Acid-Base Reactions
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2
Stomach Acidity &
Acid-Base Reactions
Copyright (c) 1999 by Harcourt Brace & Company
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Acid-Base Reactions
Section 18.4
QUESTION: You titrate 100. mL of a 0.025 M solution
of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
HBz + NaOH ---> Na+ + Bz- + H2O
C6H5CO2H = HBz
Benzoate ion = BzCopyright (c) 1999 by Harcourt Brace & Company
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3
Acid-Base Reactions
Section 18.4
The product of the titration of benzoic acid,
the benzoate ion, Bz-, is the conjugate base
of a weak acid. The final solution is basic.
Bz- + H2O
HBz + OH-
+
+
Kb = 1.6 x 10-10
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4
5
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Strategy — find the conc. of the
conjugate base Bz- in the solution AFTER
the titration, then calculate pH.
This is a two-step problem
1.
stoichiometry of acid-base reaction
2.
equilibrium calculation
Copyright (c) 1999 by Harcourt Brace & Company
All rights reserved
6
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
STOICHIOMETRY PORTION
1. Calc. moles of NaOH req’d
(0.100 L HBz)(0.025 M) = 0.0025 mol HBz
This requires 0.0025 mol NaOH
2.
Calc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.025 L
25 mL of NaOH req’d
Copyright (c) 1999 by Harcourt Brace & Company
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7
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
STOICHIOMETRY PORTION
25 mL of NaOH req’d
3. Moles of Bz- produced = moles HBz =
0.0025 mol
4. Calc. conc. of BzThere are 0.0025 mol of Bz- in a TOTAL
SOLUTION VOLUME of 125 mL
[Bz-] = 0.0025 mol / 0.125 L = 0.020 M
Copyright (c) 1999 by Harcourt Brace & Company
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8
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
EQUILIBRIUM PORTION
Bz- + H2O
HBz + OH[Bz-]
[HBz]
initial
change
equilib
Copyright (c) 1999 by Harcourt Brace & Company
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Kb = 1.6 x 10-10
[OH-]
9
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
EQUILIBRIUM PORTION
Bz- + H2O
HBz + OH[Bz-]
[HBz]
initial
0.020
0
change -x
+x
equilib 0.020 - x x
Copyright (c) 1999 by Harcourt Brace & Company
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Kb = 1.6 x 10-10
[OH-]
0
+x
x
10
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
EQUILIBRIUM PORTION
Bz- + H2O
equilib
HBz + OH[Bz-]
[HBz]
0.020 - x x
Kb = 1.6 x 10-10
[OH-]
x
2
x
Kb  1.6 x 10-10 =
0.020 - x
Solving in the usual way, we find
x = [OH-] = 1.8 x 10-6, pOH = 5.75, and pH = 8.25
Copyright (c) 1999 by Harcourt Brace & Company
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11
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
What is the pH at the half-way point?
HBz + H2O
H3O+ + Bz-
Ka = 6.3 x 10-
5
[H3O+] = { [HBz] / [Bz-] } Ka
At the half-way point, [HBz] = [Bz-], so
[H3O+] = Ka = 6.3 x 10-5
pH = 4.20
Copyright (c) 1999 by Harcourt Brace & Company
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Titrations
pH
Copyright © 1999 by Harcourt Brace & Company
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work should be mailed to: Permissions Department,
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Titrant volume, mL
13
Acid-Base Titrations
Adding NaOH from the buret to acetic acid in the flask, a weak
acid. In the beginning the pH increases very slowly.
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14
Acid-Base Titrations
Additional NaOH is added. pH rises as equivalence point is
approached.
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15
Acid-Base Titrations
Additional NaOH is added. pH increases and then levels off
as NaOH is added beyond the equivalence point.
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16
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Equivalence
point
pH of solution of
benzoic acid, a
weak acid
Copyright (c) 1999 by Harcourt Brace & Company
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17
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
EQUILIBRIUM PORTION
Bz- + H2O
HBz + OH[Bz-]
equilib
0.020 - x
Kb = 1.6 x 10-10
[HBz] [OH-]
x
x
2
Solving in the usual way, we find
x
-10 =
K

1.6
x
10
-6
x = [OH ] = 1.8 x b10 , pOH = 5.75, and pH = 8.25
0.020 - x
Copyright (c) 1999 by Harcourt Brace & Company
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18
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Half-way
point
Copyright (c) 1999 by Harcourt Brace & Company
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19
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH
What is the pH at the half-way point?
HBz + H2O
H3O+ + Bz-
Ka = 6.3 x 10-5
[H3O+] = { [HBz] / [Bz-] } Ka
At the half-way point, [HBz] = [Bz-], so
[H3O+] = Ka = 6.3 x 10-5
pH = 4.20
Copyright (c) 1999 by Harcourt Brace & Company
All rights reserved