Transcript Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria and
Solubility Equilibria
Chapter 16
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Homogeneous and Heterogeneous Solution
Equilibria
Homogeneous equilibria: one phase
• Weak acids and bases in solution do not ionize
completely
• At equilibrium in solution
–
–
–
–
non-ionized acid HA, H+ ions and conjugate base Aexample: CH3COOH, H+ and CH3COOnon-ionized base B, OH- ions and conjugate acid BH+
example: NH3, OH- and NH4+
Heterogeneous equilibria: more than one phase
– solution and solid
– precipitation of slightly soluble substances
– effect of concentration and pH
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Buffer Effect
The effect of addition of acid or base to …
acid added
base added
an unbuffered solution
acid added
or a buffered solution
base added
The common ion effect is the shift in equilibrium caused by the
addition of a compound having an ion in common with the
dissolved substance.
The presence of a common ion suppresses the ionization of
a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CH3COONa (s)
Na+ (aq) + CH3COO- (aq)
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
common
ion
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Consider mixture of salt NaA and weak acid HA.
NaA (s)
Na+ (aq) + A- (aq)
HA (aq)
H+ (aq) + A- (aq)
[H+]
Ka [HA]
=
[A-]
-log [H+] = -log Ka - log
[HA]
[A-]
-]
[A
-log [H+] = -log Ka + log
[HA]
[A-]
pH = pKa + log
[HA]
[H+][A-]
Ka =
[HA]
Henderson-Hasselbalch
equation
[conjugate base]
pH = pKa + log
[acid]
pKa = -log Ka
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What is the pH of a solution containing 0.30 M HCOOH and
0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
Common ion effect
0.30 – x  0.30
0.52 + x  0.52
HCOOH pKa = 3.77
H+ (aq) + HCOO- (aq)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
[HCOO-]
pH = pKa + log
[HCOOH]
[0.52]
= 4.01
pH = 3.77 + log
[0.30]
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A buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon
the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq)
Add strong base
OH- (aq) + CH3COOH (aq)
CH3COOH (aq)
CH3COO- (aq) + H2O (l)
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HCl
HCl + CH3COO-
H+ + ClCH3COOH + Cl-
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Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) HF is a weak acid and F- is its conjugate base
buffer solution
(b) HBr is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is its conjugate acid
buffer solution
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Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of 0.050
M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq)
[NH3]
pH = pKa + log
[NH4+]
start (moles)
end (moles)
H+ (aq) + NH3 (aq)
pKa = 9.25
0.029
0.001
NH4+ (aq) + OH- (aq)
0.028
0.0
[0.30]
pH = 9.25 + log
= 9.17
[0.36]
0.024
H2O (l) + NH3 (aq)
0.025
final volume = 80.0 mL + 20.0 mL = 100 mL
[NH4
+]
0.028
0.025
=
[NH3] =
0.10
0.10
[0.25]
pH = 9.25 + log
= 9.20
[0.28]
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Relationship between buffer capacity and pH change
Change in pH after addition of base to an acetic acid buffer system
Buffer Capacity and Buffer Range
Buffer capacity is the ability to resist pH change.
The more concentrated the components of a buffer, the greater
the buffer capacity.
A buffer has the highest capacity when the component
concentrations are equal.
Buffer range is the pH range over which the buffer acts effectively.
The pH of a buffer is distinct from its buffer capacity.
Buffers have a usable range within ± 1 pH unit of the pKa of
its acid component.
Buffer Capacity and Buffer Range
•
•
•
pH of a buffer made of equal volumes of 1.0 M HA and 1.0 M ApH of a buffer made of equal volumes of 0.1 M HA and 0.1 M ApH is the same
a) at 1.0 M of each
pH = pKa + log[base]/[acid] = pKa + log [1.0M]/[1.0M] = pKa
b) At 0.1 M each
pH = pKa + log [0.1M]/[0.1M] = pKa
c) At 0.01 M each
pH = pKa + log [0.01M]/[0.01M] = pKa
Buffer Capacity and Buffer Range
•
•
•
Buffer capacity of a buffer made of equal volumes of 1.0 M HA and 1.0 M ABuffer capacity of a buffer made of equal volumes of 0.1 M HA and 0. 1 M ABuffer capacities differ
CH3COO-(aq) + H3O+(aq)
CH3COOH(aq) + H2O(l)
Buffer made of 1.000 M HA and 1.000 M A•
•
[HA]init = [A-]init = 1.000 M
add 0.010 mol OH-
 [A-]init/[HA]init = 1.000
– [HA]final = 0.990 mol/L
– [A-]final = 1.010 mol/L
– [A-]final/ [HA]final = 1.010/0.990 = 1.02
•
Apply Henderson-Hasselbalch equation
– pH = = pKa + log[base]/[acid] = 4.74 + log (1.02) = 4.75
– Change in ratio of concentrations = (1.02-1.00)/1.00 x 100% = 2%
– Change in pH = (4.75 – 4.74)/4.74 x 100% = 0.2%
Buffer Capacity and Buffer Range
CH3COO-(aq) + H3O+(aq)
CH3COOH(aq) + H2O(l)
Buffer made of 0.100 M HA and 0.100 M A• [HA]init = [A-]init = 0.100 M
• add 0.010 mol OH-
 [A-]init/[HA]init = 1.000
– [HA]final = 0.090 mol/L
– [A-]final = 0.110 mol/L
– [A-]final/ [HA]final = 0.110/0.090 = 1.22
• Apply Henderson-Hasselbalch equation
– pH = 4.74 + log (1.22) = 4.83
– Change in ratio of concentrations = (1.22-1.00)/1.00 x 100% = 22%
– Change in pH = (4.83-4.74)/4.74 x 100% = 1.9%
Buffer Capacity and Buffer Range
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
• Buffer made of 0.250 M HA and 1.750 M A• [HA]init = 0.250
• [A-]init = 1.750 M
• add 0.010 mol OH-
 [A-]init/[HA]init = 7.00
– [HA]final = 0.240 mol/L
– [A-]final = 1.760 mol/L
– [A-]final/ [HA]final = 1.760/0.240 = 7.33
• Apply Henderson-Hasselbalch equation
– pH = 4.74 + log (7.33) = 5.61
– Change in ratio of concentrations = (7.33-7.00)/7.00 x 100% = 4.7%
– Change in pH = (5.61-5.58)/5.58 x 100% = 0.5%
Titrations (Review)
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
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Alternative Method of Equivalence Point Detection
monitor pH
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Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq)
OH- (aq) + H+ (aq)
H2O (l) + NaCl (aq)
H2O (l)
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Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq)
CH3COONa (aq) + H2O (l)
CH3COOH (aq) + OH- (aq)
CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7):
CH3COO- (aq) + H2O (l)
OH- (aq) + CH3COOH (aq)
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Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq)
H+ (aq) + NH3 (aq)
NH4Cl (aq)
NH4+ (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l)
NH3 (aq) + H+ (aq)
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Calculate the pH of an 0.150 M ammonium/ 0.075 M
ammonia buffer solution
Ka (NH4+) = 5.6 x 10-10
What is the pH after adding 25 mL 0.025 M HCl to 250
mL of the ammonium/ammonia buffer solution
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Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M
NaOH solution. What is the pH at the equivalence point ?
start (moles)
0.01
0.01
HNO2 (aq) + OH- (aq)
0.0
0.0
NO2- (aq) + H2O (l)
end (moles)
0.01
0.01
Final volume = 200 mL
[NO2-] =
= 0.05 M
0.200
NO2- (aq) + H2O (l)
OH- (aq) + HNO2 (aq)
Initial (M)
Change (M)
0.05
0.00
0.00
-x
+x
+x
x
x
Equilibrium (M) 0.05 - x
[OH-][HNO2]
x2
-11
=
2.2
x
10
Kb =
=
[NO2-]
0.05-x
0.05 – x  0.05 x  1.05 x
10-6 =
[OH-]
pOH = 5.98
pH = 14 – pOH = 8.02
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Acid-Base Indicators
HIn (aq)
H+ (aq) + In- (aq)
[HIn]
 10 Color of acid (HIn) predominates
[In ]
[HIn]
-) predominates
Color
of
conjugate
base
(In

0.10
[In-]
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Solutions of Red Cabbage Extract
pH
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Acid-Base Indicators
pH
HInd
H+ + Ind-
The titration curve of a strong acid with a strong base.
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Which indicator(s) would you use for a titration of HNO2 with
KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of weak acid.
At equivalence point, pH > 7
Use cresol red or phenolphthalein
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