SOLUBILITY

Saturated SolutionSaturated SolutionBaSOBaSO4(s)4(s) Ba Ba2+2+

(aq)(aq) + SO + SO442-2-

(aq)(aq)

Equilibrium expresses the degree of Equilibrium expresses the degree of

solubility of solid in water.solubility of solid in water.KKspsp = solubility product constant = solubility product constant

KKspsp = K = Keqeq [BaSO [BaSO44]](s) (s)

KKspsp = [Ba = [Ba2+2+] [SO] [SO442-2-] = 1.1 x 10] = 1.1 x 10-10-10

KKspsp represents the amount of dissolution represents the amount of dissolution

(how much solid dissolved into ions), the (how much solid dissolved into ions), the smaller the Ksmaller the Kspsp value, the smaller the amount value, the smaller the amount

of ions in solution (more solid is present).of ions in solution (more solid is present).

Table 1 Solubility-Product Constants (KTable 1 Solubility-Product Constants (Kspsp) of ) of

Selected Ionic Compounds at 25Selected Ionic Compounds at 2500CC

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

3 x 10-34

1.0 x 10-10

4.1 x 10-15

3.6 x 10-8

1.6 x 10-8

4.7 x 10-29

8 x 10-48

Mercury (I) iodide, Hg2I2

3.9 x 10-6

SOLUBILITY

11. Write the solubility product . Write the solubility product expression for each of the following:expression for each of the following:

a) Caa) Ca33(PO(PO44))22

b) Hgb) Hg22ClCl22 c) HgClc) HgCl2.2.

22. In a particular sample, the . In a particular sample, the concentration of silver ions was 1.2 concentration of silver ions was 1.2 x10x10-6 -6 M and the concentration of M and the concentration of bromide was 1.7x10bromide was 1.7x10-6 -6 M. What is M. What is the value of Kthe value of Kspsp for AgBr? for AgBr?

Solubility vs. Solubility ProductSolubility vs. Solubility Product

Solubility: The quantity of solute that Solubility: The quantity of solute that dissolves to form a saturated dissolves to form a saturated solution. (solution. (gg//LL))

Molar Solubility: (n Molar Solubility: (n solutesolute/L /L saturated solutionsaturated solution))

KKspsp: The equilibrium between the : The equilibrium between the

ionic solid and the saturated solution.ionic solid and the saturated solution.

Interconverting solubility and Ksp

SOLUBILITYSOLUBILITYOF OF

COMPOUNDCOMPOUND(g/L)(g/L)

MOLAR MOLAR SOLUBILITYSOLUBILITY

OFOFCOMPOUNDCOMPOUND

(mol/L)(mol/L)

MOLAR MOLAR CONCENTRATIONCONCENTRATION

OF OF IONSIONS

KKspsp

Sample Problem 1Sample Problem 1 Determining Ksp from Solubility

PROBLEM: (a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?

(b) When lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the Ksp of PbF2.

PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression.

Ksp = [Pb2+][SO42-]

= 1.40x10-4M PbSO4

Ksp = [Pb2+][SO42-] = (1.40x10-4)2 =

SOLUTION: PbSO4(s) Pb2+(aq) + SO42-(aq)(a)

1000mL

L

4.25x10-3g

100mL soln 303.3g PbSO4

mol PbSO4

1.96x10-8

Sample Problem 1Sample Problem 1 Determining Ksp from Solubility

continued

(b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2

= 2.6x10-3 M

Ksp = (2.6x10-3)(5.2x10-3)2 =

0.64g

L soln 245.2g PbF2

mol PbF2

7.0x10-8

Sample Problem 2Sample Problem 2 Determining Solubility from Ksp

PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.

PLAN: Write out a dissociation equation and Ksp expression; Find the molar solubility (S) using a table.

SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2

-Initial

Change

Equilibrium

-

-

0 0

+S + 2S

S 2S

Ksp = (S)(2S)2 S =

6.5x10 6

43 = 1.2x10x-2M

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (M)

Solubility vs. Solubility ProductSolubility vs. Solubility Product

11. A student finds that the . A student finds that the solubility of BaFsolubility of BaF22 is 1.1 g in l.00 L is 1.1 g in l.00 L of water. What is the value of Ksp?of water. What is the value of Ksp?

22. Exactly 0.133 mg of AgBr will . Exactly 0.133 mg of AgBr will dissolve in 1.00 L of water. What dissolve in 1.00 L of water. What is the value of Ksp for AgBr?is the value of Ksp for AgBr?

33. Calomel (Hg. Calomel (Hg22ClCl22) was once used ) was once used in medicine. It has a Ksp = 1.3 x in medicine. It has a Ksp = 1.3 x 1010-18-18. What is the solubility of . What is the solubility of HgHg22ClCl2 2 in g/L?in g/L?

Relationship Between KRelationship Between Kspsp and Solubility at 25 and Solubility at 2500CC

No. of Ions Formula Cation:Anion Ksp Solubility (M)

2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4

2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4

2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5

3 Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2

3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3

3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4

3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5

Ion-Product Expression (Qsp)& Solubility Product Constant (Ksp)

At equilibrium Qsp = [Mn+]p [Xz-]q = Ksp

For the hypothetical compound, MpXq

Solubility and Common Ion effectSolubility and Common Ion effect

CaFCaF2(s)2(s) Ca Ca2+2+(aq)(aq) + 2F + 2F--

(aq)(aq)

The addition of CaThe addition of Ca2+2+ or F or F- - shifts the shifts the equilibrium. According to Le Chatelier’s equilibrium. According to Le Chatelier’s Principle, more solid will form thus Principle, more solid will form thus reducing the solubility of the solid.reducing the solubility of the solid.

Solubility of a salt decreases when Solubility of a salt decreases when the solute of a common ion is added.the solute of a common ion is added.

The effect of a common ion on solubility

PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO4

2-(aq)

CrO42- added

Sample Problem 3 Calculating the Effect of a Common Ion on Solubility

PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.

PLAN: Set up a reaction equation and table for the dissolution of Ca(OH)2. The Ca(NO3)2 will supply extra [Ca2+] and will relate to the molar solubility of the ions involved.

SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration(M)

Initial

Change

Equilibrium

-

-

-

0.10 0

+S +2S

0.10 + S 2S

Ksp = 6.5x10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2 S << 0.10

S = = 4.0x10-3 Check the assumption:

4.0%

0.10M

4.0x10-3 x 100 =

6.5x10 6

4

Solubility and Common Ion effectSolubility and Common Ion effect

CaFCaF2(s)2(s) Ca Ca2+2+(aq)(aq) + 2F + 2F--

(aq)(aq)

11. The K. The Kspsp of the above equation is of the above equation is 3.2 x 103.2 x 10-11-11. (a) Calculate the . (a) Calculate the molar solubility in pure water. (b) molar solubility in pure water. (b) Calculate the molar solubility in Calculate the molar solubility in 3.5 x 103.5 x 10-4-4 M Ca(NO M Ca(NO33))22..

2. What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 x 10-2 mol of NaCl?

CRITERIA FOR PRECIPITATION OF DISSOLUTION

BaSO4(s) Ba2+(aq) + SO4

2-(aq)

Equilibrium can be established from either direction.

Q (the Ion Product) is used to determine whether or not precipitation will occur.

Q < K solid dissolvesQ = K equilibrium (saturated solution)

Q > K ppt

Sample Problem 3Sample Problem 3 Predicting Whether a Precipitate Will Form

PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF?

PLAN: Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations.

SOLUTION: CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11

mol Ca2+ = 0.100L(0.30mol/L) = 0.030mol [Ca2+] = 0.030mol/0.300L = 0.10M

mol F- = 0.200L(0.060mol/L) = 0.012mol [F-] = 0.012mol/0.300L = 0.040M

Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6x10-4

Q is >> Ksp and the CaF2 WILL precipitate.

CRITERIA FOR PRECIPITATION OF DISSOLUTION

1. Calcium phosphate has a Ksp of 1x10-26, if a sample contains 1.0x10-3 M Ca2+ & 1.0x10-8 M PO4

3- ions, calculate Q and predict whether Ca3(PO4)2 will precipitate?

2.Exactly 0.400 L of 0.50 M Pb2+ & 1.60 L of 2.5 x 10-8 M Cl- are mixed together to form 2.00L. Calculate Q and predict if a ppt will occur.

Ksp = 1.6 x 10-5

EFFECT OF pH ON SOLUBILITY

CaF2 Ca2+ + 2F-

2F- + 2H+ 2HF CaF2 + 2H+ Ca2+ + 2HF

Salts of weak acids are more soluble in acidic solutions. Thus shifting the solubility to the right. Salts with anions of strong acids are largely unaffected by pH.

Sample Problem 4Sample Problem 4 Predicting the Effect on Solubility of Adding Strong Acid

PROBLEM: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:

(a) Lead (II) bromide (b) Copper (II) hydroxide (c) Iron (II) sulfide

PLAN: Write dissolution equations and consider how strong acid would affect the anion component.

Br- is the anion of a strong acid.

No effect.

SOLUTION: (a) PbBr2(s) Pb2+(aq) + 2Br-(aq)

(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)

OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility.

(c) FeS(s) Fe2+(aq) + S2-(aq) S2- is the anion of a weak acid and will react with water to produce OH-.

Both weak acids serve to increase the solubility of FeS.

FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)

EFFECT OF pH ON SOLUBILITY

1. Consider the two slightly soluble salts BaF2 and AgBr.Which of these two would have its solubility more affected by the addition of a strong acid? Would the solubility of that salt increase or decrease.

2. What is the molar solubility of silver chloride in 1.0 L of solution that contains 2.0 x 10-2 mol of HCl?

3 STEPS TO DETERMINING THE ION CONCENTRATIONAT EQUILIBRIUM

I. Calculate the [Ion]i that occurs after dilution but before the reaction starts.

II. Calculate the [Ion] when the maximum amount of solid is formed.

- we will determine the limiting reagent and assume all of that ion is used up to make the solid.- The [ ] of the other ion will be the stoichiometric equivalent.

III. Calculate the [Ion] at equilibrium*.

*Since we assume the reaction went to completion, yet by definition a slightly soluble can’t, we must account for some of the solid re-dissolving back into solution.

1. When 50.0 mL of 0.100 M AgNO3 and 30 mL of 0.060 M Na2CrO4 are mixed, a precipitate of silver chromate is formed. The solubility product is 1.9 x 10-12. Calculate the [Ag+] and [CrO4

2-] remaining in solution at equilibrium.

2. Suppose 300 mL of 8 x 10-6 M solution of KCl is added to 800 mL of 0.004 M solution of AgNO3. Calculate [Ag+] and [Cl-] remaining in solution at equilibrium.