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CH160 General Chemistry II
Lecture Presentation
Applications of Acid-Base Equilibria
Chapter 17
Sections 1-4
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Chapter 17
1
Common Ion Effect
• Consider the ionization of weak acid HA:
HA + H2O <=> H3O+ + A• What affect will adding salt NaA to the
solution have on the acid ionization and
solution pH?
NaA  Na+ + A-
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Chapter 17
2
Common Ion Effect
• Consider the ionization of weak acid HA:
HA + H2O <=> H3O+ + AA- has 2 sources: HA
and NaA. Adding NaA
increases [A-].
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Chapter 17
3
Common Ion Effect
• Consider the ionization of weak acid HA:
HA + H2O <=> H3O+ + AA- is a “common
ion”
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Chapter 17
4
Example 1
(1a on the Example Problem Handout)
• Calculate the percent ionization of the acid
and the pH of the solution that contains
0.500M HC2H3O2 (Ka = 1.8 x 10-5) and
0.200 M NaC2H3O2.
• (ans.: pH = 4.34, 0.0090%)
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Chapter 17
5
Calculations Using Ka
• Basic Steps for Weak Acid Calculations Using Ka
– Write balanced chemical equation and the expression
for Ka
• Look up value for Ka
– For each chemical species involved in the equilibrium
(except H2O), write:
• Initial concentration
• Equilibrium concentration
– Let the change in the [H3O+] be the variable “x”
– Substitute the equilibrium concentrations into Ka and
solve for x using either
• quadratic approach
• simplified approach
– Calculate pH, equilibrium concentrations, % ionization,
etc., as specified in the problem.
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Chapter 17
6
0.5 M CH3COOH +
0.200 M CH3COONa
(problem 1a)
0.5 M
CH3COOH
pH = 4.3
(problem 8a)
+ CH3COONa
pH = 2.5
0.6%
0.009%
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Chapter 17
7
Common Ion Effect
• What affect does NaA have on weak acid
HA ionization?:
HA + H2O <=> H3O+ + AWhich way does equilibrium shift?
What do the results of problems 8a
and 1a tell us? Does this agree
with LeChatelier’s principle?
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Chapter 17
8
Common Ion Effect
• What affect does NaA have on weak acid
HA ionization?:
HA + H2O <=> H3O+ + AIn presence of NaA, HA
ionization shifts left.
“common-ion effect”
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Chapter 17
9
Buffer Solutions
• What is a buffer solution?
– solution with ability to resist pH changes upon
addition of small amounts of either acid or
base
• Requirements
– must contain an acid to neutralize added OHions
– must contain a base to neutralize added H3O+
ions
– acidic and basic species in buffer must not
neutralize each other.
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Chapter 17
11
Buffer Action
• How do buffers work? Consider buffer
with weak acid HA and salt NaA:
– Addition of acid
A- + H+  HA
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Chapter 17
12
Buffer Action
• How do buffers work? Consider buffer
with weak acid HA and salt NaA:
– Addition of acid (small amount)
A- + H+  HA
[A-] decreases
slightly
[HA] increases
slightly
Added acid is neutralized.
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Chapter 17
13
Buffer Action
• How do buffers work? Consider buffer
with weak acid HA and salt NaA:
– Addition of base
HA + OH-  H2O + A-
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Chapter 17
14
Buffer Action
• How do buffers work? Consider buffer
with weak acid HA and salt NaA:
– Addition of base
HA + OH-  H2O + A[HA] decreases
slightly
[A-] increases
slightly
Added base is neutralized.
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Chapter 17
15
Buffer Action
• Two important properties of buffer
solutions:
– Buffer capacity
• Amount of acid or base the buffer can react with
before giving a significant pH change (1 pH unit)
• Determined by how much buffer acid and base are
used to make buffer
– pH
• Determined by Ka and relative amounts of buffer
acid and base present
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Chapter 17
16
Calculation of Buffer pH
• Calculating pH for buffer containing both
weak acid HA and salt NaA. The major
equilibrium is:
– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
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Chapter 17
17
Calculation of Buffer pH
• Calculating pH for buffer containing both
weak acid HA and salt NaA.
– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
A- has 2 sources:
HA and NaA
(This seems
familiar! Didn’t we
just do this?)
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Chapter 17
18
Calculation of Buffer pH
• Calculating pH for buffer containing both
weak acid HA and salt NaA.
– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
Neither CHA nor CNaA change much
since only a very small amount of HA
ionizes.
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Chapter 17
19
Calculation of Buffer pH
• Calculating pH for buffer containing both
weak acid HA and salt NaA.
– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
• Since % ionization is small: CA-  [A-] and
CHA  [HA]
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Chapter 17
20
Calculation of Buffer pH
• Calculating pH for buffer containing both
weak acid HA and salt NaA.
– HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA]
• Since % ionization is small: CA-  [A-] and
CHA  [HA]
Solving for [H3O+] gives:
[H3O+] = KaCHA/CApH = -log[H3O+]
(This is just a common ion effect problem.)
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Chapter 17
21
Calculation of Buffer pH
• We can also take the –log of our
equation:
[H3O+] = KaCHA/CA-log[H3O+] = -log(KaCHA/CA-)
-log[H3O+] = -logKa - log CHA/CApH = pKa - log CHA/CA-
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Chapter 17
22
Example 2
(2a on Example Problem Handout)
• Calculate the pH of a buffer solution
that contains 0.25 M sodium acetate,
NaC2H3O2, and 0.35 M acetic acid,
HC2H3O2 (Ka = 1.8 x 10-5).
• (ans.: pH = 4.60)
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Chapter 17
23
Calculating pH Changes in
Buffers
• How do we calculate buffer pH after
adding acid (H3O+) or base (OH-)?
HA/NaA
Buffer
H3O+ + A-  HA + H2O
+ H3O+ or OH-
or
OH- + HA  A- + H2O
Neutralization Rxn
pH
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Calculate new
[H3O+] from [H3O+]
= KaCHA/CAChapter 17
Calculate new [HA]
& [A-]
24
Calculating pH Changes in
Buffers
• How do we calculate buffer pH after
adding acid (H3O+) or base (OH-)?
HA/NaA
Buffer
H3O+ + A-  HA + H2O
+ H3O+ or OH-
or
OH- + HA  A- + H2O
Neutralization Rxn
This much is a stoichiometry
problem. (Oh, oh! General
Chemistry I stuff here.).
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Chapter 17
Calculate new [HA]
& [A-]
25
Calculating pH Changes in
Buffers
• How do we calculate buffer pH after
adding acid (H3O+) or base (OH-)?
This part is an equilibrium
calculation.
pH
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Calculate new
[H3O+] from [H3O+]
= KaCHA/CAChapter 17
26
Example 3
(3a on Example Problem Handout)
• Calculate the pH of the solution formed
and the change in pH observed when (a)
0.050 moles of HCl and (b) 0.025 moles of
NaOH are added to 500 mL of the buffer in
example (2a). (c) Calculate the change in
pH that occurs when 0.050 moles HCl are
added to 500 mL H2O.
(ans.: (a) pH = 4.27, pH = -0.33, (b) pH = 4.74, pH = +0.14, (c) pH
= -6)
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Chapter 17
27
Buffer Preparation
• What if I need to make a buffer solution of
known pH? Select:
– Buffer system.
• Often pKa of buffer acid is close to desired pH.
– Relative amounts of buffer acid and base.
• Buffer capacity increases with concentrations.
• Buffer effectiveness best with concentrations on
same order of magnitude.
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Chapter 17
28
Example 4
(4 on Example Problem Handout)
• Starting with 1.0L of 0.100 M CH3COOH
(Ka = 1.8 x 10-5), how many grams of
sodium acetate, CH3COONa (FW =
82.034 g/mol), to give a buffer with a pH of
4.40? (Assume no volume change.)
• (ans.: 3.7 g)
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Chapter 17
29
Quantitative Acid-Base Chemistry
• How do we calculate the pH of a solution
formed by mixing an acid solution with a base
solution?
• Consider addition of 0.1 M strong base, MOH,
to 0.1 M strong acid, HX
– pH changes can be observed from titration curve
• pH vs. volume standard
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Chapter 17
30
Strong Acid-Strong Base Titration
HX + NaOH  NaX + H2O
Titration of 25 mL of 0.1 M HCl with 0.1 M
NaOH
14
12
10
pH
pH
8
Rxn is complete =
equivalence point. pH
= 7.0 since only neutral
NaX present.
6
4
2
0
0
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5
10
15
20
25
30
35
mL NaOH
40
volume
0.1 M
Chapter
17NaOH
45
50
55
31
Calculating pH in Acid-Base
Reactions
• How do we calculate pH after adding a
strong base to a strong acid?
+ MOH
HX
Strong
acid
pH
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HX + MOH  MX + H2O
Neutralization Rxn
Calculate new [H3O+]
from [HX] or [MOH]
Chapter 17
Calculate [HX] or
[MOH] left (ignore
neutral MX)
32
Calculating pH in Acid-Base
Reactions
• How do we calculate pH after adding a strong
base to a strong acid? Considerations:
• Stoichiometry
– Limiting reagent
• At what point in rxn does calculation take place?
–
–
–
–
Initial
Pre-equivalence
Equivalence
Post-equivalence
• Does dilution occur?
– If mixing 2 solutions: Vtotal = V1 + V2
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Chapter 17
33
Strong Acid-Strong Base Titration
HX + NaOH  NaX + H2O
Titration of 25 mL of 0.1 M HCl with 0.1 M
NaOH
-NaOH in excess
14
-NaOH/NaX left
12
-[OH-] = CMOH
10
pH
pH
-All HX & MOH
consumed
8
6
Only HX
4
[H3O+] =
CHX
2
-HX in excess
-NaX left (neutral)
-HX/NaX left
-pH = 7.0
-[H3O+] = CHX
0
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0
5
10
15
25 30 35 40
mL
Chapter NaOH
17
20
volume 0.1 M NaOH
45
50
55
34
Example 5
(5 of Example Problem Handout)
• Calculate the pH for a solution prepared
by mixing 25.00 mL of 0.100 M HCl(aq)
with a) 10.00 mL b) 25.00 mL and c) 35.00
mL of 0.100 M NaOH(aq). (ans.: a) 1.37, b) 7.0,
c) 12.22)
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Chapter 17
35
Weak Acid-Strong Base Titration
HA + NaOH  NaA + H2O
pH
pH
Titration of 25 mL of 0.1 M HOAc with 0.1 M
NaOH
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14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Rxn is complete =
equivalence point.
pH > 7.0 since only
basic NaA present.
0
5
10
15
20
25
Chapter 17
30
mL NaOH
35
volume 0.1 M NaOH
40
45
50
38
Weak Acid-Strong Base Titration
HA + NaOH  NaA + H2O
Titration of 25 mL of 0.1 M HOAc with 0.1 M
NaOH
pH
pH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
-NaOH in excess
-NaOH/NaA left
-HA in excess
-[OH-]  CMOH
-HA/NaA left = buffer
-All HA & MOH consumed
-[H3O+] = Ka(CHA/CNaA)
-NaA left (weak base)
-[OH-]  (KbCNaA)1/2
-pH > 7.0
-only HA
[H3O+]  (KaCHA)1/2
0
5
10
15
20
25
30
35
40
45
50
volume 0.1 M NaOH
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Chapter 17
39
Strong Acid-Weak Base Titration
HX + B  HB+ + XTitration of 25 mL of 0.1 M ammonia with 0.1
M HCl
pH
pH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Rxn is complete =
equivalence point. pH
< 7.0 since only acidic
HB+ present.
0
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5
10
15
20
25
30
35
volum e 0.1 M HCl
Chapter 17
40
45
50
40