Aqueous Ionic Equilibrium
16.1
The Danger of Antifreeze
16.2
Buffers: Solutions that Resist pH Change
16.3
Buffer Effectiveness: Buffer Range and Buffer Capacity
16.5
Solubility Equilibria and the Solubility Product Constant
16.6
Precipitation
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slide 16-1
16.1 The Danger of Antifreeze
HOCH2CH2OH(aq)
metabolism
HOCH2COOH (aq)
HOCH2COO– (aq) + H3O+ (aq)
Buffer in Blood
H2CO3 (aq) + HO– (aq)
HCO3– (aq) + H2O (l)
HCO3– (aq) + H3O+ (aq)
H2CO3 (aq) + H2O (l)
Acidosis
excess H+
Hb (aq) + O2 (g) + H2O (l)
HbO2 (aq) + H3O+ (aq)
equilibrium shift
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16.2 Buffers: Solutions That Resist pH Change
FIGURE 16.2
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A Buffer Solution
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slide 16-3
Acetic Acid as a Buffer
CH3CO2H (aq) + NaOH (aq)
CH3CO2– (aq) + HCl (aq)
CH3CO2Na (aq) + H2O (l)
CH3CO2H (aq) + H+ (aq)
Buffers resist pH change.
A buffer contains significant amounts of both a
weak acid and its conjugate base.
The weak acid neutralizes added base.
The conjugate base neutralizes added acid.
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Calculating the pH of a Buffer Solution
FIGURE 16.2
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The Common Ion Effect
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slide 16-5
ERXAMPLE 16.1
Calculating the pH of a Buffer Solution
Calculate the pH of a buffer solution that is
0.100 M in CH3CO2H and 0.100 M in CH3CO2Na.
CH3CO2H (aq) + H2O (l)
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CH3CO2– (aq) + H3O+ (l)
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slide 16-6
EXAMPLE 16.1
continued
CH3CO2H (aq) + H2O (l)
Ka =
[H3O+][A–]
[HA]
=
CH3CO2– (aq) + H3O+ (l)
x (0.100+x)
=
(0.100–x)
x (0.100)
(0.100)
=x
x << 0.1
x = 1.8 x 10–5
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% ionization =
1.8 x 10–5
0.100
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x 100% = 0.018%
slide 16-7
EXAMPLE 16.1
continued
CH3CO2H (aq) + H2O (l)
CH3CO2– (aq) + H3O+ (l)
x = 1.8 x 10–5
[H3O+] = 1.8 x 10–5
pH = –log[H3O+] = –log(1.8 x 10–5) = 4.73
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The Henderson-Hasselbalch Equation
HA (aq) + H2O (l)
Ka =
[H3O+][A–]
[HA]
when [HA] = [A–]
A– (aq) + H3O+ (l)
[H3
O+ ]
= Ka
[HA]
[A–]
and x << [HA] or [A–]
[H3O+] = Ka
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Ka =
[H3O+][A–]
[HA]
[H3
O+ ]
= Ka
[HA]
[A–]
Calculate pH by taking the negative logarithm
pH = –log [H3O+] = –log Ka – log
pH = pKa + log
pH = pKa + log
[HA]
[A–]
[A–]
[HA]
[base]
[acid]
The Henderson-Hasselbalch Equation
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slide 16-10
Calculating pH Changes in a Buffer Solution
HA (aq) + H2O (l)
A– (aq) + H3O+ (l)
Assume 1L 0.1 M in CH3CO2H and 0.1 M in CH3CO2–.
Add 0.025 mol HCl.
The stoichiometry calculation
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HA (aq) + H2O (l)
A– (aq) + H3O+ (l)
The equilibrium calculation
Ka =
[H3O+][A–]
[HA]
=
x (0.075+x)
(0.125–x)
=
x (0.075)
(0.125)
x << 0.125 or 0.0075
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Ka =
[H3O+][A–]
=
[HA]
x (0.075+x)
(0.125–x)
=
x (0.075)
(0.125)
x << 0.125 or 0.0075
x=
x = 3.0 x 10–5
Ka (0.125)
(0.075)
% ionization =
= 1.8 x 10–5 (1.67)
3.0 x 10–5
0.125
x 100% = 0.002%
pH = –log[H3O+] = –log(3.0 x 10–5) = 4.52
compared to 4.74 before addition
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Notice if you are working within a solution, the amounts of acid
and base in moles may be substituted for concentrations:
nHA
[HA]
[A–]
=
Vsolution
nA–
=
nHA
nA–
Vsolution
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FIGURE 16.3
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Buffering Action
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Buffers Containing a Base and Its Conjugate Acid
Kw = Ka x Kb
pKw = pKa + pKb
14 = pKa + pKb
pKa = 14 – pKb
pH = pKa + log
FIGURE 16.4
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[base]
[acid]
Buffer Containing a Weak Base
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slide 16-16
16.3 Buffer Effectiveness:
Buffer Range and Buffer Capacity
Relative Amounts of Acid and Base
[HA] ≈ [A–]
Consider two solutions:
pKa = 5.00
[HA] + [A–] = cHA = 0.200 M
Solution I
[HA] = 0.100 M
[A–] = 0.100 M
Solution II
[HA] = 0.180 M
[A–] = 0.020 M
Add 0.010 mol of NaOH to each solution.
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Solution I: 0.10 mol HA and 0.10 mol A–; initial pH = 4.05
pH = pKa + log
% change =
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[base]
[acid]
= 5.00 + log
0.110
0.090
= 5.09
5.09 – 5.00
x 100% = 1.8%
5.00
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slide 16-18
Solution II: 0.180 mol HA and 0.020 mol A–; initial pH = 4.05
pH = pKa + log
% change =
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[base]
[acid]
= 5.00 + log
0.170
0.030
= 4.25
4.25 – 4.05
x 100% = 5.0%
4.05
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slide 16-19
Absolute Concentrations of the Acid and
Conjugate Base
Solution I: 0.50 mol HA and 0.50 mol A–; initial pH = 5.00
pH = pKa + log
[base]
[acid]
% change =
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= 5.00 + log
0.49
0.51
= 5.02
5.02 – 5.00
x 100% = 0.40%
5.00
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slide 16-20
Solution II: 0.050 mol HA and 0.050 mol A–; initial pH = 5.00
pH = pKa + log
% change =
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[base]
[acid]
= 5.00 + log
0.060
0.040
= 5.18
5.18 – 5.00
x 100% = 3.6%
5.00
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slide 16-21
Buffer Range
pH = pKa + log
pH = pKa + log
[base]
[acid]
[base]
[acid]
= 5.00 + log
= 5.00 + log
1
10
10
1
= 4.0
= 6.0
The effective range for a buffering
system is one pH unit on either side of
pKa
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slide 16-22
Buffer Capacity
Buffer capacity is the amount of acid or base that you can
add to a buffer without causing a large change in pH.
buffer capacity increases with increasing absolute
concentrations of the buffer components
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Indicators are compounds which change colour according to pH changes
Example: Phenolphthalein
FIGURE 16.12 Phenolphthalein
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slide 16-24
pH = pKa + log
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[In–]
[HIn]
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slide 16-25
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16.5 Solubility Equilibria and the Solubility
Product Constant
CaF2 (s)
Ca2+ (aq) + 2 F– (aq)
Ksp = [Ca2+][F–]2
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Ksp and Molar Solubility
molar solubility = solubility expressed in mol
L
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≠ Ksp
slide 16-28
Ag+ (aq) + Cl– (aq)
AgCl (s)
Ksp = [Ag+][Cl–]
Ksp = S x S
S=
Ksp =
1.77 x 10–10
= 1.33 x 10–5 M
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Ksp and Relative Solubility
Ksp = [Fe2+][CO32–]
Ksp = [Mg+][HO–]2
Ksp = S x S
Ksp = S x (2S)2
S=
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2
Ksp
S=
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3
/ Ksp /4
slide 16-30
The Effect of a Common Ion on Solubility
equilibrium shifts left
CaF2 (s)
Ca2+ (aq) + 2 F– (aq)
Common Ion
0.100 M F–
In general, the solubility of an ionic compound is lower in
a solution containing a common ion than in pure water.
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EXAMPLE 16.10
Calculating Molar Solubility in the Presence
of a Common Ion
What is the molar solubility of CaF2 in a solution containing
0.100 M NaF?
Solution
CaF2 (s)
Ca2+ (aq) + 2 F– (aq)
Ksp = [Ca2+][F–]2
Ksp = (S)(0.100 + 2S)2
Ksp ≈ (S)(0.100)2
S = 1.46 x 10–8
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The Effect of an Uncommon Ion on Solubility
(Salt Effect)
2 Ag+ (aq) + SO42– (aq)
Ag2SO4 (s)
Na+ (aq) + SO42– (aq)
Na+ (aq)/SO42– (aq)
The effect of adding high concentrations of uncommon ions is to
slightly increase the solubility of an insoluble salt and is called
the uncommon ion effect or salt effect.
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The Effect of pH on Solubility
high HO–
equilibrium shifts left
Mg(OH)2 (s)
Mg2+ (aq) + 2 HO– (aq)
equilibrium shifts right
high H3O+
reacts with
HO–
In general, the solubility of an ionic compound with a strongly basic or
weakly basic anion increases with increasing acidity (decreasing pH).
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16.6 Precipitation
Ca2+ (aq) + 2 F– (aq)
CaF2 (s)
Ksp = [Ca2+][F–]2
Q = [Ca2+][F–]2
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Q < Ksp
no precipitation
Q = Ksp
saturated solution
Q > Ksp
supersaturated solution
precipitation occurs
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slide 16-35
EXAMPLE 16.13
Finding the Minimum Required Reagent
Concentration for Selective Precipitation
The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M
and [Ca2+] = 0.011 M) can be separated by selective precipitation with
KOH. What minimum [OH–] triggers the precipitation of the Mg2+ ion?
Solution Ksp (Mg(OH)2) = 2.06 x 10–13 Ksp (Ca(OH)2) = 4.68 x 10–6
Q = [Mg2+][HO–]2 = (0.059)[HO–]2
Q = Ksp saturated solution and precipitation just begins
2.06 x 10–13 = (0.059)[HO–]2
[HO–] = 1.9 x 10–6 M
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slide 16-36
EXAMPLE 16.14
Finding the Concentrations of Ions Left in
Solution after Selective Precipitation
You add potassium hydroxide to the solution in Example 16.13. When
the [HO–] reaches 1.9 x 10–6 M (as you just calculated), magnesium
hydroxide begins to precipitate out of solution. As you continue to add
KOH, the magnesium hydroxide continues to precipitate. However, at
some point, the [HO–] becomes high enough to begin to precipitate the
calcium ions as well. What is the concentration of Mg2+ when Ca2+ begins
to precipitate?
Solution Ksp (Mg(OH)2) = 2.06 x 10–13 Ksp (Ca(OH)2) = 4.68 x 10–6
What is [HO–] when Ca2+ begins to ppt?
Q = [Ca2+][HO–]2 = (0.011)[HO–]2 = Ksp
4.68 x 10–6 = (0.011)[HO–]2
[HO–] = 2.06 x 10–2
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slide 16-37
EXAMPLE 16.14
continued
Ksp (Mg(OH)2) = 2.06 x 10–13 Ksp (Ca(OH)2) = 4.68 x 10–6
[HO–] = 2.06 x 10–2 M when Ca2+ just begins to ppt.
What is [Mg2+] when Ca2+ begins to ppt?
Ksp = [Mg2+][HO–]2
2.06 x 10–13 = [Mg2+] (2.06 x 10–2)2
[Mg2+]
=
2.06 x 10–13
(2.06 x 10–2)2
[Mg2+] = 4.9 x 10–10 M
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End of Chapter 16
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