Applications of
Aqueous Equilibria
Buffered Solutions
• A solution that resists a change in
pH when either hydroxide ions or
protons are added.
• Buffered solutions contain
either:
– A weak acid and its salt
– A weak base and its salt
Acid/Salt Buffering Pairs
The salt will contain the anion of the acid,
and the cation of a strong base (NaOH, KOH)
Weak Acid
Formula
of the acid
Hydrofluoric
HF
Formic
HCOOH
Benzoic
C6H5COOH
Acetic
Carbonic
Propanoic
Hydrocyanic
CH3COOH
H2CO3
HC3H5O2
HCN
Example of a salt of the
weak acid
KF – Potassium fluoride
KHCOO – Potassium formate
NaC6H5COO – Sodium benzoate
NaH3COO – Sodium acetate
NaHCO3 - Sodium bicarbonate
NaC3H5O2 - Sodium propanoate
KCN - potassium cyanide
Base/Salt Buffering Pairs
The salt will contain the cation of the base,
and the anion of a strong acid (HCl, HNO3)
Formula of
the base
Example of a salt of the weak
acid
NH3
NH4Cl - ammonium chloride
Methylamine
CH3NH2
CH3NH2Cl – methylammonium chloride
Ethylamine
C2H5NH2
C2H5NH3NO3 - ethylammonium nitrate
Aniline
C6H5NH2
C6H5NH3Cl – aniline hydrochloride
Base
Ammonia
Pyridine
C5H5N
C5H5NHCl – pyridine hydrochloride
Weak Acid/Strong Base Titration
13
12
11
10
9
Endpoint is above
pH 7
pH
8
7
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Strong Acid/Strong Base Titration
13
12
11
10
9
pH
8
7
A solution that is
0.10 M HCl is
titrated with
0.10 M NaOH
Endpoint is at
pH 7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Strong Acid/Strong Base Titration
13
12
A solution that is
0.10 M NaOH is
titrated with
0.10 M HCl
11
10
9
pH
8
7
Endpoint is at
pH 7
It is important to
recognize that
titration curves are
not always
increasing from left
to right.
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters HCl (0.10 M)
30.00
35.00
40.00
45.00
Strong Acid/Weak Base Titration
13
12
11
10
9
pH
8
7
6
5
Endpoint is below
pH 7
4
A solution that is
0.10 M HCl is
titrated with
0.10 M NH3
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NH3 (0.10 M)
30.00
35.00
40.00
45.00
Titration of an Unbuffered Solution
13
12
11
A solution that is
0.10 M HC2H3O2
is titrated with
0.10 M NaOH
10
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Titration of a Buffered Solution
13
12
11
A solution that is
0.10 M HC2H3O2 and
0.10 M NaC2H3O2 is
titrated with 0.10 M
NaOH
10
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Comparing Results
Buffered
13
13
12
12
11
11
10
10
9
9
8
8
7
pH
pH
Unbuffered
6
6
5
5
4
4
3
3
2
1
0.00
7
2
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
1
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
milliliters NaOH (0.10 M)
In what ways are the graphs different?
In what ways are the graphs similar?
40.00
45.00
Comparing Results
pH
Gra ph
Buffered
Unbuffered
mL 0.10 M NaOH
Buffer capacity
• The best buffers have a ratio
[A-]/[HA] = 1
• This is most resistant to change
• True when [A-] = [HA]
• Make pH = pKa (since log1=0)
General equation
• Ka = [H+] [A-]
[HA]
• so [H+] = Ka [HA]
[A-]
• The [H+] depends on the ratio [HA]/[A-]
• taking the negative log of both sides
• pH = -log(Ka [HA]/[A-])
• pH = -log(Ka)-log([HA]/[A-])
• pH = pKa + log([A-]/[HA])
This is called the
Henderson-Hasselbalch Equation
pH
 [A  ] 
 pKa  log 
  pKa

[
HA
]


 [base ] 

 log 

[
acid
]


 [ BH  ] 
 [acid ] 
pOH  pKb  log 
  pKb  log 

 [base] 
 [ B] 
Using the
Henderson-Hasselbalch Equation
• pH = pKa + log([A-]/[HA])
• pH = pKa + log(base/acid)
• Calculate the pH of the following
mixtures
• 0.75 M lactic acid (HC3H5O3) and 0.25
M sodium lactate (Ka = 1.4 x 10-4)
• 0.25 M NH3 and 0.40 M NH4Cl
• (Kb = 1.8 x 10-5)
Prove they’re buffers
• What would the pH be if 0.020 mol of
HCl is added to 1.0 L of both of the
following solutions.
– 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
– 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)
To do this I must introduce the BAAM table!
• What would the pH be if 0.050 mol of
solid NaOH is added to each of the
proceeding.
Buffer capacity
• The pH of a buffered solution is
determined by the ratio [A-]/[HA].
• As long as this doesn’t change much the
pH won’t change much.
• The more concentrated these two are
the more H+ and OH- the solution will be
able to absorb.
• Larger concentrations bigger buffer
capacity.
Buffer Capacity
• Calculate the change in pH that occurs
when 0.010 mol of HCl(g) is added to
1.0L of each of the following:
• 5.00 M HAc and 5.00 M NaAc
• 0.050 M HAc and 0.050 M NaAc
• Ka= 1.8x10-5
Summary
• Strong acid and base just stoichiometry.
• Determine Ka, use for 0 mL base
• Weak acid before equivalence point
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak acid at equivalence point Kb
• Weak base after equivalence - leftover
strong base.
Summary
• Determine Ka, use for 0 mL acid.
• Weak base before equivalence point.
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak base at equivalence point Ka.
• Weak base after equivalence - leftover
strong acid.
Selection of Indicators
Solubility Equilibria
Will it all dissolve, and
if not, how much?
• All dissolving is an equilibrium.
• If there is not much solid it will all
dissolve.
• As more solid is added the solution will
become saturated.
• Solid
dissolved
• The solid will precipitate as fast as it
dissolves .
• Equilibrium
Watch out
• Solubility is not the same as solubility
product.
• Solubility product is an equilibrium
constant.
• it doesn’t change except with
temperature.
• Solubility is an equilibrium position for how
much can dissolve.
• A common ion can change this.
Ksp Values for Some Salts at 25C
Name
Formula
Ksp
Barium carbonate
BaCO3
2.6 x 10-9
Barium chromate
BaCrO4
Barium sulfate
Formula
Ksp
Lead(II) bromide
PbBr2
6.6 x 10-6
1.2 x 10-10
Lead(II) chloride
PbCl2
1.2 x 10-5
BaSO4
1.1 x 10-10
Lead(II) iodate
Pb(IO3)2
3.7 x 10-13
Calcium carbonate
CaCO3
5.0 x 10-9
Lead(II) iodide
PbI2
8.5 x 10-9
Calcium oxalate
CaC2O4
2.3 x 10-9
Lead(II) sulfate
PbSO4
1.8 x 10-8
Calcium sulfate
CaSO4
7.1 x 10-5
Magnesium carbonate
MgCO3
6.8 x 10-6
Copper(I) iodide
CuI
1.3 x 10-12
Magnesium hydroxide
Mg(OH)2
5.6 x 10-12
Copper(II) iodate
Cu(IO3)2
6.9 x 10-8
Silver bromate
AgBrO3
5.3 x 10-5
Copper(II) sulfide
CuS
6.0 x 10-37
Silver bromide
AgBr
5.4 x 10-13
Iron(II) hydroxide
Fe(OH)2
4.9 x 10-17
Silver carbonate
Ag2CO3
8.5 x 10-12
FeS
6.0 x 10-19
Silver chloride
AgCl
1.8 x 10-10
Fe(OH)3
2.6 x 10-39
Silver chromate
Ag2CrO4
1.1 x 10-12
Lead(II) bromide
PbBr2
6.6 x 10-6
Silver iodate
AgIO3
3.2 x 10-8
Lead(II) chloride
PbCl2
1.2 x 10-5
Silver iodide
AgI
8.5 x 10-17
Lead(II) iodate
Pb(IO3)2
3.7 x 10-13
Strontium carbonate
SrCO3
5.6 x 10-10
Lead(II) iodide
PbI2
8.5 x 10-9
Strontium fluoride
SrF2
4.3 x 10-9
Lead(II) sulfate
PbSO4
1.8 x 10-8
Strontium sulfate
SrSO4
3.4 x 10-7
ZnS
2.0 x 10-25
Iron(II) sulfide
Iron(III) hydroxide
Name
Zinc sulfide
Solving Solubility Problems
For the salt AgI at 25C, Ksp = 1.5 x 10-16
AgI(s)  Ag+(aq) + I-(aq)
I
O
O
C
+x
+x
E
x
x
1.5 x 10-16 = x2
x = solubility of AgI in mol/L = 1.2 x 10-8 M
Solving Solubility Problems
For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5
PbCl2(s)  Pb2+(aq) + 2Cl-(aq)
I
O
O
C
+x
+2x
E
x
2x
1.6 x 10-5 = (x)(2x)2 = 4x3
x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M
Relative solubilities
• Ksp will only allow us to compare the
solubility of solids the that fall apart into
the same number of ions.
• The bigger the Ksp of those the more
soluble.
• If they fall apart into different number of
pieces you have to do the math.
The Common Ion Effect
• When the salt with the anion of a weak
acid is added to that acid,
• It reverses the dissociation of the acid.
• Lowers the percent dissociation of the
acid.
• The same principle applies to salts with
the cation of a weak base..
• The calculations are the same as last
chapter.
Solving Solubility with a Common Ion
For the salt AgI at 25C, Ksp = 1.5 x 10-16
What is its solubility in 0.05 M NaI?
AgI(s)  Ag+(aq) + I-(aq)
I
O
0.05
C
+x
0.05+x
E
x
0.05+x
1.5 x 10-16 = (x)(0.05+x)  (x)(0.05)
x = solubility of AgI in mol/L = 3.0 x 10-15 M
Precipitation and Qualitative Analysis
pH and solubility
• OH- can be a common ion.
• More soluble in acid.
• For other anions if they come from a weak
acid they are more soluble in a acidic
solution than in water.
• CaC2O4
Ca+2 + C2O4-2
• H+ + C2O4-2
HC2O4• Reduces C2O4-2 in acidic solution.
Precipitation
•
•
•
•
•
Ion Product, Q =[M+]a[Nm-]b
If Q>Ksp a precipitate forms.
If Q<Ksp No precipitate.
If Q = Ksp equilibrium.
A solution of 750.0 mL of 4.00 x 10-3M
Ce(NO3)3 is added to 300.0 mL of
2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp=
1.9 x 10-10M)precipitate and if so, what is
the concentration of the ions?
Selective Precipitations
• Used to separate mixtures of metal ions in
solutions.
• Add anions that will only precipitate certain
metals at a time.
• Used to purify mixtures.
• Often use H2S because in acidic solution
Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will
precipitate.
Selective Precipitation
• In Basic adding OH-solution S-2 will
increase so more soluble sulfides will
precipitate.
• Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3,
Al(OH)3
Selective precipitation
• Follow the steps first with insoluble
chlorides (Ag, Pb, Ba)
• Then sulfides in Acid.
• Then sulfides in base.
• Then insoluble carbonate (Ca, Ba, Mg)
• Alkali metals and NH4+ remain in solution.
Complex Ions
A Complex ion is a charged species composed
of:
1. A metallic cation
2. Ligands – Lewis bases that have a
lone electron pair that can form a
covalent bond with an empty orbital
belonging to the metallic cation
NH3, CN-, and H2O
are Common Ligands
N
H
H
H
-
C
N
O
H
H
The addition of each ligand has
its own equilibrium
• Usually the ligand is in large excess.
• And the individual K’s will be large so we
can treat them as if they go to equilibrium.
• The complex ion will be the biggest ion in
solution.
Coordination Number
 Coordination number refers to the
number of ligands attached to the cation
 2, 4, and 6 are the most common
coordination numbers
Coordination
Example(s)
number
2
Ag(NH3)2+
4
CoCl42-
6
Co(H2O)62+
Cu(NH3)42+
Ni(NH3)62+
Complex Ions and Solubility
AgCl(s)  Ag+ + Cl-
Ksp = 1.6 x 10-10
Ag+ + NH3  Ag(NH3)+
K1 = 1.6 x 10-10
Ag(NH3)+ NH3  Ag(NH3)2+
AgCl + 2NH3  Ag(NH3)2+ + Cl-
K2 = 1.6 x 10-10
K = KspK1K2

3 2

[ Ag ( NH ) ][Cl ]
K  2.8 x 10 
2
[ NH 3 ]
3