E. John McMurry Robert C. Fay
General Chemistry: Atoms First Chapter 15 Applications of Aqueous Equilibria
Lecture Notes
Alan D. Earhart Southeast Community College Lincoln, NE
Copyright 2010 Pearson Prentice Hall, Inc.
Neutralization Reactions
Strong Acid-Strong Base HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)
Assuming complete dissociation: H1+(aq) + OH1-(aq) or H3O1+(aq) + OH1-(aq) (net ionic equation) After neutralization: pH = 7
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H2O(l) 2H2O(l)
Neutralization Reactions
Weak Acid-Strong Base CH3CO2H(aq) + NaOH(aq) H2O(l) + NaCH3CO2(aq)
Assuming complete dissociation: CH3CO2H(aq) + OH1-(aq) H2O(l) + CH3CO21-(aq)
(net ionic equation)
After neutralization: pH > 7
Copyright 2010 Pearson Prentice Hall, Inc. Chapter 10/3
Neutralization Reactions
Strong Acid-Weak Base HCl(aq) + NH3(aq) NH4Cl(aq)
Assuming complete dissociation: H1+(aq) + NH3(aq) or H3O1+(aq) + OH1-(aq) NH41+(aq) H2O(l) + NH41+(aq)
(net ionic equation) After neutralization: pH < 7
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Neutralization Reactions
Weak Acid-Weak Base CH3CO2H(aq) + NH3(aq) NH4CH3CO2(aq)
CH3CO2H(aq) + NH3(aq)
NH41+(aq) + CH3CO21-(aq)
(net ionic equation)
After neutralization: pH = ?
Copyright 2010 Pearson Prentice Hall, Inc. Chapter 10/5
The Common-Ion Effect
Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium. CH3CO2H(aq) + H2O(l) H3O1+(aq) + CH3CO21-(aq)
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Chapter 10/6
The Common-Ion Effect
The pH of 0.10 M acetic acid is 2.89. Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. CH3CO2H(aq) + H2O(l) I C E 0.10 -x 0.10 - x Ka = [H3O1+][CH3CO21-] [CH3CO2H]
Copyright 2010 Pearson Prentice Hall, Inc. Chapter 10/7
H3O1+(aq) + CH3CO21-(aq) 0 +x x 0.10 +x 0.10 + x
The Common-Ion Effect
1.8 x 10-5 = (x)(0.10 + x) (0.10 - x) x(0.10) 0.10
x = [H3O1+] = 1.8 x 10-5 M pH =-log([H3O1+]) = -log(1.8 x 10-5) = 4.74 Why is there a difference in pH?
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Chapter 10/8
The Common-Ion Effect
Le Chtelier's Principle CH3CO2H(aq) + H2O(l) H3O1+(aq) + CH3CO21-(aq)
The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left.
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Chapter 10/9
Chapter 10/10
The Common-Ion Effect
Chapter 10/11
Buffer Solutions
Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH. Weak acid + Conjugate base For Example: CH3CO2H + CH3CO21HF + F1NH41+ + NH3 H2PO41- + HPO42-
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Chapter 10/12
Buffer Solutions
CH3CO2H(aq) + H2O(l) Weak acid H3O1+(aq) + CH3CO21-(aq) Conjugate base (NaCH3CO2)
Addition of OH1- to a buffer: CH3CO2H(aq) + OH1-(aq) Addition of H3O1+ to a buffer: CH3CO21-(aq) + H3O1+(aq)
100% 100%
H2O(l) + CH3CO21-(aq)
H2O(l) + CH3CO2H(aq)
Chapter 10/13
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Buffer Solutions
Buffer Solutions
Chapter 10/15
The Henderson-Hasselbalch Equation
CH3CO2H(aq) + H2O(l) Weak acid Acid(aq) + H2O(l) H3O1+(aq) + CH3CO21-(aq) Conjugate base H3O1+(aq) + Base(aq)
Ka =
[H3O1+][Base] [Acid]
[H3O1+] = Ka
[Acid] [Base]
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Chapter 10/16
The Henderson-Hasselbalch Equation
[H3O1+] = Ka [Acid] [Base] [Acid] [Base]
-log([H3O1+]) = -log(Ka) - log
pH = pKa + log [Acid]
[Base]
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Chapter 10/17
pH Titration Curves
Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a standard solution of another substance, whose concentration is known. buret standard solution (known concentration)
Erlenmeyer flask
unknown concentration solution An indicator is added which changes color once the reaction is complete
Chapter 10/18
Strong Acid-Strong Base Titrations
Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together.
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Chapter 10/19
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Chapter 10/20
Strong Acid-Strong Base Titrations
1. Before the Addition of Any NaOH
0.100 M HCl
Strong Acid-Strong Base Titrations
2. Before the Equivalence Point H3O1+(aq) + OH1-(aq) Excess H3O1+
100%
2H2O(l)
Strong Acid-Strong Base Titrations
3. At the Equivalence Point
The quantity of H3O1+ is equal to the quantity of added OH1-. pH = 7
Strong Acid-Strong Base Titrations
4. Beyond the Equivalence Point
Excess OH1-
Strong Acid-Strong Base Titrations
Chapter 10/25
Weak Acid-Strong Base Titrations
Weak Acid-Strong Base Titrations
1. Before the Addition of Any NaOH CH3CO2H(aq) + H2O(l) 0.100 M CH3CO2H H3O1+(aq) + CH3CO21-(aq)
Weak Acid-Strong Base Titrations
2. Before the Equivalence Point CH3CO2H(aq) + OH1-(aq) Excess CH3CO2H Buffer region
100%
H2O(l) + CH3CO21-(aq)
Weak Base Acid-Strong Titrations
3. At the Equivalence Point CH3CO21-(aq) + H2O(l) pH > 7 Notice how the pH at the equivalence point is shifted up versus the strong acid titration. OH1-(aq) + CH3CO2H(aq)
Weak Acid-Strong Base Titrations
4. Beyond the Equivalence Point
Excess OH1Notice that the strong and weak acid titration curves correspond after the equivalence point.
Weak Acid-Strong Base Titrations
Weak Base-Strong Acid Titrations
1. Before the Addition of Any HCl
0.100 M NH3
Weak Base-Strong Acid Titrations
2. Before the Equivalence Point NH3(aq) + H3O1+(aq)
100%
NH41+(aq) + H2O(l)
Excess NH3 Buffer region
Weak Base-Strong Acid Titrations
3. At the Equivalence Point NH41+(aq) + H2O(l) H3O1+(aq) + NH3(aq)
pH < 7
Weak Base-Strong Acid Titrations
4. Beyond the Equivalence Point
Excess H3O1+
Solubility Equilibria
A saturated solution of calcium fluoride in contact with solid CaF2 contains constant equilibrium concentrations of Ca2+(aq) and F1-(aq) because at equilibrium the ions crystallize at the same rate as the solid dissolves. CaF2(s) Ca2+(aq) + 2F1-(aq)
Ksp = [Ca2+][F1-]2
Chapter 10/37
Solubility Equilibria
MmXx(s) mMn+(aq) + xXy-(aq)
Ksp = [Mn+]m [Xy-]x Ksp is the solubility product constant.
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Chapter 10/38
Measuring Ksp and Calculating Solubility from Ksp
If the concentrations of Ca2+(aq) and F1-(aq) in a saturated solution of calcium fluoride are known, Ksp may be calculated. CaF2(s) Ca2+(aq) + 2F1-(aq) [F1-] = 4.1 x 10-4 M
[Ca2+] = 2.0 x 10-4 M
Ksp = [Ca2+][F1-]2 = (2.0 x 10-4)(4.1 x 10-4)2 = 3.4 x 10-11 (at 25 C)
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Chapter 10/39
Measuring Ksp and Calculating Solubility from Ksp
Chapter 10/40
Measuring Ksp and Calculating Solubility from Ksp
Solubility and the Common-Ion Effect Calculate the molar solubility of MgF2 in water at 25 C. MgF2(s) E Mg2+(aq) + 2F1-(aq) x 2x
Ksp = 7.4 x 10-11 = [Mg2+][F1-]2 = (x)(2x)2 4x3 = 7.4 x 10-11 x = [Mg2+] = Molar solubility = 2.6 x 10-4 M
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Factors That Affect Solubility
Solubility and the Common-Ion Effect Calculate the molar solubility of MgF2 in 0.10 M NaF at 25 C. MgF2(s) E Mg2+(aq) + 2F1-(aq) x 0.10 + 2x
Ksp = 7.4 x 10-11 = [Mg2+][F1-]2 = (x)(0.10 + 2x)2 7.4 x 10-11 = (x)(0.10 + 2x)2 (x)(0.10)2 x = [Mg2+] = Molar solubility = 7.4 x 10-9 = 7.4 x 10-9 M (0.10)2
Chapter 10/42
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Factors That Affect Solubility
Solubility and the Common-Ion Effect MgF2(s) Mg2+(aq) + 2F1-(aq)
Molar solubility: 2.6 x 10-4 M Molar solubility in 0.10 M NaF: 7.4 x 10-9 M Why does the solubility decrease in the presence of a common ion? Le Chtelier's Principle
Copyright 2010 Pearson Prentice Hall, Inc. Chapter 10/43
Factors That Affect Solubility
Solubility and the Common-Ion Effect MgF2(s) Mg2+(aq) + 2F1-(aq)
Factors That Affect Solubility
Solubility and the pH of the Solution CaCO3(s) + H3O1+(aq) Ca2+(aq) + HCO31-(aq) + H2O(l)
Factors That Affect Solubility
Solubility and the Formation of Complex Ions Ag1+(aq) + NH3(aq) Ag(NH3)1+(aq) + NH3(aq) Ag1+(aq) + 2NH3(aq) Ag(NH3)1+(aq) Ag(NH3)21+(aq) Ag(NH3)21+(aq) K1 = 2.1 x 103 K2 = 8.1 x 103 Kf = 1.7 x 107 (at 25 C) Kf is the formation constant. How is it calculated?
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Factors That Affect Solubility
Solubility and the Formation of Complex Ions K1 = [Ag(NH3)1+] [Ag ][NH3]
1+
K2 =
[Ag(NH3)21+] [Ag(NH3)1+][NH3]
Kf = K1K2 =
[Ag(NH3)21+] [Ag1+][NH3]2
= (2.1 x 103)(8.1 x 103) = 1.7 x 107
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Chapter 10/47
Factors That Affect Solubility
Solubility and the Formation of Complex Ions Ag1+(aq) + 2NH3(aq) Ag(NH3)21+(aq)
Factors That Affect Solubility
Solubility and Amphoterism Aluminum hydroxide is soluble both in strongly acidic and in strongly basic solutions. In acid: In base: Al(OH)3(s) + 3H3O1+(aq) Al(OH)3(s) + OH1-(aq) Al3+(aq) + 6H2O(l) Al(OH)41-(aq)
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Chapter 10/50
Factors That Affect Solubility
Solubility and Amphoterism
Chapter 10/51
Factors That Affect Solubility
Solubility and Amphoterism
Chapter 10/52
Precipitation of Ionic Compounds
CaF2(s) Ca2+(aq) + 2F1-(aq) Ksp = [Ca2+][F1-]2 IP = Qc = [Ca2+]t [F1-]t2 IP is called the ion-product.
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Chapter 10/53
Precipitation of Ionic Compounds
Compare the IP to Ksp If IP > Ksp: The solution is supersaturated and precipitation will occur. If IP = Ksp: The solution is saturated and equilibrium exists already. If IP < Ksp: The solution is unsaturated and precipitation will not occur.
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Chapter 10/54
Separation of Ions by Selective Precipitation
MS(s) + 2H3O1+(aq) M2+(aq) + H2S(aq) + 2H2O(l) Kspa = [M2+][H2S] [H3O1+]2
Chapter 10/55
Separation of Ions by Selective Precipitation
MS(s) + 2H3O1+(aq) M2+(aq) + H2S(aq) + 2H2O(l) Kspa = [M2+][H2S] [H3O1+]2 [M2+]t [H2S]t [H3O1+]t2
Qc =
Adjusting the pH changes Qc (ion-product).
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Chapter 10/56
Qualitative Analysis