Acid-Base Equilibria
Chapter 16
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
16.2
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
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
Adding more acid creates a shift left IF
enough acetate ions are present
16.3
Which of the following are buffer systems? (a) KF/HF
(b) KCl/HCl, (c) Na2CO3/NaHCO3
(a) KF 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
16.3
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 =
H+ (aq) + HCOO- (aq)
x = 1.038 X 10 -4
pH = 3.98
[HCOOH]
16.2
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]
[H+][A-] pK = -log K
Ka =
a
a
[HA]
Henderson-Hasselbach
equation
[conjugate base]
pH = pKa + log
[acid]
[A-]
pH = pKa + log
[HA]
16.2
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
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]
HCOOH pKa = 3.77
16.2
HCl
HCl + CH3COO-
H+ + ClCH3COOH + Cl-
16.3
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 =
1.8 X
10-5 
(.30 – x)
0.36x
0.30
x = 1.5 X 10-5
pOH = 4.82
pH= 9.18
16.3
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?
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
start (M)
end (M)
Ka=
0.01
0.29
NH4+ (aq) + OH- (aq)
0.28
0.0
[H+] [NH3]
= 5.6 X 10-10
[NH4+]
[H+] 0.25
0.28
0.24
H2O (l) + NH3 (aq)
0.25
[H+] = 6.27 X 10 -10
= 5.6 X
10-10
pH = 9.20
16.3
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]
16.3
Chemistry In Action: Maintaining the pH of Blood
16.3
Titrations
In a titration a solution of accurately known concentration is
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)
4.7
Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq)
OH- (aq) + H+ (aq)
H2O (l) + NaCl (aq)
H2O (l)
100%
ionization!
No equilibrium
16.4
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)
16.4
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)
16.4
Acid-Base Indicators
16.5
pH
16.5
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
16.5
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
Final volume = 200 mL
NO2- (aq) + H2O (l)
Initial (M)
Change (M)
0.01
= 0.05 M
0.200
OH- (aq) + HNO2 (aq)
[NO2-] =
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
pOH = 5.98
0.05 – x  0.05 x  1.05 x 10-6 = [OH-]
pH = 14 – pOH = 8.02
Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation
bonded to one or more molecules or ions.
Co2+ (aq) + 4Cl- (aq)
Co(H2O)2+
6
CoCl42- (aq)
CoCl24
16.10
16.10
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.
• Keywords such as "excess" and "concentrated" of any
solution may indicate complex ions. AgNO3 + HCl forms
the white precipitate, AgCl. With excess, concentrated
HCl, the complex ion, AgCl2-, forms and the solution
clears.
Coordination Number
• Total number of bonds from the ligands
to the metal atom.
• Coordination numbers generally range
between 2 and 12, with 4
(tetracoordinate) and 6
(hexacoordinate) being the most
common.
Some Coordination Complexes
molecular
formula
Lewis
base/ligand
Lewis acid donor
atom
coordination
number
Ag(NH3)2+
NH3
Ag+
N
2
[Zn(CN)4]2- CN-
Zn2+
C
4
[Ni(CN)4]2-
CN-
Ni2+
C
4
[PtCl6] 2-
Cl-
Pt4+
Cl
6
Ni2+
N
6
[Ni(NH3)6]2+ NH3