chapter 17 sections 4-7 solubility equilibria © 2012 pearson education, inc

Chapter 17 sections 4-7

Solubility Equilibria

© 2012 Pearson Education, Inc.

AqueousEquilibria

Solubility Product ConstantsConsider the equilibrium that exists in a saturated

solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42(aq)


The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42]

where the equilibrium constant, Ksp, is called the solubility product constant.

AqueousEquilibria

Sample Exercise 17.9 Writing Solubility-Product (Ksp) Expressions

Practice ExerciseGive the solubility-product-constant expressions and Ksp values (from Appendix D) for (a) barium carbonate, (b) silver sulfate.

Write the expression for the solubility-product constant for CaF2, and look up the corresponding Ksp value in Appendix D.

AqueousEquilibria

Solubility Product Constants

• Ksp is not the same as solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).


AqueousEquilibria

A. AgI is the most soluble of the three.

B. AgBr is the most soluble of the three.

C. AgCl is the most soluble of the three.

D. All of the compounds are totally insoluble based on their small Ksp

values.

AqueousEquilibria

Solubility Product Constants• For solids dissolving to form aqueous solutions.

Bi2S3(s) 2Bi3+(aq) + 3S2(aq)

Ksp = solubility product constant

and

Ksp = [Bi3+]2[S2]3

•“molar Solubility” = s = concentration of Bi2S3 that dissolves, which equals ½ [Bi3+] and 1/3[S2].

• Note: Ksp is constant (at a given temperature)

• s is variable (especially with a common ion present)

AqueousEquilibria

Sample Exercise 17.10 Calculating Ksp from Solubility

Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 104 M. Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving Ag+ or CrO4

2– ions in the solution, calculate Ksp for this compound.

Practice ExerciseA saturated solution of Mg(OH)2 in contact with undissolved Mg(OH)2(s) is prepared at 25 C. The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH– ions, calculate Ksp for this compound.

AqueousEquilibria

Sample Exercise 17.11 Calculating Solubility from Ksp

The Ksp for CaF2 is 3.9 1011 at 25 C . Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.

Practice ExerciseThe Ksp for LaF3 is 2 1019. What is the solubility of LaF3 in water in moles per liter?

AqueousEquilibria

Factors Affecting Solubility

• The Common-Ion Effect– If one of the ions in a solution equilibrium

is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease:

BaSO4(s) Ba2+(aq) + SO42(aq)


AqueousEquilibria

The Common-Ion Effect

• Consider a solution of acetic acid:

• If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO(aq)


AqueousEquilibria


“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”


AqueousEquilibria


Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

Ka for HF is 6.8 104.

[H+] [F][HF]

Ka = = 6.8 104


AqueousEquilibria


Because HCl, a strong acid, is also present, the initial [H+] is not 0, but rather 0.10 M.

[HF], M [H+], M [F], M

Initially 0.20 0.10 0

Change

At equilibrium

HF(aq) H+(aq) + F(aq)


+x

x

+x-x

0.10 + x 0.100.20 x 0.20

AqueousEquilibria


= x

1.4 103 = x

(0.10) (x)(0.20)6.8 104 =

(0.20) (6.8 104)(0.10)


AqueousEquilibria


• Therefore, [F] = x = 1.4 103

[H3O+] = 0.10 + x = 0.10 + 1.4 103 = 0.10 M

• So,

pH = log (0.10)

pH = 1.00


AqueousEquilibria

Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility

Calculate the molar solubility of CaF2 at 25 C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF.

Practice ExerciseFor manganese(II) hydroxide,Mn(OH)2, Ksp = 1.6 1013. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH.

AqueousEquilibria

Factors Affecting Solubility• pH

– If a substance has a basic anion, it will be more soluble in an acidic solution.

– Substances with acidic cations are more soluble in basic solutions.


AqueousEquilibria

Sample Exercise 17.13 Predicting the Effect of Acid on SolubilityWhich of these substances are more soluble in acidic solution than in basic solution: (a) Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)?

Practice ExerciseWrite the net ionic equation for the reaction between an acid and (a) CuS, (b) Cu(N3)2.

AqueousEquilibria

Factors Affecting Solubility• Complex Ions

– Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.


AqueousEquilibria

Factors Affecting Solubility

• Complex Ions– The formation

of these complex ions increases the solubility of these salts.


AqueousEquilibria

Sample Exercise 17.14 Evaluating an Equilibrium Involving a Complex Ion

Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when NH3 is added.

Practice ExerciseCalculate [Cr3+] in equilibrium with Cr(OH)4

when 0.010 mol of Cr(NO3)3 is dissolved in 1 L of solution buffered at pH 10.0.

AqueousEquilibria

Factors Affecting Solubility• Amphoterism

– Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

– Examples of such cations are Al3+, Zn2+, and Sn2+.


AqueousEquilibria

A. Amphoteric substances are acids whereas amphiprotic substances are acids.

B. Amphoteric substances are bases whereas amphiprotic substances are acids.

C. Amphoteric substances can act both as an acid and a base whereas amphiprotic substances are acids.

D. Amphoteric substances can act both as an acid and a base whereas amphiprotic substances can accept or donate a proton.

AqueousEquilibria

Will a Precipitate Form?

• In a solution,– If Q = Ksp, the system is at equilibrium

and the solution is saturated.

– If Q < Ksp, more solid can dissolve until Q = Ksp.

– If Q > Ksp, the salt will precipitate until Q = Ksp.


AqueousEquilibria

Sample Exercise 17.15 Predicting Whether a Precipitate FormsDoes a precipitate form when 0.10 L of 8.0 103 M Pb(NO3)2 is added to 0.40 L of 5.0 103 M Na2SO4?

Practice ExerciseDoes a precipitate form when 0.050 L of 2.0 102 M NaF is mixed with 0.010 L of 1.0 102 M Ca(NO3)2?

AqueousEquilibria

Precipitation

•After precipitation occurs, the equilibrium concentration of the remaining solution can be determined.

•First calculate concentration if all solid precipitates.•Then determine how much solid dissolves.

AqueousEquilibria

Example

Consider a solution formed by mixing 100.0 mL of 0.0500 M Pb(NO3)2 and 200.0 mL of 0.100 M NaI.

The Ksp=1.4 x 10-8 for PbI2

AqueousEquilibria

Solution

Calculate the initial concentration:

MmL

mLmmolmLPb 2

02 1067.1

)0.300(

)/0500.0)(0.100(][

MmL

mLmmolmLI 2

0 1067.6)0.300(

)/100.0)(0.200(][

occurwillionprecipitatKQ

Q

MMIPbQ

sp

5

222200

2

1043.7

)1067.6)(1067.1(][][

AqueousEquilibria

Solution

Pb2+ I-

Preliminary 1.67 x 10-2 M 6.67 x 10-2 M

Reacted

Initial 0

- 1.67 x 10-2 M - 3.34 x 10-2 M

3.33 x 10-2 M

To calculate the concentration of ions in solution, first assume that the precipitation reaction goes to completion

AqueousEquilibria

Solution

Pb2+ I-

Initial 0 mmol 0.0333M

Change

Equilibrium

+x + 2x

+x 0.0333 + 2x

Then determine the ‘bounce back’ or the amount of precipitate that re-dissolves.

because the Ksp is small, x<< 0.0333

0.0333 M

5

8222

103.1

104.1)0333.0)((]][[

x

xIPbK sp

+ 1.3x10-5 M + 2(1.3 x 10-5 M)

1.3 x10-5 M + 0.0333 M

AqueousEquilibria

Example

•A solution is prepared by mixing 150.0 mL of 1.00 x 10-2 M Mg(NO3)2 and 250.0 mL of 1.00 x 10-1 M NaF. Calaculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9)

AqueousEquilibria

Solution

Calculate the initial concentration:

MmL

mLmmolmLMg 3

02 1075.3

)0.400(

)/0100.0)(0.150(][

MmL

mLmmolmLF 2

0 1025.6)0.400(

)/100.0)(0.250(][

occurwillionprecipitatKQ

Q

MMFMgQ

sp

5

223200

2

1046.1

)1025.6)(1075.3(][][

AqueousEquilibria

Solution

Mg2+ F-

Preliminary 3.75 x 10-3 M 6.25 x 10-2 M

Reacted

Initial 0

- 3.75 x 10-3 M - 7.50 x 10-3 M

5.50 x 10-2 M

Assume that the precipitation reaction goes to completion

AqueousEquilibria

Solution

Mg2+ F-

Initial 0 M 0.0550 M

Change

Equilibrium

+x + 2x

+x 0.0550 + 2x

because the Ksp is small, x<< 0.0550

0.0550 M

6

9222

101.2

104.6)0550.0.0)((]][[

x

xFMgK sp

+ 2.1x10-6M + 2(2.1 x 10-6 M)

2.1 x10-6 M + 0.0550 M

Then determine the ‘bounce back’ or the amount of precipitate that re-dissolves.

AqueousEquilibria

Selective Precipitation

• Used to separate mixtures of metals in solution.

• Use anion that forms precipitate with one metal, but not another

AqueousEquilibria

Example

•A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, which compound will precipitate first? Specify the concentration of I- needed to begin precipitation.

•PbI2 (Ksp = 1.4 x 10-8)

•CuI (Ksp = 5.3 x 10-12)

AqueousEquilibria

Solution

For PbI2:

Ksp= [Pb2+][I-]2 = 1.4 x 10-8

[Pb2+]0 = 2.0 x 10-3 M, therefore,

[Pb2+][I-]2 = (2.0 x 10-3 M)[I-]2= 1.4 x 10-8

[I-]= 2.6 x 10-3

AqueousEquilibria

Solution

For CuI:

Ksp= [Cu+][I-] = 3.5 x 10-12

[Cu+]0 = 1.0 x 10-4 M, therefore,

[Cu+][I-] = (1.0 x 10-4 M)[I-]= 3.5 x 10-12

[I-]= 5.3 x 10-8

AqueousEquilibria

Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation

A solution contains 1.0 102 M and 2.0 102 M Pb2+. When Cl is added, both AgCl (Ksp = 1.8 1010) and PbCl2 (Ksp = 1.7 105 M) can precipitate. What concentration of Cl is necessary to begin the precipitation of each salt? Which salt precipitates first?

Practice ExerciseA solution consists of 0.050 M Mg2+ and Cu2+. Which ion precipitates first as OH is added? What concentration of OH is necessary to begin the precipitation of each cation? [Ksp = 1.8 1011 for Mg(OH)2, and Ksp = 4.8 1020 for Cu(OH)2.]

AqueousEquilibria

A. ZnS and CuS both precipitate and separating the two ions is nolonger feasible using given procedures.

B. ZnS precipitates and separating the two ions is still feasible usinggiven procedures.

C. CuS precipitates and separating the two ions is still feasible usinggiven procedures.

D. No change in observations and ability to separate the two ions.

AqueousEquilibria

Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.


AqueousEquilibria

A. No, both would precipitate in step 1 and subsequently are not easily separated.

B. No, both would precipitate in step 2 and subsequently are not easily separated.

C. Yes, ZnS precipitates in step 1 and CuS in step 4.

D. Yes, CuS precipitates in step 2 and Zn2+ remains in solution.

AqueousEquilibria

A. The solution definitely contains either Pb2+ or Hg2

2+ cation.

B. The solution definitely contains the Ag+ cation.

C. The solution must contain one or more of the following cations: Cu2+, Bi3+, or Cd2+.

D. The solution must contain one or more of the following cations: Ag+, Pb2+ or Hg2

2+.

chapter 17 sections 4-7 solubility equilibria © 2012 pearson education, inc

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