philip dutton university of windsor, canada n9b 3p4 prentice-hall © 2002 general chemistry...
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Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 17: Additional Aspects of Acid-Base Equilibria
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 2 of 42
Contents
17-1 The Common-Ion Effect in Acid-Base Equilibria
17-2 Buffer Solutions
17-3 Acid-Base Indicators
17-4 Neutralization Reactions and Titration Curves
17-5 Solutions of Salts of Polyprotic Acids
17-6 Acid-Base Equilibrium Calculations: A Summary
Focus On Buffers in Blood
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 3 of 42
17-1 The Common-Ion Effect in Acid-Base Equilibria
• The Common-Ion Effect describes the effect on an equilibrium by a second substance that furnishes ions that can participate in that equilibrium.
• The added ions are said to be common to the equilibrium.
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 4 of 42
Solutions of Weak Acids and Strong Acids
• Consider a solution that contains both 0.100 M CH3CO2H and 0.100 M HCl.
CH3CO2H + H2O ↔ CH3CO2- + H3O+
HCl + H2O ↔ Cl- + H3O+
(0.100-x) M x M x M
0.100 M 0.100 M
[H3O+] = (0.100 + x) M essentially all due to HCl
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 5 of 42
Acetic Acid and Hydrochloric Acid
0.1 M HCl 0.1 M CH3CO2H 0.1 M HCl +0.1 M CH3CO2H
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 6 of 42
Example 17-1Demonstrating the Common-Ion Effect:
A Solution of a weak Acid and a Strong Acid.
(a) Determine [H3O+] and [CH3CO2-] in 0.100 M CH3CO2H.
(b) Then determine these same quantities in a solution that is 0.100 M in both CH3CO2H and HCl.
CH3CO2H + H2O → H3O+ + CH3CO2-
Recall Example 17-6 (p 680):
[H3O+] = [CH3CO2-] = 1.333·10-3 M
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 7 of 42
Example 17-1
CH3CO2H + H2O → H3O+ + CH3CO2-
Initial concs.
weak acid 0.100 M 0 M 0 M
strong acid 0 M 0.100 M 0 M
Changes -x M +x M +x M
Eqlbrm conc. (0.100 - x) M (0.100 + x) M x M
Assume x << 0.100 M, 0.100 – x ≈0.100 + x ≈ 0.100 M
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 8 of 42
Example 17-1
Eqlbrm conc. (0.100 - x) M (0.100 + x) M x M
Assume x << 0.100 M, 0.100 – x ≈0.100 + x ≈ 0.100 M
CH3CO2H + H2O → H3O+ + CH3CO2-
[H3O+] [CH3CO2-]
[CH3CO2H]Ka=
x · (0.100 + x)
(0.100 - x)=
x · (0.100)
(0.100)= = 1.8·10-5
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Example 17-1: Conclusion
General Chemistry: Chapter 18 Slide 9 of 42
In the presence of the strong acid, the concentration of the weak acid anion [A-] is very much less than it is when only the weak acid (HA) is present.
[CH3CO2-] = 1.8·10-5 M compared to 1.3·10-3 M.
In the presence of the strong acid, the concentration of the [OH-] ions originating from the water self ionization is much less than it is in the pure water.
Le Chatellier’s Principle
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 10 of 42
Suppression of Ionization of a Weak Acid
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 11 of 42
Suppression of Ionization of a Weak Base
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 12 of 42
Solutions of Weak Acids and Their Salts
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 13 of 42
Solutions of Weak Bases and Their Salts
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 14 of 42
17-2 Buffer Solutions
• Two component systems that change pH only slightly on addition of acid or base.– The two components must not neutralize each other but
must neutralize strong acids and bases.
• A weak acid and it’s conjugate base.• A weak base and it’s conjugate acid
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 15 of 42
Buffer Solutions
• Consider [CH3CO2H] = [CH3CO2-] in a solution.
[H3O+] [CH3CO2-]
[CH3CO2H]Ka= = 1.8·10-5
= 1.8·10-5 [CH3CO2H]
[CH3CO2-]
Ka[H3O+] =
pH = -log[H3O+] = -logKa = -log(1.8·10-5) = 4.74
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 16 of 42
How A Buffer Works
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 17 of 42
The Henderson-Hasselbalch Equation
• A variation of the ionization constant expression.• Consider a hypothetical weak acid, HA, and its
salt NaA:
HA + H2O ↔ A- + H3O+[H3O+] [A-]
[HA]Ka=
[H3O+] = Ka
[HA]
[A-]
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 18 of 42
Henderson-Hasselbalch Equation
pKa + log [HA]
pH =[A-]
pKa + log [acid]
pH =[conjugate base]
-log[H3O+] = -logKa - [HA]
[A-]log
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 19 of 42
Henderson-Hasselbalch Equation
• Only useful when you can use initial concentrations of acid and salt.– This limits the validity of the equation.
• Limits can be met by:
0.1 < [HA]
< 10[A-]
[A-] > 10·Ka and [HA] > 10·Ka
pKa + log [acid]
pH=[conjugate base]
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 20 of 42
Example 17-5Preparing a Buffer Solution of a Desired pH.
What mass of NaC2H3O2 must be dissolved in 0.300 L of 0.25 M HC2H3O2 to produce a solution with pH = 5.09? (Assume that the solution volume is constant at 0.300 L)
HC2H3O2 + H2O ↔ C2H3O2- + H3O+
Equilibrium expression:
[H3O+] [HC2H3O2]
Ka=[C2H3O2
-]= 1.8·10-5
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 21 of 42
Example 17-5
[H3O+] [HC2H3O2]
Ka=[C2H3O2
-]= 1.8·10-5
[H3O+] = 10-5.09 = 8.13·10-6 M
[HC2H3O2] = 0.25 M
Solve for [C2H3O2-]
[H3O+]
[HC2H3O2]= Ka
[C2H3O2-] = 0.5535 M
8.13·10-6
0.25= 1.80·10-5
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 22 of 42
Example 17-5
1 mol NaC2H3O2
82.0 g NaC2H3O2
mass C2H3O2- = 0.300 L
[C2H3O2-] = 0.5535 M
1 L
0.5535 mol
1 mol C2H3O2-
1 mol NaC2H3O2· ·
· = 13.62 g NaC2H3O2
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 23 of 42
Six Methods of Preparing Buffer Solutions
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 24 of 42
Calculating Changes in Buffer Solutions
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 25 of 42
Buffer Capacity and Range
• Buffer capacity is the amount of strong acid or base that a buffer can neutralize before its pH changes appreciably.– Maximum buffer capacity exists when [HA] and [A-]
are large and approximately equal to each other.
• Buffer range is the pH range over which a buffer effectively neutralizes added acids and bases.– Practically, range is 2 pH units around pKa
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 26 of 42
17-3 Acid-Base Indicators
• Color of some substances (e.g. weak acids) depends on the pH.
HIn + H2O ↔ In- + H3O+
>90% acid form the color appears to be the acid color
>90% base form the color appears to be the base color
Intermediate color is seen in between these two states.
Complete color change occurs over 2 pH units.
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 27 of 42
Indicator Colors and Ranges
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Several indicators in a large pH trange
Csonka Gábor Általános kémia -sav-bázis 2 Dia 28 /34
pH = 0
Benzil-orange
Bromcresol-purple
Thymolftaleine
Brom-phenol blue
Congo-redThymolblue
pH = 14
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Two indicators, pH = pK ± 1, in narrow range
Csonka Gábor Általános kémia -sav-bázis 2 Dia 29 /34
Thymol blue
Cresol red
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 30 of 42
17-4 Neutralization Reactions and Titration Curves
• Equivalence point:– The point in the reaction at which both acid and base have been
consumed.– Neither acid nor base is present in excess.
• End point:– The point at which the indicator changes color.
• Titrant:– The known solution added to the solution of unknown
concentration.
• Titration Curve:– The plot of pH vs. titrant volume.
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 31 of 42
The millimole
• Typically:– Volume of titrant added is less than 50 mL.– Concentration of titrant is less than 1 mol/L.– Titration uses less than 1/1000 mole of acid and base.
L/1000
mol/1000= M =
L
mol
mL
mmol=
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 32 of 42
Titration of a Strong Acid with a Strong Base
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 33 of 42
Titration of a Strong Acid with a Strong Base
• The pH has a low value at the beginning.• The pH changes slowly
– until just before the equivalence point.
• The pH rises sharply – perhaps 6 units per 0.1 mL addition of titrant.
• The pH rises slowly again.• Any Acid-Base Indicator will do.
– As long as color change occurs between pH 4 and 10.
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 34 of 42
Titration of a Strong Base with a Strong Acid
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 35 of 42
Titration of a Weak Acid with a Strong Base
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 36 of 42
Titration of a Weak Acid with a Strong Base
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 37 of 42
Titration of a Weak Polyprotic Acid
H3PO4 ↔ H2PO4- ↔ HPO4
2- ↔ PO43-
NaOHNaOH
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 38 of 42
17-5 Solutions of Salts of Polyprotic Acids
• The third equivalence point of phosphoric acid can only be reached in a strongly basic solution.
• The pH of this third equivalence point is not difficult to caluclate.– It corresponds to that of Na3PO4 (aq) and PO4
3- can ionize only as a base.
PO43- + H2O → OH- + HPO4
2-
Kb = Kw/Ka3 = 2.4·10-2
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 39 of 42
Example 17-9
Kb = 2.4·10-2 PO43- + H2O → OH- + HPO4
2-
Initial concs. 1.0 M 0 M 0 M
Changes -x M +x M +x M
Eqlbrm conc. (1.00 - x) M x M x M
Determining the pH of a Solution Containing the Anion (An-) of a Polyprotic Acid.
Sodium phosphate, Na3PO4, is an ingredient of some preparations used to clean painted walls before they are repainted. What is the pH of 1.0 M Na3PO4?
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 40 of 42
Example 17-9
x2 + 0.024x – 0.024 = 0 x = 0.1434 M
pOH = +0.85 pH = 13.15
[OH-] [HPO42-]
[PO43-]
Kb=x · x
(1.00 - x)= = 2.4·10-2
It is more difficult to calculate the pH values of NaH2PO4 and Na2HPO4 because two equilibria must be considered simultaneously.
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 41 of 42
Concentrated Solutions of Polyprotic Acids
• For solutions that are reasonably concentrated (> 0.1 M) the pH values prove to be independent of solution concentrations.
for H2PO4-
for HPO42-
pH = 0.5 (pKa1 + pKa2) = 0.5 (2.15 + 7.20) = 4.68
pH = 0.5 (pKa2 + pKa3) = 0.5 (7.20 + 12.38) = 9.79
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 42 of 42
17-6 Acid-Base Equilibrium Calculations:A Summary
• Determine which species are potentially present in solution, and how large their concentrations are likely to be.
• Identify possible reactions between components and determine their stoichiometry.
• Identify which equilibrium equations apply to the particular situation and which are most significant.
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 43 of 42
Focus On Buffers in Blood
CO2(g) + H2O ↔ H2CO3(aq)
H2CO3(aq) + H2O(l) ↔ HCO3-(aq)
Ka1 = 4.4·10-7 pKa1 = 6.4
pH = 7.4 = 6.4 +1.0
pH = pKa1 + log [H2CO3]
[HCO3-]
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 44 of 42
Buffers in Blood
• 10/1 buffer ratio is somewhat outside maximum buffer capacity range but…
• The need to neutralize excess acid (lactic) is generally greater than the need to neutralize excess base.
• If additional H2CO3 is needed CO2 from the lungs can be utilized.
• Other components of the blood (proteins and phosphates) contribute to maintaining blood pH.
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Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 45 of 42
Chapter 18 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.