Chapter 16 Solubility and Complex Ion Equilibria AP*
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Transcript of Chapter 16 Solubility and Complex Ion Equilibria AP*
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Chapter 16
Solubility and Complex Ion Equilibria
AP*
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AP Learning Objectives
LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec 16.1-16.3)
LO 6.21 The student can predict the solubility of a salt, or rank the solubility of salts, given the relevant Ksp values. (Sec 16.1)
LO 6.22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values. (Sec 16.1)
LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, pH) that influence the solubility. (Sec 16.1-16.2)
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Section 16.1Solubility Equilibria and the Solubility Product
AP Learning Objectives, Margin Notes and References
Learning Objectives LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or
environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.
LO 6.21 The student can predict the solubility of a salt, or rank the solubility of salts, given the relevant Ksp values.
LO 6.22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values. LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions,
pH) that influence the solubility.
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Section 16.1Solubility Equilibria and the Solubility Product
Solubility Equilibria Because ionic compounds are strong
electrolytes, they dissociate completely to the extent that they dissolve.
When an equilibrium equation is written, the solid is the reactant and the ions in solution are the products.
The equilibrium constant expression is called the solubility-product constant. It is represented as Ksp.
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Section 16.1Solubility Equilibria and the Solubility Product
Solubility ProductFor example:
BaSO4(s) Ba⇌ 2+(aq) + SO42–(aq)
The equilibrium constant expression is Ksp = [Ba2+][SO4
2]Another example:Ba3(PO4)2(s) 3 Ba⇌ 2+(aq) + 2 PO4
3–(aq)The equilibrium constant expression is
Ksp = [Ba2+]3[PO43]2
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 6
Solubility Equilibria
Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature. (not affected by equilibrium position)
Solubility – an equilibrium position.
Bi2S3(s) 2Bi3+(aq) + 3S2–(aq)2 33+ 2
sp = Bi S K
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Section 16.1Solubility Equilibria and the Solubility Product
Solubility vs. Solubility Product Ksp is not the same as solubility. Solubility is the quantity of a substance that dissolves
to form a saturated solution (reach equilibrium) Common units for solubility: Grams per liter (g/L) Moles per liter (mol/L)
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Section 16.1Solubility Equilibria and the Solubility Product
Calculating Solubility from Ksp The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. What
is its molar solubility? Follow the same format as before:1)CaF2(s) Ca⇌ 2+(aq) + 2 F–(aq)
2) Ksp = [Ca2+][F–]2 = 3.9 × 10–11
3) CaF2(s) [Ca2+](M) [F–](M)
Initial concentration
(M)
---
0
0
Change in concentration
(M)
---
+s
+2s
Equilibrium concentration
(M)
---
s
2s
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Section 16.1Solubility Equilibria and the Solubility Product
Example (completed)4) Solve: Substitute the equilibrium
concentration values from the table into the solubility-product equation:3.9 × 10–11 = (s) (2s)2 = 4s3
s = 2.1 × 10–4 M(If you want the answer in g/L, multiply by molar mass; this would give 0.016 g/L.)
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 10
In comparing several different salts at a given temperature, does a higher Ksp value always mean a higher solubility? Explain. If yes, explain and verify. If no, provide a counter-example.
CONCEPT CHECK! Relative SolubilitiesCONCEPT CHECK! Relative Solubilities
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 11
EXERCISE!EXERCISE!
Which of these expressions correctly expresses the solubility-product constant for Ag3PO4 in water? (a) [Ag][PO4], (b) [Ag+][PO4
3–], (c) [Ag+]3[PO43–], (d)
[Ag+][PO43–]3, (e) [Ag+]3[PO4
3–]3.
Writing Solubility-Product (Ksp) Expressions
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 12
EXERCISE!EXERCISE!
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 × 10–4 M. Assuming that the Ag2CrO4 solution is saturated and that there are no other important equilibria involving Ag+ or CrO4
2– ions in the solution, calculate Ksp for this compound.
Calculating Ksp from Solubility
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 13
Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10
1.3×10-5 M
Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18
1.6×10-5 M
EXERCISE!EXERCISE!
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Section 16.1Solubility Equilibria and the Solubility Product
Factors Affecting Solubility The Common-Ion Effect
If one of the ions in a solution equilibrium is already dissolved in the solution, the solubility of the salt will decrease.
If either calcium ions or fluoride ions are present, then calcium fluoride will be less soluble.
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Section 16.1Solubility Equilibria and the Solubility Product
Calculating Solubility with a Common Ion
What is the molar solubility of CaF2 in 0.010 M Ca(NO3)2?
Follow the same format as before:1) CaF2(s) Ca⇌ 2+(aq) + 2 F–(aq)
2) Ksp = [Ca2+][F–]2 = 3.9 × 10–11
3) CaF2(s) [Ca2+](M) [F–](M)
Initial concentration
(M)
---
0.010
0
Change in concentration
(M)
---
+s
+2s
Equilibrium concentration
(M)
---
0.010 + s
2s
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Section 16.1Solubility Equilibria and the Solubility Product
Example (completed)
4) Solve: Substitute the equilibrium concentration values from the table into the solubility-product equation:3.9 × 10–11 = (0.010 + s)(2s)2 (We assume that s << 0.010, so that 0.010 + s = 0.010!)3.9 × 10–11 = (0.010)(2s)2 s = 3.1 × 10–5 M
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 17
Calculate the solubility of AgCl in: Ksp = 1.6 × 10–10
a) 100.0 mL of 4.00 x 10-3 M calcium chloride. 2.0×10-8 M
b) 100.0 mL of 4.00 x 10-3 M calcium nitrate. 1.3×10-5 M
EXERCISE!EXERCISE!
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Section 16.1Solubility Equilibria and the Solubility Product
© 2015 Pearson Education, Inc.
Factors Affecting Solubility pH: If a substance has a basic anion, it will be
more soluble in an acidic solution.
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 19
How does the solubility of silver chloride in water compare to that of silver chloride in an acidic solution (made by adding nitric acid to the solution)?
Explain.
The solubilities are the same.
CONCEPT CHECK!CONCEPT CHECK!
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 20
How does the solubility of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)?
Explain.
The silver phosphate is more soluble in an acidic solution.
CONCEPT CHECK!CONCEPT CHECK!
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Section 16.1Solubility Equilibria and the Solubility Product
Copyright © Cengage Learning. All rights reserved 21
How does the Ksp of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)?
Explain.
The Ksp values are the same.
CONCEPT CHECK!CONCEPT CHECK!
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Section 16.2Precipitation and Qualitative Analysis
AP Learning Objectives, Margin Notes and References
Learning Objectives LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or
environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.
LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, pH) that influence the solubility.
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Section 16.2Precipitation and Qualitative Analysis
Precipitation (Mixing Two Solutions of Ions)
Copyright © Cengage Learning. All rights reserved 23
Will a Precipitate Form? To decide, we calculate the reaction quotient, Q,
and compare it to the solubility product constant, Ksp.
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid can dissolve, so no precipitate forms.
If Q > Ksp, a precipitate will form.
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Section 16.2Precipitation and Qualitative Analysis
Does a precipitate form when 0.10 L of 8.0 × 10–3 M Pb(NO3)2 is added to 0.40 L of 5.0 × 10–3 M Na2SO4? (Ksp of 6.3 × 10–7 )
Predicting Whether a Precipitate Forms
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Section 16.2Precipitation and Qualitative Analysis
Copyright © Cengage Learning. All rights reserved 25
What are the equilibrium concentrations of the remaining ions once precipitation has occurred?
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Section 16.2Precipitation and Qualitative Analysis
Selective Precipitation (Mixtures of Metal Ions)
Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture.
Example: Solution contains Ba2+ and Ag+ ions. Adding NaCl will form a precipitate with Ag+ (AgCl),
while still leaving Ba2+ in solution.
Copyright © Cengage Learning. All rights reserved 26
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Section 16.2Precipitation and Qualitative Analysis
A solution contains 1.0 × 10–2 M Ag+ and 2.0 × 10–2 M Pb2+. When Cl– is added, both AgCl(Ksp = 1.8 × 10–10) and PbCl2(Ksp = 1.7 × 10–5) can precipitate. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?
Selective Precipitation
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Section 16.2Precipitation and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
At a low pH, [S2–] is relatively low and only the very insoluble HgS and CuS precipitate.
When OH– is added to lower [H+], the value of [S2–] increases, and MnS and NiS precipitate.
Copyright © Cengage Learning. All rights reserved 28
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Section 16.2Precipitation and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
Copyright © Cengage Learning. All rights reserved 29
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Section 16.2Precipitation and Qualitative Analysis
Separating the Common Cations by Selective Precipitation
Copyright © Cengage Learning. All rights reserved 30
Ksp values can be compare directly for ions that produce the same number of ions in solution.
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Section 16.2Precipitation and Qualitative Analysis
31
potassium. sodium lithium
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Section 16.3Equilibria Involving Complex Ions
AP Learning Objectives, Margin Notes and References
Learning Objectives LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or
environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.
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Section 16.3Equilibria Involving Complex Ions
Complex Ion Equilibria
Charged species consisting of a metal ion surrounded by ligands. Ligand: Lewis base (a molecule with a lone pair that
can be donated to an empty orbital on the metal form a covalent bond)
The number of ligands attached to the metal is called the coordination number. (commonly 6,4,or 2)
Copyright © Cengage Learning. All rights reserved 33
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Section 16.3Equilibria Involving Complex Ions
Complex Ion Equilibria
Formation (stability) constant. Equilibrium constant for each step of the formation of
a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution.
Copyright © Cengage Learning. All rights reserved 34
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Section 16.3Equilibria Involving Complex Ions
Complex Ion Equilibria
Be2+(aq) + F–(aq) BeF+(aq) K1 = 7.9 × 104
BeF+(aq) + F–(aq) BeF2(aq) K2 = 5.8 × 103
BeF2(aq) + F–(aq) BeF3–
(aq) K3 = 6.1 × 102
BeF3–
(aq) + F–(aq) BeF42–
(aq) K4 = 2.7 × 101
Copyright © Cengage Learning. All rights reserved 35
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Section 16.3Equilibria Involving Complex Ions
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Section 16.3Equilibria Involving Complex Ions
© 2015 Pearson Education, Inc.
How Complex Ion Formation Affects Solubility
Silver chloride is insoluble. It has a Ksp of 1.6 × 10–10.
In the presence of NH3, the solubility greatly increases because Ag+ will form complex ions with NH3.
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Section 16.3Equilibria Involving Complex Ions
Complex Ions and Solubility
Two strategies for dissolving a water–insoluble ionic solid. If the anion of the solid is a good base, the solubility
is greatly increased by acidifying the solution. In cases where the anion is not sufficiently basic, the
ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation.
Copyright © Cengage Learning. All rights reserved 38
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Section 16.3Equilibria Involving Complex Ions
© 2015 Pearson Education, Inc.
Amphoterism and Solubility• Amphoteric oxides and hydroxides are soluble in strong
acids or base, because they can act either as acids or bases.
• Examples are oxides and hydroxides of Al3+, Cr3+, Zn2+, and Sn2+.
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Section 16.3Equilibria Involving Complex Ions
Calculate the solubility of silver chloride in 10.0 M ammonia given the following information:
Ksp (AgCl) = 1.6 × 10–10
Ag+ + NH3 AgNH3+ K = 2.1 × 103
AgNH3+ + NH3 Ag(NH3)2
+ K = 8.2 × 103
0.48 MCalculate the concentration of NH3 in the final equilibrium mixture.
9.0 MCopyright © Cengage Learning. All rights reserved 40
CONCEPT CHECK!CONCEPT CHECK!