Aqueous Equilibria
Chapter 17
HW problems: 3, 5, 14, 15, 16, 23, 24, 27a, 28a, 31, 37, 43, 45, 51, 57
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
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! (and should NOT neutralize
EACH OTHER!!!!)
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
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
Adding more acid creates a shift left IF
enough acetate ions are present
Which of the following are buffer systems? (a) KF/HF
(b) KCl/HCl, (c) Na2CO3/NaHCO3
(a) HF is a weak acid and F- is its conjugate base
buffer solution
(b) HCl is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution
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)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
Ka for HCOOH = 1.8 x 10 -4
[H+] [HCOO-]
Ka =
[HCOOH]
H+ (aq) + HCOO- (aq)
x = 1.038 X 10 -4
pH = 3.98
OR…… Use the Henderson-Hasselbach equation
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-] pK = -log K
Ka =
a
a
[HA]
Henderson-Hasselbach
equation
[conjugate base]
pH = pKa + log
[acid]
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]
HCl
HCl + CH3COO-
H+ + ClCH3COOH + Cl-
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?
NH3 (aq) + H2O (l)
Kb =
[NH4+] [OH-]
[NH3]
Initial
Change
End
NH4+ (aq) + OH- (aq)
= 1.8 X 10-5
0.30
-x
0.30 - x
0.36
+x
0
+x
0.36 + x
x
(.36 + x)(x)
1.8 X 10-5 =
(.30 – x)
0.36x
x = 1.5 X 10-5
1.8 X 10-5 ≅
0.30
pOH = 4.82
pH= 9.18
What is the pH after the addition of 20.0 mL of 0.050 M
NaOH to 80.0 mL of the buffer solution?
final volume = 80.0 mL + 20.0 mL = 100 mL
NH4+
0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M
OH-
0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M
NH3
0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M
0.01
0.24
0.29
NH4+ (aq) + OH- (aq)
H2O (l) + NH3 (aq)
end (M)
0.28
0.0
0.25
Equilibrium Calculation NH4+ (aq)
H+ (aq) + NH3 (aq)
start (M)
[H+] [NH3]
Ka=
[NH4+]
[H+] 0.25
= 5.6 X 10-10
0.28
[H+] = 6.27 X 10 -10
= 5.6 X 10-10
pH = 9.20
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+]
H+ (aq) + NH3 (aq)
pKa = 9.25
[0.30]
pH = 9.25 + log
= 9.17
[0.36]
final volume = 80.0 mL + 20.0 mL = 100 mL
start (M)
end (M)
0.01
0.29
NH4+ (aq) + OH- (aq)
0.28
0.0
0.24
H2O (l) + NH3 (aq)
0.25
[0.25]
pH = 9.25 + log
= 9.20
[0.28]
Chemistry In Action: Maintaining the pH of Blood
Titrations
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 the
endpoint (hopefully close to the equivalence point)
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)
Titration Curve
A titration curve is a plot of pH vs. the amount of titrant
added. Typically the titrant is a strong (completely)
dissociated acid or base. Such curves are useful for
determining endpoints and dissociation constants of weak
acids or bases.
Features of the Strong Acid-Strong Base
Titration Curve
1. The pH starts out low, reflecting the high [H3O+] of the
strong acid and increases gradually as acid is neutralized by
the added base.
2. Suddenly the pH rises steeply. This occurs in the immediate
vicinity of the equivalence point. For this type of titration
the pH is 7.0 at the equivalence point.
3. Beyond this steep portion, the pH increases slowly as more
base is added.
Sample Calculation: Strong Acid-Strong
Base Titration Curve
Consider the titration of 40.0 mL of 0.100 M HCl with 0.100 M
NaOH.
Region 1. Before the equivalence point, after adding 20.0 mL of
0.100 M NaOH. (Half way to the equivalence point.)
Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 mol H3O+
- Moles of OH- added = 0.0200 L x 0.100 M =0.00200 mol OH
amount
(mol)
of
H
O
remaining

3
[H 3O ] 
original volume of acid  volume of added base
0.00200mol
[H 3O ] =
= 3.33E - 2
0.0600L
+
pH=1.477
Sample Calculation: Strong Acid-Strong
Base Titration Curve (Cont. I)
Region 2. At the equivalence point, after adding 40.0 mL of 0.100
M NaOH.
Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 mol H3O+
- Moles of OH- added = 0.0400 L x 0.100 M =0.00400 mol OH
amount
(mol)
of
H
O
remaining

3
[H 3O ] 
original volume of acid  volume of added base
An equal number of moles of NaOH and HCl have reacted, leaving
only a solution of their salt (NaCl). The pH=7.00 because the cation
of the strong base and the anion of the strong acid have no effect on
pH.
Sample Calculation: Strong Acid-Strong
Base Titration Curve (cont. II)
Region 3. After the equivalence point, after adding 50.0 mL of 0.100
M NaOH. (Now calculate excess OH-)
Total moles of OH- = 0.0500 L x 0.100 M = 0.00500 mol OH-Moles of H3O+ consumed = 0.0400 L x 0.100 M =0.00400 mol

amount
(mol)
of
OH
remaining

[OH ] 
original volume of acid  volume of added base
0.00100mol
[OH ] =
= .0111
0.0900L
-
pOH=1.954
pH=12.046
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)
The four Major Differences Between a Strong
Acid-Strong Base Titration Curve and a Weak
Acid-Strong Base Titration Curve
1. The initial pH is higher.
2. A gradually rising portion of the curve,
called the buffer region, appears before
the steep rise to the equivalence point.
3. The pH at the equivalence point is greater
than 7.00.
4. The steep rise interval is less pronounced.
Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq)
H+ (aq) + NH3 (aq)
NH4Cl (aq)
NH4Cl (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l)
NH3 (aq) + H+ (aq)
The four Major Differences Between a Weak
Acid-Strong Base Titration Curve and a Weak
Base-Strong Acid Titration Curve
1. The initial pH is above 7.00.
2. A gradually decreasing portion of the curve,
called the buffer region, appears before a
steep fall to the equivalence point.
3. The pH at the equivalence point is less than
7.00.
4. Thereafter, the pH decreases slowly as excess
strong acid is added.
Features of the Titration of a
Polyprotic Acid with a Strong Base
1. The loss of each mole of H+ shows up as
separate equivalence point (but only if the
two pKas are separated by more than 3 pK
units).
2. The pH at the midpoint of the buffer region
is equal to the pKa of that acid species.
3. The same volume of added base is required
to remove each mole of H+.
Acid-Base Indicators
The titration curve of a strong acid with a strong base.
16.5
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
Finding the Equivalence Point
(calculation method)
• Strong Acid vs. Strong Base
– 100 % ionized! pH = 7 No equilibrium!
• Weak Acid vs. Strong Base
– Acid is neutralized; Need Kb for conjugate base
equilibrium
• Strong Acid vs. Weak Base
– Base is neutralized; Need Ka for conjugate acid
equilibrium
• Weak Acid vs. Weak Base
– Depends on the strength of both; could be conjugate
acid, conjugate base, or pH 7
Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of
a 0.10 M NaOH solution. What is the pH at the
equivalence point ?
start (moles)
0.01
0.01
HNO2 (aq) + OH- (aq)
NO2- (aq) + H2O (l)
end (moles)
0.0
0.0
0.01
NO2- (aq) + H2O (l)
0.01
= 0.05 M
0.200
OH- (aq) + HNO2 (aq)
0.05
0.00
0.00
-x
+x
+x
x
x
Final volume = 200 mL
Equilibrium Calculation
Initial (M)
Change (M)
[NO2-] =
Equilibrium (M) 0.05 - x
[OH-][HNO2]
x2
-11
=
2.2
x
10
Kb =
=
[NO2-]
0.05-x
x  1.05 x 10-6 = [OH-]
pOH = 5.98
pH = 14 – pOH = 8.02
Solubility Equilibria
• Looking at dissolution or precipitation reactions of
ionic compounds
• Ksp (solubility product) is the equilibrium constant for the
equilibrium reaction between a solid and its
saturated ions
– Indicates how soluble the solid is when in water
• Solids, liquids and solvents do not appear in
the equilibrium equation (just like before)
Solubility and Ksp
• Solubility is how much actually dissolves while
Ksp is the equilibrium constant.
– pH, temperature, and the presence of other ions
in solution affect solubility
– Only temperature affects Ksp
• Can use Ksp to calculate solubility but need to
be careful…not always correct!
Factors that Affect Solubility(4)
1. Common Ion Effect
•
•
presence of other ions reduces solubility
presence of 2nd solute that gives a common ion
reduces solubility
2. pH
•
dissolving increases when pH is made more acidic
(more of the basic ion X- will be made to react with extra H+ ions)
•
the more basic the anion, the more its solubility is
affected by pH
3. Formation of a Complex Ion
4. Amphoterism
Complex Ion Equilibria and Solubility
Where they “smush “ it all together
A complex ion is an ion containing a central metal cation
bonded to one or more molecules or ions.
Co2+ (aq) + 4Cl- (aq)
2-
CoCl4 (aq)
•Metal ions can act as Lewis acids
•Can react with water
Co(H2O)2+
6
CoCl24
•Can react with other Lewis bases
(must be able interact more than water can)
•Keq becomes Kf (formation constant)
•Kf’s size shows stability of complex
ion in solution
Complex Ion Formation
• These are usually formed from a transition
metal surrounded by ligands (polar
molecules or negative ions).
• As a "rule of thumb" you place twice the
number of ligands around an ion as the
charge on the ion... example: the dark blue
Cu(NH3)42+ (ammonia is used as a test for
Cu2+ ions), and Ag(NH3)2+.
• Memorize the common ligands.
Common Ligands
Ligands
Names used in the ion
H2O
NH3
aqua
ammine
OHClBrCNSCN-
hydroxy
chloro
bromo
cyano
thiocyanato (bonded through
sulphur)
isothiocyanato (bonded through
nitrogen)
Names
• Names: ligand first, then cation
Examples:
– tetraamminecopper(II) ion: Cu(NH3)42+
– diamminesilver(I) ion: Ag(NH3)2+.
– tetrahydroxyzinc(II) ion: Zn(OH)4 2-
• The charge is the sum of the parts
(2+) + 4(-1)= -2.
When Complexes Form
• Aluminum also forms complex ions as do some
post transitions metals. Ex: Al(H2O)63+
• Transitional metals, such as Iron, Zinc and
Chromium, can form complex ions.
• The odd complex ion, FeSCN2+, shows up once
in a while
• Acid-base reactions may change NH3 into NH4+
(or vice versa) which will alter its ability to act as
a ligand.
• Visually, a precipitate may go back into solution
as a complex ion is formed. For example, Cu2+ +
a little NH4OH will form the light blue precipitate,
Cu(OH)2. With excess ammonia, the complex,
Cu(NH3)42+, forms.
Factors that Affect Solubility (cont.)
4. Amphoterism
•
Some insoluble, metal hydroxides will dissolve in
strong acids or strong bases
-
•
will dissolve in acid because of basic anion
Will dissolve in base because a complex ion forms
Solubility varies with the metal bonded
Precipitation & Separation of Ions
Q > Ksp get precipitation until Q = Ksp
Q = Ksp at equilibrium
Q < Ksp solid dissolves until Q = Ksp
• Selective Precipitation of ions
– Separate ions in an aqueous solution using a
reagent that forms a precipitate 1 or few ions in
solution
– Ex. Ag+ + Cu2+ + HCl
AgCl + Cu2+ + H+