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) ⇌ HPO4

2-(aq) + H2O• This reaction reduces PO4

3- 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 S2O3

2-.

• 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

3-2

2

23

-23

net 10x 2.7 )2 - (1

]NH[

]Cl][)[Ag(NH

S

SK

Solubility Exercise #1

• Solubility of AgCl in 1.0 NH3 (continued):

• 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.

0.052 10x 2.7 )2 - (1

3- S

S

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-, SO4

2-, CO32-, PO4

3-, 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)

134

32

243

f 10x 1.1 ]NH][Cu[

])[Cu(NH

K

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)

(2) Ag(NH3)+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq);

3

3

3f1 10x 2.1

]][NH[Ag

])[Ag(NH

K

3

33

232f 10x 8.2

]][NH)[Ag(NH

])[Ag(NH

K

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);

7f2f12

3

23f 10x 1.7 x

]][NH[Ag

])[Ag(NH

KKK

Stepwise complex ion formation for Cu(NH3)42+

Individual equilibrium steps:1. Cu2+(aq) + NH3(aq) ⇌ Cu(NH3)2+(aq); K1 = 1.9 x 104

2. Cu(NH3)2+(aq) + NH3(aq) ⇌ Cu(NH3)22+(aq); K2 = 3.9 x 103

3. Cu(NH3)22+(aq) + NH3(aq) ⇌ Cu(NH3)3

2+(aq); K3 = 1.0 x 103

4. Cu(NH3)32+(aq) + NH3(aq) ⇌ Cu(NH3)4

2+(aq); K4 = 1.5 x 102

Combining equilibrium:• Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)4

2+(aq);

• Kf = K1 x K2 x K3 x K4 = 1.1 x 1013

]][NH[Cu

])[Cu(NH

43

2

243

f

K

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+ K = 2.1 x 103

AgNH3+ + NH3 Ag(NH3)2

+ 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