble Ionic Compounds
Any “insoluble” ionic compound is actually slightly soluble in aqueous solution: very small amount of such a compound that dissolves will dissociate completely.
For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions.
PbF2(s) Pb2+(aq) + 2F-(aq)
Qsp = Qc[PbF2] = [Pb2+][F-]2
Qc =
[Pb2+][F-]2
[PbF2]

4. Qsp and Ksp
Qsp is called the ion-product expression for a slightly soluble ionic compound.
For any slightly soluble compound MpXq:
MpXq(s) pMn+(aq) + qXz-(aq)
Qsp = [Mn+]p[Xz-]q
When the solution is saturated, the system is at equilibrium, and Qsp = Ksp, the solubility product constant.
The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation).

5. Writing Ion-Product Expressions
Write the ion-product expression at equilibrium for each compound:
(a) magnesium carbonate (b) iron(II) hydroxide
(c) calcium phosphate (d) aluminum hydroxide
Ksp = [Mg2+][CO32-]
Ksp = [Fe2+][OH-]2
Ksp = [Ca2+]3[PO43-]2
Ksp = [Al3+][OH-]3

6. Solubility-Product Constants (Ksp) of Selected Ionic Compounds at 25°C
Name, Formula
Ksp
Aluminum hydroxide, Al(OH)3
Cobalt(II) carbonate, CoCO3
Iron(II) hydroxide, Fe(OH)2
Lead(II) fluoride, PbF2
Lead(II) sulfate, PbSO4
Silver sulfide, Ag2S
Zinc iodate, Zn(IO3)2
3x10-34
1.0x10-10
4.1x10-15
3.6x10-8
1.6x10-8
4.7x10-29
8x10-48
Mercury(I) iodide, Hg2I2
3.9x10-6

7. PROBLEM: Determining Ksp from Solubility
Lead(II) sulfate (PbSO4) is a key component in lead-acid car batteries. Its solubility in water at 25°C is 4.25x10-3 g/100 mL solution. What is the Ksp of PbSO4?
[PbSO4(aq)] = 1.40x10-4 M [Pb2+] = [SO42-] = 1.40x10-4 M Ksp = 1.96x10-8

8. Determining Ksp from Solubility
When lead(II) fluoride (PbF2) is shaken with pure water at 25°C, the solubility is found to be 0.64 g/L. Calculate Ksp of PbF2.
[PbF2(aq)] = 2.6 x10-3 M [Pb2+] = 2.6 x10-3 M [F-] = 5.2 x10-3 M Ksp = 7.0 x10-8

9. Determining Solubility from Ksp
PROBLEM:
Calculate the molar solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.
S = =
= 1.2x10-2 M

10. Practice: Correlate Ksp with molar solubility for ionic compound
(Ksp = xsm)
(a) magnesium carbonate (b) iron(II) hydroxide
(c) calcium phosphate (d) aluminum hydroxide

11. Relationship Between Ksp and Solubility at 25°C
No. of Ions
Formula
Cation/Anion
Ksp
Solubility (M) 2
MgCO3
1/1
3.5x10-8
1.9x10-4
PbSO4
1.6x10-8
1.3x10-4
BaCrO4
2.1x10-10
1.4x10-5 3
Ca(OH)2
1/2
6.5x10-6
1.2x10-2
BaF2
1.5x10-6
7.2x10-3
CaF2
3.2x10-11
2.0x10-4
Ag2CrO4
2/1
2.6x10-12
8.7x10-5
The higher the Ksp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas.

12. Predicting Formation of a Precipitate
For a saturated solution of ionic salt, Qsp = Ksp.
For equilibrium MpXq(s) pMn+(aq) + qXz-(aq)
Shift to the right = ___________
Shift to the left = _____________
When two solutions containing the ions of slightly soluble salts are mixed,
If Qsp = Ksp, the solution is saturated and no change will occur.
If Qsp > Ksp, a precipitate will form
If Qsp =< Ksp, no precipitate will form

13. How to Predict Whether a Precipitate Will Form?
Find [ions] after two solutions are mixed. Consider the volume change if applicable.
M2 = M1 x V1/V2
Calculate solubility quotient Qsp from the formula of possible products
If calculated Qsp is larger than Ksp, precipitation will form

14. Predicting Whether a Precipitate Will Form
A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate form when L of 0.30 M Ca(NO3)2 is mixed with L of M NaF? Ksp(CaF2) = 3.2x10-11.
[Ca2+] = 0.10 M, [F-] = M Q(CaF2) = 1.6x10-4

15. Selective Precipitation
Selective precipitation is used to separate a solution containing a mixture of ions.
After adding a precipitating ion that form precipitate with both ions, Qsp increases from 0.
Once Qsp > Ksp(less soluble compd), precipitate of that compound forms.
As long as Qsp < Ksp(more soluble compd), the less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound.

16. Selective Precipitation
To completely separate two metal ions, the concentration of the precipitating ion should results in
Ksp(more soluble precipitate)  Qsp > Ksp(less soluble precipitate)
The maximal concentration of the precipitating ion can be determined from the Ksp of the more soluble precipitate.

17. Separating Ions by Selective Precipitation
A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.
[OH-] = 5.6x10-5 M

18. The effect of a common ion on solubility
Add Na2CrO4(aq)
saturated PbCrO4(aq)
More PbCrO4(s)
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
Why? The added common ion CrO42- causes the equilibrium to shift to the left. Solubility decreases and solid PbCrO4 precipitates.

19. Calculating the Effect of a Common Ion on Solubility
In the previous practice, we calculated the solubility of Ca(OH)2 in water as M. What is its solubility in 0.10 M KNO3? Ksp of Ca(OH)2 is 6.5x10-6.
Using approximation, s = 6.5x10-4 M

20. Effect of pH on Solubility
Changes in pH affects the solubility of many slightly soluble ionic compounds.
The addition of H3O+ will increase the solubility of a salt that contains the anion of a weak acid.
CaCO3(s) Ca2+(aq) + CO32-(aq)
CO32-(aq) + H3O+(aq) → HCO3-(aq) + H2O(l)
HCO3-(aq) + H3O+(aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)
The net effect of adding H3O+ to CaCO3 is the removal of CO32- ions, which causes an equilibrium shift to the right. More CaCO3 will dissolve.

21. Test for the presence of a carbonate.
When a carbonate mineral is treated with HCl, bubbles of CO2 form.

22. Predicting the Effect on Solubility of Adding Strong Acid
Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of each ionic compound:
(a) lead(II) bromide (b) copper(II) hydroxide (c) iron(II) phosphate

23. Limestone cave in Nerja, Málaga, Spain.
Limestone is mostly CaCO3 (Ksp = 3.3x10-9).
Ground water rich in CO2 trickles over CaCO3, causing it to dissolve. This gradually carves out a cave.
Water containing HCO3- and Ca2+ ions drips from the cave ceiling. The air has a lower PCO2 than the soil, causing CO2 to come out of solution. A shift in equilibrium results in the precipitation of CaCO3 to form stalagmites and stalactites.
CO2(g) CO2(aq)
CO2(aq) + 2H2O(l) H3O+(aq) + HCO3-(aq)
CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)

24. Formation of acidic precipitation.
Chemical Connections
Formation of acidic precipitation.
Since pH affects the solubility of many slightly soluble ionic compounds, acid rain has far-reaching effects on many aspects of our environment.

25. Formation of Complex ion affects Solubility equilibrium
Recall in Chapter 21 metal ion forms complex ions with ligands: Mx+ + n L→ MLnx+
Starting at equilibrium where Qsp = Ksp, formation of complex ion from the metal ion will affect solubility equilibrium by reducing [Mx+].
After adding a reagent that form complex ion with metal cation Mx+, [Mx+] , so Qsp .
Since Qsp < Ksp, the equilibrium shifts towards right, causing more dissolving.

26. Examples of Formation of Complex ion affecting Solubility equilibrium
Addition of ammonia helps insoluble silver chloride to dissolve: Ag+ + 2 NH3 → Ag(NH3)2+
Addition of excess hydroxide ion helps insoluble aluminum hydroxide to dissolve:
Al(OH)3(s) + 3 OH-(aq) → Al(OH)63-(aq)