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18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
Chapter 18Reaction Rates and Equilibrium
18.1 Rates of Reaction18.2 The Progress of Chemical Reactions18.3 Reversible Reactions and Equilibrium
18.4 Solubility Equilibrium
18.5 Free Energy and Entropy
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
How is it possible to ingest a poison without being harmed?
CHEMISTRY & YOUCHEMISTRY & YOU
Barium sulfate, a poison, can absorb X-rays, so tissues coated with the liquid will appear as light areas in the X-ray images.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Solubility Product ConstantThe Solubility Product Constant
The Solubility Product Constant
What is the relationship between the solubility product constant and the solubility of a compound?
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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• Most ionic compounds containing alkali metals are soluble in water. • For example, more than 35 g of sodium
chloride will dissolve in only 100 g of water.
The Solubility Product ConstantThe Solubility Product Constant
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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• By contrast, some ionic compounds are insoluble in water. • For example, compounds that contain
phosphate, sulfite, or carbonate ions tend not to dissolve in water.
The Solubility Product ConstantThe Solubility Product Constant
• Most ionic compounds containing alkali metals are soluble in water. • For example, more than 35 g of sodium
chloride will dissolve in only 100 g of water.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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This table provides some general rules for the solubility of ionic compounds in water.
Interpret DataInterpret Data
Solubility of Ionic Compounds in WaterCompounds Solubility Exceptions
Salts of Group 1A metals and ammonia
Soluble Some lithium compounds
Ethanoates, nitrates, chlorates, and perchlorates
Soluble Few exceptions
Sulfates Soluble Compounds of Pb, Ag, Hg, Ba, Sr, and Ca
Chlorides, bromides, and iodides
Soluble Compounds of Ag and some compounds of Hg and Pb
Sulfides and hydroxides Most are insoluble
Alkali metal sulfides and hydroxides are soluble. Compounds of Ba, Sr, and Ca are slightly soluble.
Carbonates, phosphates, and sulfites
Insoluble Compounds of the alkali metals and of ammonium ions
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Most insoluble ionic compounds will actually dissolve to some extent in water.
• These compounds are said to be slightly soluble in water.
The Solubility Product ConstantThe Solubility Product Constant
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Solubility Product ConstantThe Solubility Product Constant
When the “insoluble” compound silver chloride is mixed with water, a very small amount of silver chloride dissolves in the water.
An equilibrium is established between the solid and the dissolved ions in the saturated solution.
AgCl(s) Ag+(aq) + Cl–(aq)
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Solubility Product ConstantThe Solubility Product Constant
When the “insoluble” compound silver chloride is mixed with water, a very small amount of silver chloride dissolves in the water.
AgCl(s) Ag+(aq) + Cl–(aq)
You can write an equilibrium-constant expression for this process.
Keq = [Ag+] [Cl–]
[AgCl]
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Solubility Product ConstantThe Solubility Product Constant
To compare the solubility of salts, it is useful to have a constant that reflects only the concentrations of the dissolved ions.• This constant is called the solubility
product constant (Ksp).
• It is equal to the product of the concentrations of the ions each raised to a power equal to the coefficient of the ion in the dissociation equation.
Ksp = [A]a [B]b
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Solubility Product ConstantThe Solubility Product Constant
The smaller the numerical value of the solubility product constant, the lower the solubility of the compound.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Solubility Product ConstantThe Solubility Product Constant
The table below lists the Ksp values for some ionic compounds that are slightly soluble in water.
Solubility Product Constants (Ksp) at 25oC
Ionic compound Ksp Ionic compound Ksp Ionic compound Ksp
Halides
AgCl
AgBr
AgI
PbCl2PbBr2
PbI2
PbF2
CaF2
1.8 10–10
5.0 10–13
8.3 10–17
1.7 10–5
2.1 10–6
7.9 10–9
3.6 10–8
3.9 10–11
Sulfates
PbSO4
BaSO4
CaSO4
6.3 10–7
1.1 10–10
2.4 10–5
Hydroxides
Al(OH)3
Zn(OH)2
Ca(OH)2
Mg(OH)2
Fe(OH)2
3.0 10–34
3.0 10–16
6.5 10–6
7.1 10–12
7.9 10–16Sulfides
NiS
CuS
Ag2S
ZnS
FeS
CdS
PbS
4.0 10–20
8.0 10–37
8.0 10–51
3.0 10–23
8.0 10–19
1.0 10–27
3.0 10–28
Carbonates
CaCO3
SrCO3
ZnCO3
Ag2CO3
BaCO3
4.5 10–9
9.3 10–10
1.0 10–10
8.15 10–12
5.0 10–9
Chromates
PbCrO4
Ag2CrO4
1.8 10–14
1.2 10–12
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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What is the concentration of lead ions and chromate ions in a saturated solution of lead(II) chromate at 25oC?
(Ksp = 1.8 10–14)
Sample Problem 18.6Sample Problem 18.6
Finding the Ion Concentrations in a Saturated Solution
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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KNOWNS
UNKNOWN[Pb2+] = ? M
[CrO42–] = ? M
Analyze List the knowns and the unknowns.1
Sample Problem 18.6Sample Problem 18.6
Write the expression for Ksp. Then modify it so that there is a single unknown.
Ksp = 1.8 10–14
PbCrO4(s) Pb2+(aq) + CrO42–(aq)
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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• Start with the general expression for the solubility product constant.
Calculate Solve for the unknowns.2
Sample Problem 18.6Sample Problem 18.6
The exponent for each ion is 1.Ksp = [A]a [B]b
Ksp = [Pb2+] [CrO42–] = 1.8 10–14
• Use the chemical equation to write the correct expression for Ksp for the reaction.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Substitute [Pb2+] for [CrO42–]
in the expression to get an equation with one unknown.
Calculate Solve for the unknowns.2
Sample Problem 18.6Sample Problem 18.6
At equilibrium, [Pb2+] = [CrO4
2–].
Ksp = [Pb2+] [Pb2+] = [Pb2+]2 = 1.8 10–14
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Solve for [Pb2+].
Calculate Solve for the unknowns.2
Sample Problem 18.6Sample Problem 18.6
[Pb2+] = 1.8 10–14
[Pb2+] = [CrO42–] = 1.3 10–7 M
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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• Calculate [Pb2+] [CrO42–] to evaluate the
answers.
• The result is 1.7 10–14, which is close to the value of Ksp.
• The result varies slightly from the actual value because the answers were rounded to two significant figures.
Evaluate Does the result make sense?3
Sample Problem 18.6Sample Problem 18.6
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Which of the following compounds is least soluble at 25oC?
A. PbF2
B. ZnS
C. SrCO3
D. CaSO4
Ksp at 25oC
Ionic compound
Ksp
CaSO4 2.4 10–5
PbF2 3.6 10–8
SrCO3 9.3 10–10
ZnS 3.0 10–23
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Which of the following compounds is least soluble at 25oC?
A. PbF2
B. ZnS
C. SrCO3
D. CaSO4
Ksp at 25oC
Ionic compound
Ksp
CaSO4 2.4 10–5
PbF2 3.6 10–8
SrCO3 9.3 10–10
ZnS 3.0 10–23
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Common Ion EffectThe Common Ion Effect
The Common Ion Effect
How can you predict whether precipitation will occur when two solutions are mixed?
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Common Ion EffectThe Common Ion Effect
In a saturated solution of lead(II) chromate, an equilibrium is established between the solid lead(II) chromate and its ions in solution.
What would happen if you added some lead nitrate to this solution?
PbCrO4(s) Pb2+(aq) + CrO42–(aq)
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Common Ion EffectThe Common Ion Effect
PbCrO4(s) Pb2+(aq) + CrO42–(aq)
• Lead(II) nitrite is soluble in water, so adding Pb(NO3)2 causes the concentration of lead ion to increase.
• The addition of lead ions is a stress on the equilibrium.
• Applying Le Châtelier’s principle, the stress can be relieved if the reaction shifts to the left.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Common Ion EffectThe Common Ion Effect
The yellow solid in the test tube, which is PbCrO4, cannot dissolve because the solution is saturated with Pb2+ and CrO4
2– ions.
When some Pb(NO3)2 is added, the excess lead ions combine with the chromate ions in solution to form additional solid PbCrO4.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Common Ion EffectThe Common Ion Effect
• A common ion is an ion that is found in both ionic compounds in a solution.
• The lowering of the solubility of an ionic compound as a result of the addition of a common ion is called the common ion effect.
In this example, the lead ion is a common ion.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Ksp of BaSO4 is 1.1 10–10. Why can a patient ingest the toxic BaSO4 without being harmed?
CHEMISTRY & YOUCHEMISTRY & YOU
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Ksp of BaSO4 is 1.1 10–10. Why can a patient ingest the toxic BaSO4 without being harmed?
CHEMISTRY & YOUCHEMISTRY & YOU
Barium sulfate is not very soluble. Therefore, very little of it can dissolve in bodily fluids and be absorbed as a toxin.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Small amounts of silver bromide can be added to the lenses used for eyeglasses. The silver bromide causes the lenses to darken in the presence of large amounts of UV light. The Ksp
of silver bromide is 5.0 10–13. What is the concentration of bromide ion in a 1.00-L saturated solution of AgBr to which 0.020 mol of AgNO3 is added?
Sample Problem 18.7Sample Problem 18.7
Finding Equilibrium Ion Concentrations in the Presence of a Common Ion
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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KNOWNS UNKNOWN[Br–] = ? M
Analyze List the knowns and the unknown.1
Sample Problem 18.7Sample Problem 18.7
Use one unknown to express both [Ag+] and [Br–]. Let x be the equilibrium concentration of bromide ion and x + 0.020 be the equilibrium concentration of silver ion.
Ksp = 5.0 10–13
moles of AgNO3 added = 0.020 mol
AgBr(s) Ag+(aq) + Br–(aq)
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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• Write the expression for Ksp.
Calculate Solve for the unknown.2
Sample Problem 18.7Sample Problem 18.7
Ksp = [Ag+] [Br–]
Ksp = [Ag+] x
• Substitute x for [Br–] in the solubility product expression.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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• Rearrange the equation to solve for x.
Sample Problem 18.7Sample Problem 18.7
Based on the small value of Ksp, you can assume that x will be very small compared to 0.020. Thus, [Ag+] ≈ 0.020 M.
• Substitute the values for Ksp and [Ag+] in the expression and solve.
Ksp x = [Ag+]
(5.0 10–13)x = 0.020
[Br–] = 2.5 10–11 M
Calculate Solve for the unknown.2
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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• The concentration of Br– in a saturated solution of AgBr is 7.0 10–7 M (the square root of the Ksp).
• It makes sense that the addition of AgNO3
would lower the concentration of Br– because the presence of a common ion, Ag+, causes AgBr to precipitate from solution.
Evaluate Does the result make sense?3
Sample Problem 18.7Sample Problem 18.7
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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The Common Ion EffectThe Common Ion Effect
A precipitate will form if the product of the concentrations of two ions in the mixture is greater than the Ksp value for the compound formed from the ions.
As solutions of barium nitrate and sodium sulfate are mixed, a precipitate of BaSO4 is formed.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Does a precipitate form when 1 L of 0.034 M Na2SO4 is mixed with 1 L of 0.0012 M Ba(NO3)2? The Ksp for BaSO4 is 1.1 10–10.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Does a precipitate form when 1 L of 0.034 M Na2SO4 is mixed with 1 L of 0.0012 M Ba(NO3)2? The Ksp for BaSO4 is 1.1 10–10.
[Ba2+] [SO42–] = (0.0006 M) (0.014 M) = 8.4 10–6
The result is larger than the Ksp value for BaSO4, so a
precipitate will form.
18.4 Solubility Equilibrium >18.4 Solubility Equilibrium >
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Key Concepts and Key Concepts and Key EquationKey Equation
The smaller the value of the solubility product constant, the lower the solubility of the compound.
A precipitate will form if the product of the concentrations of two ions in the mixture is greater than the Ksp value of the compound formed from the ions.
Ksp = [A]a × [B]b
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Glossary TermsGlossary Terms
• solubility product constant: an equilibrium constant applied to the solubility of electrolytes; it is equal to the product of the concentrations of the ions each raised to a power equal to the coefficient of the ion in the dissociation equation
• common ion: an ion that is common to both salts in a solution; in a solution of silver nitrate and silver chloride, Ag+ would be a common ion
• common ion effect: a decrease in the solubility of an ionic compound caused by the addition of a common ion