Solubility Equilibrium
• In saturated solutions dynamic equilibrium
exists between undissolved solids and ionic
species in solutions
• Solids continue to dissolve and ion-pairs
continue to form solids.
• The rate of dissolution process is equal to the
rate of precipitation.
Solubility Product Constant
•
•
General expression:
MmXn(s) ⇄ mMn+(aq) + nXm-(aq)
•
Solubility product, Ksp = [Mn+]m[Xm-]n
Solubility and Solubility Products
Examples:
• AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
•
Ksp = [Ag+][Cl-] = 1.6 x 10-10
• If s is the solubility of AgCl, then:
•
•
•
[Ag+] = s and [Cl-] = s
Ksp = (s)(s) = s2 = 1.6 x 10-10
s = 1.3 x 10-5 mol/L
Solubility and Solubility Products
• Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq)
•
Ksp = [Ag+]2[CrO42-] = 9.0 x 10-12
• If s is the solubility of Ag2CrO4, then:
•
•
•
[Ag+] = 2s and [CrO42-] = s
Ksp = (2s)2(s) = 4s3 = 9.0 x 10-12
s = 1.3 x 10-4 mol/L
Solubility and Solubility Products
• More Examples:
• Ca(IO3)2(s) ⇌ Ca2+(aq) + 2 IO3-(aq)
•
Ksp = [Ca2+][IO3-]2 = 7.1 x 10-7
• If the solubility of Ca(IO3)2(s) is s mol/L, then:
•
Ksp = 4s3 = 7.1 x 10-7
•
s = 5.6 x 10-3 mol/L
Solubility and Solubility Products
• Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-(aq)
•
Ksp = [Mg2+][OH-]2 = 8.9 x 10-12
• If the solubility of Mg(OH)2 is s mol/L, then:
• [Mg2+] = s mol/L and [OH-] = 2s mol/L,
•
Ksp = (s)(2s)2 = 4s3 = 8.9 x 10-12
•
s = 1.3 x 10-4 mol/L
Solubility and Solubility Products
• More Examples:
• Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq)
•
Ksp = [Ag+]3[PO43-] = 1.8 x 10-18
• If the solubility of Ag3PO4 is s mol/L, then:
•
Ksp = (3s)3(s) = 27s4 = 1.8 x 10-18
•
s = 1.6 x 10-5 mol/L
Solubility and Solubility Products
•
Cr(OH)3(s) ⇌ Cr3+(aq) + 3 OH-(aq)
•
Ksp = [Cr3+][OH-]3 = 6.7 x 10-31
• If the solubility is s mol/L, then:
• Ksp = [Cr3+][OH-]3 = (s)(3s)3 = 27s4 = 6.7 x 10-31
•
s = 1.3 x 10-8 mol/L
Solubility and Solubility Products
• More Examples:
• Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43-(aq)
•
Ksp = [Ca2+]3[PO43-]2 = 1.3 x 10-32
• If the solubility is s mol/L, then:
• [Ca2+] = 3s, and [PO43-] = 2s
• Ksp = (3s)3(2s)2 = 108s5 = 1.3 x 10-32
• s = 1.6 x 10-7 mol/L
Factors that affect solubility
• Temperature
– Solubility generally increases with temperature;
• Common ion effect
– Common ions reduce solubility
• Salt effect
– This slightly increases solubility
• pH of solution
– pH affects the solubility of ionic compounds in which the
anions are conjugate bases of weak acids;
• Formation of complex ion
– The formation of complex ion increases solubility
Common Ion Effect
• Consider the following solubility equilibrium:
• AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10;
• The solubility of AgCl is 1.3 x 10-5 mol/L at 25oC.
• If NaCl is added, equilibrium shifts left due to
increase in [Cl-] and some AgCl will precipitate out.
• For example, if [Cl-] = 1.0 x 10-2 M,
• Solubility of AgCl = (1.6 x 10-10)/(1.0 x 10-2)
•
= 1.6 x 10-8 mol/L
Effect of pH on Solubility
• Consider the following equilibrium:
Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq);
• If HNO3 is added, the following reaction occurs:
H3O+(aq) + PO43-(aq) ⇌ HPO42-(aq) + H2O
• This reaction reduces PO43- in solution, causing more
solid Ag3PO4 to dissolve.
• In general, the solubility of compounds such as
Ag3PO4, which anions are conjugate bases of weak
acids, increases as the pH is lowered by adding nitric
acid.
Effect of pH on Solubility
• Consider the following equilibrium:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-(aq);
• Increasing the pH means increasing [OH-] and
equilibrium will shift to the left, causing some of
Mg(OH)2 to precipitate out.
• If the pH is lowered, [OH-] decreases and equilibrium
shifts to the right, causing solid Mg(OH)2 to dissolve.
• The solubility of compounds of the type M(OH)n
decreases as pH is increased, and increases as pH is
decreased.
Formation of Complex Ions on Solubility
• Many transition metals ions have strong affinity for
ligands to form complex ions.
• Ligands are molecules, such as H2O, NH3 and CO, or
anions, such as F-, CN- and S2O32-.
• Complex ions are soluble – thus, the formation of
complex ions increases solubility of slightly soluble
ionic compounds.
Effect of complex ion formation on solubility
• Consider the following equilibria:
• AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10
• Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) ; Kf = 1.7 x 107
• Combining the two equations yields:
• AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq);
• Knet = Ksp x Kf = (1.6 x 10-10) x (1.7 x 107)
= 2.7 x 10-3
• Knet > Ksp implies that AgCl is more soluble in
aqueous NH3 than in water.
Solubility Exercise #1
• Calculate the solubility of AgCl in water and in 1.0 M
NH3 solution at 25oC.
• Solutions:
Solubility in water = (Ksp)
= (1.6 x 10-10) = 1.3 x 10-5 mol/L
Solubility Exercise #1
• Solubility of AgCl in 1.0 NH3:
•
AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq)

• [Initial], M
1.0
0.0
0.0
• [Change]
-2S
+S
+S
• [Equilm.]
(1 – 2S)
S
S

K net 
[Ag(NH

-
) ][ Cl ]
3 2
[ NH 3 ]
2

S
2
(1 - 2 S )
2
 2.7 x 10
-3
Solubility Exercise #1
• Solubility of AgCl in 1.0 NH3 (continued):
S

2.7 x 10
-3
 0.052
(1 - 2 S )
•
S = 0.052 – 0.104S;
•
S = 0.052/1.104 = 0.047 mol/L
• AgCl is much more soluble in NH3 solution than in
water.
Predicting Formation of Precipitate
• Qsp = Ksp  saturated solution, but no precipitate
• Qsp > Ksp  saturated solution, with precipitate
• Qsp < Ksp  unsaturated solution,
• Qsp is ion product expressed in the same way as Ksp
for a particular system.
Predicting Precipitation
• Consider the following case:
20.0 mL of 0.025 M Pb(NO3)2 is added to 30.0 mL of 0.10
M NaCl. Predict if precipitate of PbCl2 will form.
(Ksp for PbCl2 = 1.6 x 10-5)
Predicting Precipitation
•
•
•
•
•
•
Calculation:
[Pb2+] = (20.0 mL x 0.025 M)/(50.0 mL) = 0.010 M
[Cl-] = (30.0 mL x 0.10 M)/(50.0 mL) = 0.060 M
Qsp = [Pb2+][Cl-]2 = (0.010 M)(0.060 M)2
= 3.6 x 10-5
Qsp > Ksp  precipitate of PbCl2 will form.
Practical Applications of Solubility Equilibria
• Qualitative Analyses
– Isolation and identification of cations and/or anions in
unknown samples
• Synthesis of Ionic Solids of commercial interest
• Selective Precipitation based on Ksp
Qualitative Analysis
• Separation and identification of cations, such as Ag+,
Ba2+, Cr3+, Fe3+, Cu2+, etc. can be carried out based
on their different solubility and their ability to form
complex ions with specific reagents, such as HCl,
H2SO4, NaOH, NH3, and others.
• Separation and identification of anions, such as Cl-,
Br-, I-, SO42-, CO32-, PO43-, etc., can be accomplished
using reagents such as AgNO3, Ba(NO3)2 under
neutral or acidic conditions.
Selective Precipitation
(Mixtures of Metal Ions)
• Use a reagent whose anion forms a precipitate
with only one or a few of the metal ions in the
mixture.
• Example:
 Solution contains Ba2+ and Ag+ ions.
 Adding NaCl will form a precipitate with Ag+
(AgCl), while still leaving Ba2+ in solution.
Separation of Cu2+ and Hg2+ from Ni2+ and
Mn2+ using H2S
• At a low pH, [S2–] is relatively low and only
the very insoluble HgS and CuS precipitate.
• When OH– is added to lower [H+], the value
of [S2–] increases, and MnS and NiS
precipitate.
Separation of Cu2+ and Hg2+ from Ni2+ and
Mn2+ using H2S
Separating the Common Cations by
Selective Precipitation
Synthesis of Ionic Solids
• Chemicals such as AgCl, AgBr, and AgI that are
important in photography are prepared by
precipitation method.
• AgNO3(aq) + KBr(aq)  AgBr(s) + KNO3(aq)
Selective Precipitation
• Compounds with different solubility can be
selectively precipitated by adjusting the concentration
of the precipitating reagents.
• For example, AgCl has a much lower Ksp than PbCl2
• If Ag+ and Pb2+ are present in the same solution, the
Ag+ ion can be selectively precipitated as AgCl,
leaving Pb2+ in solution.
Complex Ion Equilibria
• Complex ions are ions consisting central metal ions
and ligands covalently bonded to the metal ions;
• Ligands can be neutral molecules such as H2O, CO,
and NH3, or anions such as Cl-, F-, OH-, and CN-;
• For example, in the complex ion [Cu(NH3)4]2+, four
NH3 molecules are covalently bonded to Cu2+.
Formation of Complex Ions
• In aqueous solutions, metal ions form complex ions
with water molecules as ligands.
• If stronger ligands are present, ligand exchanges
occur and equilibrium is established.
• For example:
Cu2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4]2+(aq)
Kf 
[Cu(NH
[ Cu
2
)
3 4
2
][ NH 3 ]
]
4
 1.1 x 10
13
Stepwise Formation of Complex Ion
•
•
•
At molecular level, ligand molecules or ions combine with
metal ions in stepwise manner;
Each step has its equilibrium and equilibrium constant;
For example:
(1) Ag+(aq) + NH3(aq) ⇌ Ag(NH3)+(aq)
K f1 
[Ag(NH

) ]
3

 2.1 x 10
3
[Ag ][NH 3 ]
(2) Ag(NH3)+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq);
Kf2 

[Ag(NH
[Ag(NH
) ]
3 2

) ][NH 3 ]
3
 8.2 x 10
3
Stepwise Formation of Complex Ion
Individual equilibrium steps:
(1) Ag+(aq) + NH3(aq) ⇌ Ag(NH3)+(aq); Kf1 = 2.1 x 103
(2) Ag(NH3)+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq); Kf2 = 8.2 x 103
Combining (1) and (2) yields:
•
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq);
Kf 
[Ag(NH


) ]
3 2
[Ag ][NH 3 ]
2
 K f1 x K f2  1.7 x 10
7
Stepwise complex ion formation for Cu(NH3)42+
Individual equilibrium steps:
1. Cu2+(aq) + NH3(aq) ⇌ Cu(NH3)2+(aq);
2. Cu(NH3)2+(aq) + NH3(aq) ⇌ Cu(NH3)22+(aq);
3. Cu(NH3)22+(aq) + NH3(aq) ⇌ Cu(NH3)32+(aq);
4. Cu(NH3)32+(aq) + NH3(aq) ⇌ Cu(NH3)42+(aq);
Combining equilibrium:
•
Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+(aq);
Kf 
•
[Cu(NH
[Cu
2
)
3 4
2
][NH 3 ]
]
4
Kf = K1 x K2 x K3 x K4 = 1.1 x 1013
K1 = 1.9 x 104
K2 = 3.9 x 103
K3 = 1.0 x 103
K4 = 1.5 x 102
Complex Ions and Solubility
• Two strategies for dissolving a water–
insoluble ionic solid.
 If the anion of the solid is a good base, the
solubility is greatly increased by acidifying the
solution.
 In cases where the anion is not sufficiently basic,
the ionic solid often can be dissolved in a solution
containing a ligand that forms stable complex
ions with its cation.
Concept Check
(a) Calculate the solubility of silver chloride in
10.0 M ammonia given the following information:
Ksp (AgCl) = 1.6 x 10–10
Ag+ + NH3
AgNH3+
AgNH3+ + NH3
Ag(NH3)2+
K = 2.1 x 103
K = 8.2 x 103
(b) Calculate the concentration of NH3 in the final
equilibrium mixture.
Answers: (a) 0.48 M;
(b) 9.0 M