chapter 17 sections 4-7 solubility equilibria © 2012 pearson education, inc
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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)
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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).
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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)
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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)
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AqueousEquilibria
The Common-Ion Effect
“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.”
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AqueousEquilibria
The Common-Ion Effect
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
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AqueousEquilibria
The Common-Ion Effect
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)
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+x
x
+x-x
0.10 + x 0.100.20 x 0.20
AqueousEquilibria
The Common-Ion Effect
= x
1.4 103 = x
(0.10) (x)(0.20)6.8 104 =
(0.20) (6.8 104)(0.10)
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AqueousEquilibria
The Common-Ion Effect
• 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
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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.
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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.
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AqueousEquilibria
Factors Affecting Solubility
• Complex Ions– The formation
of these complex ions increases the solubility of these salts.
© 2012 Pearson Education, Inc.
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+.
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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.
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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.
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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+.
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