Chapter 15: Solubility Equilibrium

Sections 15.6-15.8

15.6 Solubility Equilibria

• Solubility of various compounds plays important roles.– Sugar and table salt’s

ability to dissolve in water help us to flavor food.

– Calcium sulfate is less soluble in hot water than cold water, which allows us to coat tubes in boilers.

– The low solubility of barium sulfate allows a safe way to X-ray the gastrointestinal tract.

Aqueous Equilibria

• Assume that when an ionic solid dissolves, that it dissociates completely.

CaF2(s) → Ca2+ (aq) + 2F- (aq)

• There is a possibility for the ions to collide and re-form the solid phase.

Ca2+ (aq) + 2F- (aq)→ CaF2(s) • Finally, equilibrium is reached, and no

more solid will dissolve (saturated).CaF2(s) ↔ Ca2+ (aq) + 2F- (aq)

Solubility Product Constant

• Remember that solids are left out of equilibrium expressions

Ksp=[Ca2+][F-]2

• Ksp is the solubility product constant (sometimes

referred to as just the “solubility product”).

*Notice that the amount of solid present nor the size of particles effect the equilibrium.

Solubility vs. Solubility Product

• Solubility is an equilibrium position.

• Solubility product is an equilibrium constant and has ONE value for a particular solid at a particular temperature.

Calculating Ksp from Solubility

Calculate Ksp for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15 mol/L at 25°C.

Ksp= [Bi3+]2[S2-]3

=(2.0 x 10-15)2 (3.0 x 10-15)3 = 1.1 x 10-73

HW 15.75-15.79 odd on page 743

Bi2S3(s) ↔ 2Bi3+(aq) + 3S2- (aq)

I Not needed 0 0

C -1.0 x 10-15 = -x +2x +3x

E Not needed 2.0 x 10-15 3.0 x 10-15

Calculating Solubility from Ksp

The Ksp value from Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility.

Ksp=[Cu2+][IO3-]2=[x][2x]2=(x)(4x2)=4x3

1.4 x 10-7=4x3

X=3.3 x 10-3 mol/L= Cu(IO3)2 solubility

HW 15.81 on page 743

Cu(IO3)2 (s) ↔ Cu2+(aq) + 2 IO3- (aq)

I Not needed 0 0

C -x +x +2x

E Not needed +x +2x

Relative Solubilities• Use Ksp values to predict relative solubilities of a group of

salts.– Salts producing the same number of ions

• Ex. AgI (s), CuI (s), CaSO4 (s)• Solubility is directly proportional to Ksp because

each compound produces 2 ions.» Largest Ksp= most soluble» Smallest Ksp= least soluble

– Salts producing differing numbers of ions.• Ex. CuS (s), Ag2S (s), Bi2S3 (s)• Solubility NOT directly proportional to Ksp and you

have to calculate the solubilities.

Practice Problems

• HW 15.83,15.87 on page 743

Common Ion Effect

Consider, the solubility of solid Ag2CrO4 (Ksp=9.0x10-12) in a 0.100 M aqueous solution of AgNO3.

– What is the common ion?– What is the [Ag+]0?

– What is the [CrO4-2]0?

– What is the Ksp expression?

Ksp = [Ag+]2[CrO4]2-

9.0 x 10-12 = (0.100+2x)2 (x)*Since Ksp is small, x is considered negligible compared to 0.100 M;

therefore 0.100+2x ≈ 0.100

9.0 x 10-12 = (0.100)2 (x)

X=9.0 x 10-10 M= solubility of Ag2CrO4 (s)

[Ag+]=0.100 M and [CrO4-]=9.0 x 10-10 M

Ag2CrO4 (s) ↔ 2 Ag+ (aq) + CrO4 2- (aq)

I Not needed 0.100 M 0

C -x +2x +x

E Not needed 0.100 + 2x +x

Solubility of Ag2CrO4 in pure water = 1.3 x 10-4 M

Solubility of Ag2CrO4 in 0.100 M AgNO3=9.0x10-10 M

*Note that solubility of a solid is lowered due to the common ion effect.

Example 15.85 on page 743

HW 15.91 and 15.93 on page 743.

pH and Solubility

• If the anion (X-) is an effective base (or HX is a weak acid), the salt MX will increase solubility in an acidic solution.– Effective bases include OH-, S2-, CO3-, C2O4

2-, and CrO4

2-.

– Salts with these anions are more soluble in acidic solutions that pure water.

HW 15.95 (in reference to 15.81) on page 743

Precipitation

• Precipitation: the formation of a solid from solution

• Ion Product (Q)=defined the same as Ksp except initial concentrations are used

Ex. Ca(NO3)2 (aq) is mixed with NaF (aq). What is the ion product for CaF2?

Q=[Ca2+]0[F-]02

To predict if precipitation will occur:

• If Q>Ksp, precipitation occurs until concentrations are reduced to satisfy Ksp.

• If Q<Ksp, no precipitation occurs.

Determining Precipitation Conditions

A solution is prepared by adding 750.0 mL of 4.00 x 10-3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp=1.9 x 10-10)

precipitate from this solution?

1. Calculate initial conditions.

2. Solve for Q.

Q=[Ce3+]0[IO3-]0

3=(2.86x10-3)(5.71x10-3)3= 5.32 x 10-10

3. Compare Q to Ksp.

Q>Ksp, so Ce(IO3)3 will precipitate.

Example 15.97 on page 743 (Ksp values on page 718)

33 3

0

(0.750 )(4.00 10 / )[ ] 2.86 10

0.750 0.300

L x mol LCe x M

L L

23

3 0

(0.300 )(2.00 10 / )[ ] 5.71 10

0.750 0.300

L x mol LIO x M

L L

Precipitation and Equilibrium

Concentrations

A solution is prepared by mixing 150.0 mL of 1.00x10-2 M Mg(NO3)2 and 250.0 mL of 1.00x10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp=6.4x10-9).

1. Find Initial Concentrations

2. Find Q

3. Compare Q vs. Ksp

Q>Ksp, so solid MgF2 will form.

* The next steps are used to determine equilibrium concentrations.

2 22 3

0

(150.0 )(1.00 10 )[ ] 3.75 10

400.0

mmolMg mL x MMg x M

mLsolution mL

12

0

(250.0 )(1.00 10 )[ ] 6.25 10

400.0

mmolF mL x MF x M

mLsolution mL

2 2 3 2 2 50 0[ ] [ ] (3.75 10 )(6.25 10 ) 1.46 10Q Mg F x x x

4. Run the reaction to completion. (BRA)

5. Calculate the concentration of excess reactant.

[F-]excess=22.0 mmol / 400.0 mL = 5.50 x 10-2 M

Mg2+ + 2 F- → MgF2(s)

Before rxn

(150.0 mL) (1.00x10-2M)=1.50 mmol

(250.0 mL) (1.00x10-1M)=25.0 mmol

0

Reaction -x -2 x +x

After rxn 1.50-1.50=0 *limiting reactant

25.0-2(1.50)=22.0 mmol*excess

1.50 mmol

6. Determine concentrations at equilibrium (ICE).

Ksp= 6.4x10-9=[Mg2+][F-]2

=(x)(5.50x10-2+2x)2 ≈ (x)(5.50x10-2)2

X=2.1 x 10-6 M = [Mg2+]

[F-]=5.50x10-2 M

HW15.99 on page 743

MgF2 (s) ↔ Mg2+ (aq) + 2F- (aq)

I Not needed 0 M 5.50 x 10-2 M

C -x +x +2x

E Not needed + x 5.50 x 10-2 + 2x

Selective Precipitation• Using anions that form precipitates with only one or a few

metal ions in a mixture in order to separate the metal ions.

• Most insoluble sulfide salts can be precipitated in an acidic solutions. • Soluble sulfide salts can be precipitated by making the solution slightly

basic.

A solution contains 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp=1.4x10-8) or CuI (Ksp=5.3x10-12) precipitate first? Specify the [I-] necessary to begin precipitation of each salt.

-For PbI2: Ksp=1.4x10-8=[Pb2+][I-]2

1.4x10-8=(2.0x10-3) [I-]2

[I-]=2.6 x 10-3 M is necessary to begin precipitation

-For CuI: Ksp=5.3x10-12=[Cu+][I-] 5.3x10-12=(1.0x10-4)[I-]

[I-]=5.3x10-8 M is necessary to begin precipitation

*CuI will precipitate first since [I-] required is less.

HW 15.101 on page 744

Qualitative Analysis

Groups:1. Insoluble Chlorides (Ag+, Pb2+, Hg2

2+)

2. Sulfides insoluble in acid solution (Hg2+, Cd2+, Bi3+, Cu2+, Sn4+)

3. Sulfides insoluble in basic solution (Co2+, Zn2+, Mn2+, Ni2+, Fe2+, Cr3+, Al3+)

4. Insoluble Carbonates (Group 2A)

5. Alkali Metal and ammonium ions (flame test)

Complex Ion Equilibria

• Complex Ion: a charged species with a metal ion surrounded by ligands (Ag(NH3)+)– The ligand donates its lone electron pair to an empty

orbital of the metal ion to form a covalent bond

• Ligand: a Lewis base (H2O, NH3, Cl-, CN-)

• Coordination number: the number of ligands attached to the metal ion

• Formation constants or stability constants:– Metal ions add ligands one at a time in steps

characterized by their own equilibrium constants

Ag+ + NH3↔ Ag(NH3)+ K1=2.1x103

Ag(NH3)+ + NH3↔ Ag(NH3)2+ K2=8.2x103

– All species (Ag+, NH3, Ag(NH3)+, Ag(NH3)2+) exist

at equilibrium.

HW 15.107 on page 744

• Usually, [ligand] is much larger than [metal ion] and approximations are used to simplify problems.

• Assume both reactions go to completion

Ag+ + NH3↔ Ag(NH3)+ K1=2.1x103

Ag(NH3)+ + NH3↔ Ag(NH3)2+ K2=8.2x103

Ag+ + 2 NH3 → Ag(NH3)2+ K=K1 x K2

• From here, use BRA to find concentrations and the equilibrium constant. – Assume the ligand amount consumed is neglible

Examples 15.103a, 105, 109 on page 744

HW 15.103 b, 107on page 744

Strategies for Dissolving Water-Insoluble Ionic Solids

1. If the anion of the solid is a good base, solubility is increased with the addition of an acid

2. If the anion is NOT a good base, solids can be dissolved in a solution containing a ligand to form stable complex ions with its cation.

Aqueous Ammonia is Added to Silver Chloride (white). Silver Chloride, Insoluble in Water,

Dissolves to Form Ag(NH3)2+ (aq) and Cl-(aq)

Ex 15.111 on page 745HW 15.113 and 15.115 on page 745