Chapter 18 — Advanced Aqueous Equilibria
1
More About Chemical Equilibria: Acid–
Base & Precipitation Reactions
Chapter 18
Principles of
Chemical Reactivity:
Other Aspects of
Aqueous Equilibria
Jeffrey Mack
California State University,
Sacramento
Stomach Acidity & Acid–Base
Reactions
The Common Ion Effect
• In the previous chapter, you examined the
behavior of weak acids and bases in terms of
equilibrium involving conjugate pairs.
• The pH of a solution was found via Ka or Kb.
• What would happen if you started with a
solution of acid that was mixed with a solution
of its conjugate base?
• The change of pH when a significant ammout
of conjugate base is present is an example of
the “Common Ion Effect”.
The Common Ion Effect
The Common Ion Effect
What is the effect on the pH of a 0.25M NH 3(aq)
solution when NH4Cl is added?
First let’s find the pH of a 0.25M NH3(aq)
Solution:
NH3 (aq)  H2O
NH4  (aq)  OH (aq)
NH4+ is an ion that is COMMON to the equilibrium.
Le Chatelier predicts that the equilibrium will shift to
the left to reduce the disturbance.
This results in a reduciton of the hydroxide ion
concentration, which will lower the pH.
Hint: NH4+ is an acid!
Initial
Change
Equilibrium
[NH3]
[NH4+]
[OH-]
0.25
x
0.025  x
0
+x
x
0
+x
x
Chapter 18 — Advanced Aqueous Equilibria
The Common Ion Effect
The Common Ion Effect
First let’s find the pH of a 0.25M NH3(aq)
Solution:
Initial
Change
Equilibrium
[NH3]
[NH4+]
[OH-]
0.25
x
0.025  x
0
+x
x
0
+x
x
K b  1.8 x 105 

2
First let’s find the pH of a 0.25M NH 3(aq) Solution:
[NH4  ][OH- ]
x2

[NH3 ]
0.25 - x
K b  1.8 x 105 
Assuming x is << 0.25, x = [OH ] =
OH   0.25  1.8  105  0.0021 M
-
2
[NH4 ][OH ]
x

[NH3 ]
0.25 - x
pOH = 2.67
pH = 14.00  2.67 = 11.33 for 0.25 M NH 3
The Common Ion Effect
The Common Ion Effect
What is the pH of a solution made by adding
equal volumes of 0.25M NH3(aq) and 0.10M
NH4Cl(aq)?
Since the solutions are mixed with one
another, the effect of dilution is cancelled out.
One can use the initial concentrations of each
species in the reaction without calculating new
molarities.
This also works with mole ratios.
The Common Ion Effect
[NH3]
[NH4+]
[OH-]
0.25
x
0.025  x
0.10
+x
0.10 + x
0
+x
x
K b  1.8 x 105 
NH4  (aq)  OH (aq)
NH3 (aq)  H2O
Initial
Change
Equilibrium
[NH3]
[NH4+]
[OH-]
0.25
x
0.025  x
0.10
+x
0.10 + x
0
+x
x
Since there is more ammonia than ammonium
present, the RXN will proceed to the right.
The Common Ion Effect
What is the pH of a solution made by adding equal
volumes of 0.25 M NH3(aq) and 0.10 M NH4Cl(aq)?
Initial
Change
Equilibrium
What is the pH of a solution made by adding equal
volumes of 0.25 M NH3(aq) and 0.10 M NH4Cl(aq)?
[NH4  ][OH- ]
(0.10  x)  x

[NH3 ]
0.25  x
What is the pH of a solution made by adding equal volumes of
0.25 M NH3(aq) and 0.10 M NH4Cl(aq)?
K b  1.8 x 105 
[NH4  ][OH- ]
(0.10  x)  x

[NH3 ]
0.25  x
Assuming x is << 0.25 or 0.10
x  OH  
0.25
 1.8  10 5  4.5  10 5 M
0.10
pOH = 4.35 pH = 9.65
The pH drops from11.33 due to the common ion!
Chapter 18 — Advanced Aqueous Equilibria
Controlling pH: Buffer Solutions
HCl is added to pure
water.
HCl is added to a
solution of a weak acid
H2PO4- and its
conjugate base HPO42-.
Controlling pH: Buffer Solutions
Consider the acetic acid / acetate buffer system.
• The ability for the acid to consume OH is seen
from the reverse of the base hydrolysis:
CH3CO2 (aq)  H2O(l)
CH3CO2H(aq)  OH (aq)
K b  5.6  1010
• Krev is >> 1, indicating that the reaction is product
1
favored.
K 
 1.8  109
rev
3
Controlling pH: Buffer Solutions
• A “Buffer Solution” is an example of the
common ion effect.
• From an acid/base standpoint, buffers are
solutions that resist changes to pH.
• A buffer solution requires two components
that do not react with one another:
1. An acid capable of consuming OH
2. The acid’s conjugate base capable of
consuming H3O+
Controlling pH: Buffer Solutions
Consider the acetic acid / acetate buffer
system.
• Similarly, the conjugate base (acetate) is
readily capable of consuming H3O+
• Krev is >> 1, indicating that the reaction is
product favored.
K rev 
Kb
• An hydroxide added will immediately react with
the acid so long as it is present.
1
 5.6  10 4
Ka
• An hydronium ion added will immediately
react wit the acid so long as it is present.
Buffer Solutions
Buffer Solutions
Problem:
Problem:
What is the pH of a buffer that has [CH 3CO2H] =
0.700 M and [CH3CO2] = 0.600 M?
What is the pH of a buffer that has [CH 3CO2H] =
0.700 M and [CH3CO2] = 0.600 M?
Since the concentration of acid is greater than the
base, equilibrium will move the reaction to the right.
CH3CO2H(aq)  H2O
CH3CO2 (aq)  H3O (aq)
Chapter 18 — Advanced Aqueous Equilibria
4
Buffer Solutions
Buffer Solutions
Problem:
Problem:
What is the pH of a buffer that has [CH 3CO2H] =
0.700 M and [CH3CO2] = 0.600 M?
What is the pH of a buffer that has [CH 3CO2H] =
0.700 M and [CH3CO2] = 0.600 M?
Initial
Change
Equilibrium
[CH3CO2H]
[CH3CO2]
[H3O+]
0.700
x
0.700  x
0.600
+x
0.600 + x
0
+x
x
K a  1.8  10 5 
[H3O ]  0.600
0.700
[H3O+] = 2.1 105
Assuming that x << 0.700 and 0.600, we find:
K a  1.8  10 5 
Buffer Solutions
The expression for calculating the [H +] of the buffer
reduces to:
[H3O ] 
Orig. conc. of CH3CO2H
 Ka
Orig. conc. of CH3CO2
[H3O ] 
pH = 4.68
[H3O ]  0.600
0.700
Buffer Solutions
Similarly for a basic solution the [OH ] of the buffer
reduces to:
[OH ] 
[Acid]
 Ka
[Conj. base]
The H3O+ concentration depends only Ka and the
ratio of acid to base.
Buffer Solutions: The HendersonHasselbalch Equation


[Acid]
 log  [H3O ] 
 Ka 
[Conj. base]




[Acid]

 log[H3O ]   log K a   log 

 [conj. Base] 
 conj. Base 
pH  pK a  log 


 Acid 

The result is known as the “Henderson-Hasselbalch”
equation.
The pH of a buffer can be adjusted by manipulating
the ratio of acid to base.
Orig. conc. of CH3CO2
 Kb
Orig. conc. of CH3CO2H
[OH ] 
[Base]
 Kb
[Conj. acid]
The OH concentration depends only Kb and the ratio
of base to 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
H+ (aq) + HCOO- (aq)
0.30
0.00
-x
+x
+x
0.30 - x
x
0.52 + x
pH = pKa + log
[HCOO-]
[HCOOH]
pH = 3.77 + log
[0.52]
= 4.01
[0.30]
0.30 – x  0.30
0.52 + x  0.52
HCOOH pKa = 3.77
0.52
24
Chapter 18 — Advanced Aqueous Equilibria
Preparing a Buffer
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• A pH of 4.30 corresponds to an [H 3O+] = 5.0105 M
Preparing a Buffer
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• A pH of 4.30 corresponds to an [H 3O+] = 5.0105 M
• First choose an acid with a pKa close to the desired
pH
• Next adjust the ratio of acid to conjugate base to
achieve the desired pH.
 conj. Base 
pH  pK a  log 


 Acid 

Preparing a Buffer
5
Preparing a Buffer
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• A pH of 4.30 corresponds to an [H 3O+] = 5.0105 M
• First choose an acid with a pKa close to the desired
pH
Preparing a Buffer
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• A pH of 4.30 corresponds to an [H 3O+] = 5.0105 M
• First choose an acid with a pKa close to the desired
pH
Possible Acids
HSO4 / SO24
CH3CO2H/ CH3CO2
HCN/ CN
Ka
1.2  102
1.8  105
4.0  1010
• Acetic acid is the best choice.
Preparing a Buffer
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• A pH of 4.30 corresponds to an [H 3O+] = 5.0105 M
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• A pH of 4.30 corresponds to an [H 3O+] = 5.0105 M
[Acid]
 Ka
[Conj. base]
[Acid]
5.00  10 5 
 1.8  10 5
[Conj. base]
[Acid]
2.78

[Conj. base]
1
[Acid]
2.78

[Conj. base]
1
[H3O ] 
• Therefore, if you start with 0.100 mol of acetate ion
then add 0.278 mol of acetic acid, will result in a
solution with a pH of 4.30.
Chapter 18 — Advanced Aqueous Equilibria
Preparing a Buffer
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• Therefore, if you start with 0.100 mol of acetate ion
then add 0.278 mol of acetic acid, will result in a
solution with a pH of 4.30.
• Since both species are in the same solution (the
same volume), the mole ratios are equal to the
concentration ratios!
Preparing a Buffer
Buffer prepared from
8.4 g NaHCO3
6
Preparing a Buffer
• Suppose you wish to prepare a buffer a solution at
pH of 4.30. How would you proceed?
• So, by adding 8.20 g of sodium acetate to 2780 ml
of a 0.100 M acetic acid solution, one would make a
buffer of pH 4.30.
82.03g
 8.20g NaCH3CO2
1mol
1L
103 mL
0.278 mols CH3CO2H 

 2780 mL
0.100mol
1L
0.100 mol NaCH3CO2 
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
a) 1.00 L of pure water
b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and
[CH3CO2] = 0.600 M (pH = 4.68)
weak acid
16.0 g Na2CO3
conjugate base
What is the pH?
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
a) 1.00 L of pure water
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and
[CH3CO2] = 0.600 M (pH = 4.68)
mL  L  mols H3O+  M(H3O+)  pH
1.00 mL 
1L
1.00 mols HCl 1 mol H3O
1



 0.00100 M H3O
103 mL
1L
1 mol HCl 1.00 L
pH   log(0.00100)  3.00
1. The acid added will immediately be consumed by
the acetate ion.
2. This in turn increases the acetic acid concentration.
3. The acetic acid then will react with water to
reestablish equilibrium.
Chapter 18 — Advanced Aqueous Equilibria
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and
[CH3CO2] = 0.600 M (pH = 4.68)
1. The acid added will immediately be consumed by
the acetate ion.
7
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to
b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and
[CH3CO2] = 0.600 M (pH = 4.68)
H3O (aq)  CH3CO2 (aq)  CH3CO2H(aq)
H3O (aq)  CH3CO2 (aq)  CH3CO2H(aq)
Before
Change
After
Adding an Acid to a Buffer
What is the pH when 1.00 mL of 1.00 M HCl is added to:
b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and
[CH3CO2] = 0.600 M (pH = 4.68)
2. This in turn increases the acetic acid
concentration.
3. The acetic acid then will react with water to
reestablish equilibrium.
[H3O ] 
[HOAc]
0.701
 Ka 
 (1.8  10 5 )
[OAc - ]
0.599
[H3O ]  2.1 10 5
pH  4.68
The pH does not change! The solution absorbs the added
acid. It is a buffer!
[CH3CO2H]
0.600
 0.00100
0.599
0.700
+ 0.00100
0.701
What is the pH when 1.00 mL of 1.00 M HCl is added to
b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and
[CH3CO2] = 0.600 M (pH = 4.68)
CH3CO2H(aq)  H2O(l)  H3O (aq)  CH3CO2 (aq)
Initial
Change
Equilibrium
[H3O ] 
What is the pH when 1.00 mL of 1.00 M HCl is added to
b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and
[CH3CO2] = 0.600 M (pH = 4.68)
[CH3CO2]
Adding an Acid to a Buffer
CH3CO2H(aq)  H2O(l)  H3O (aq)  CH3CO2 (aq)
Adding an Acid to a Buffer
[H3O+]
0.00100
 0.00100
0
[CH3CO2H]
[H3O+]
[CH3CO2]
0.701
x
0.700  x
0
+x
x
0.599
+x
0.599 + x
[HOAc]
0.701
 Ka 
 (1.8 x 10 -5 )
[OAc - ]
0.599
Commercial Buffers
• The solid acid and
conjugate base in the
packet are mixed with
water to give the specified
pH.
• Note that the quantity of
water does not affect the
pH of the buffer.
Chapter 18 — Advanced Aqueous Equilibria
8
Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3
(a) KF 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
43
44
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)
pH = pKa + log
start (moles)
end (moles)
[NH3]
[NH4+]
H+ (aq) + NH3 (aq)
pKa = 9.25
0.029
0.001
NH4+ (aq) + OH- (aq)
0.028
0.0
pH = 9.25 + log
[0.30]
= 9.17
[0.36]
0.024
H2O (l) + NH3 (aq)
0.025
final volume = 80.0 mL + 20.0 mL = 100 mL
[NH4+] =
45
Acid–Base Titrations
Adding NaOH from the buret to acetic acid in
the flask, a weak acid. In the beginning the
pH increases very slowly.
0.028
0.025
[NH3] =
0.10
0.10
[0.25]
= 9.20
[0.28]
46
pH = 9.25 + log
Acid–Base Titrations
Additional NaOH is added. pH rises as
equivalence point is approached.
Chapter 18 — Advanced Aqueous Equilibria
Acid–Base Titrations
9
Titration of a Strong Acid with a
Strong Base
Additional NaOH is added. pH increases and
then levels off as NaOH is added beyond the
equivalence point.
Titration of a Strong Acid with a
Strong Base
The reaction of a strong acid and strong base
produces a salt and water.
The Net Ionic Equation is:
H3O (aq)  OH (aq)  2H2O(l)
Titration of a Weak Acid with a
Strong Base
Problem:
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
At the equivalence point, the moles of base added
equal the moles of acid titrated.


H3O   OH 


H3O   OH   K w
H3O   H3O   K w

14
H3O   K w  1.00  10
pH  7
Titration of a Weak Acid with a
Strong Base
At the equivalence
point, all of the acid is
converted to its
conjugate base.
The conjugate base
will then react with
water to reestablish
equilibrium.
The pH can be
determined from Kb.
Titration of a Weak Acid with a
Strong Base
HBz (aq) + OH(aq)  Bz(aq) + H2O(l)
Equivalence Point
(moles base = moles acid)
C6H5CO2H = HBz
Benzoate ion = Bz-
Chapter 18 — Advanced Aqueous Equilibria
Titration of a Weak Acid with a
Strong Base
10
Titration of a Weak Acid with a
Strong Base
Problem:
Problem:
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Solution:
At the equivalence point, all of the HBz is converted
to Bz- by the strong base.
The conjugate base of a weak acid (Bz-) will
hydrolyze to reform the weak acid (Kb). The pH will be
>7
This will yield the [H3O+] and pH.
Volume of OH- added to the eq. point:
Titration of a Weak Acid with a
Strong Base
HBz (aq) + OH(aq)  Bz(aq) + H2O(l)
100.0mL 
1L
0.025mol HBz 1mol OH
1L
103 mL




 25mL
103 mL
1L
1molHbz 0.100molOH
1L
The new total volume of the solution is 125 mL
Titration of a Weak Acid with a
Strong Base
Problem:
Problem:
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
Moles of OH- & Bz- at the eq. point:
HBz (aq) + OH(aq)  Bz(aq) + H2O(l)
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
[OH-] at the eq. point:
100.0 mL 
1L
0.025 mol HBz 1 mol OH 1 mol Bz 



 0.0025 mols Bz 
103 mL
1L
1 mol Hbz 1 mol OH
The concentration of Bz- at the eq. point is:
0.0025 mols Bz  103 mL

 0.020 M Bz 
125 mL
1L
Titration of a Weak Acid with a
Strong Base
Bz  (aq)  H2O(l)
Initial
Change
Equilibrium
HBz (aq)  OH (aq) K b  1.6  10 10
[Bz-]
[HBz]
[OH-]
0.020
-x
0.020 - x
0
+x
x
0
+x
x
Titration of a Weak Acid with a
Strong Base
Problem:
Problem:
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
[OH-] at the eq. point:
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
[OH-] at the eq. point:
Initial
Change
Equilibrium
[Bz-]
[HBz]
[OH-]
0.020
-x
0.020 - x
0
+x
x
0
+x
x
K b  1.6 x 1010 
x2
0.020  x
K b  1.6  10 10 
x2
0.020  x
Assuming that x << 0.020,
X = [OH ] =1.8  106
pOH = 5.75
pH = 8.25
Chapter 18 — Advanced Aqueous Equilibria
Titration of a Weak Acid with a
Strong Base
Conclusion: At the equivalence
point of the titration, unlike the
titration of a strong acid and
strong base, the pH is > 7.
This is due to the production of
the conjugate base of a week
acid.
11
Titration of a Weak Acid with a
Strong Base
Conclusion: What would the pH
equal at the half-way point of the
titration?
Equivalence point
pH = 8.25
Hint: Only ½ of the moles of
weak acid have been converted
to its conjugate base!
Half-way point
pH = ??
Titration of a Weak Acid with a
Strong Base
Titration of a Weak Acid with a
Strong Base
Problem:
Problem:
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
At the half-way point, moles of Hbz and Bz- are
equal.
This is a BUFFER SOLUTION!
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
At the half-way point, moles of Hbz and Bz- are
equal.
This is a BUFFER SOLUTION!
 conj. Base 
pH  pK a  log 


 Acid 

Titration of a Weak Acid with a
Strong Base
Problem:
100. mL of a 0.025 M solution of benzoic acid is
titrated with 0.100 M NaOH to the equivalence point.
What is the pH of the final solution?
At the half-way point, moles of Hbz and Bz- are
equal.
This is a BUFFER SOLUTION!
 conj. Base 
pH  pK a  log 
  pK a  log(1)  pK a  0  pK a

 Acid 



pH   log 6.3  10 5  4.20
Acetic Acid Titrated with NaOH
Chapter 18 — Advanced Aqueous Equilibria
Titration of a Weak Polyprotic Acid
with a Strong Base
In the case of a
titration of a weak
polyprotic acid (HnA)
there are “ n”
equivalence points.
In the case of the
diprotic oxalic acid,
(H2C2O4) there are
two equivalence
points.
Titration of a Weak Base with a
Strong Acid
12
Titration of a Weak Polyprotic Acid
with a Strong Base
The titration of a polyprotic weak acid follows the same
process a monoprotic weak acid.
As the acid is titrated, buffering occurs until the last eq. point is
reached.
K a(1)

 HA  (aq)  H2O(l)
H2 A(aq)  OH (aq) 

The pH is relative to the amounts of conjugate acids and
bases.
K a( 2 )

 A 2  (aq)  H2O(l)
HA  (aq)  OH (aq) 

At the second eq. point all of the acid has been converted to
A2-, pH is determined by Kb.
K
Kb  w
K a(2)
pH Indicators for Acid–Base
Titrations
In the case of a
titration of a weak
base, the process
follows that of a weak
acid in reverse.
There exists a region
of buffering followed
by a rapid drop in pH
at the eq. point.
pH Indicators for Acid–Base
Titrations
• An acid/base indicator is a substance that changes
color at a specific pH.
• HInd (acid) has another color than Ind (base)
• These are usually organic compounds that have
conjugated pi-bonds, often they are dyes or
compounds that occur in nature such as red
cabbage pigment or tannins in tea.
• Care must be taken when choosing an appropriate
indicator so that the color change (end point of the
titration) is close to the steep portion of the titration
curve where the equivalence point is found.
pH Indicators for Acid–Base
Titrations
Chapter 18 — Advanced Aqueous Equilibria
Natural Indicators: Red Rose Extract
in Methanol
Neutral
pH
pH<<7
pH >7
Buffer
<7
pH >>7
Solubility of Salts
13
Solubility of Salts
• Prior to this chapter, exchange
reactions which formed ionic salts
were governed by the solubility rules.
• A compound was either soluble,
insoluble or slightly soluble.
• So how do we differentiate between
these?
• The answer lies in equilibrium.
• It turns out that equilibrium governs
the solubility of inorganic salts.
Lead(II) iodide
Solubility of Salts
The extent of solubility can be measured by the
equilibrium process of the salt’s ion
concentrations in solution, Ksp.
Ksp is called the solubility constant for an ionic
compound. It is the product of the ion’s
solubilities.
For the salt:
AxBy(s)  xAy+(aq) + yBx-(aq)
Ksp = [Ay+]x[Bx-]y
Solubility of Salts
Consider the solubility of a salt MX:
If MX is added to water then:
H2O(l)
MX(s) 
 M (aq)  X (aq)
K sp  [M ][X ]
Generally speaking,
If Ksp >> 1 then MX is considered to be soluble
If Ksp << 1 then MX is considered to be insoluble
If Ksp  1 then MX is slightly soluble
Analysis of Silver Group
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
• All salts formed in this
experiment are said to
be INSOLUBLE.
• They form when
mixing moderately
concentrated solutions
of the metal ion with
chloride ions.
Chapter 18 — Advanced Aqueous Equilibria
Analysis of Silver Group
14
Analysis of Silver Group
Ag+ Pb2+ Hg22+
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
AgCl PbCl2 Hg2Cl2
• Although all salts formed in this experiment
are said to be insoluble, they do dissolve to
some SLIGHT extent.
AgCl(s) e Ag+(aq)  Cl-(aq)
• When equilibrium has been established, no
more AgCl dissolves and the solution is
SATURATED.
Analysis of Silver Group
AgCl(s)  Ag+(aq) + Cl-(aq)
When solution is SATURATED, expt. shows
that [Ag+] = 1.67 x 10-5 M.
This is equivalent to the SOLUBILITY of AgCl.
What is [Cl-]?
[Cl ] = [Ag+] = 1.67 x 10-5 M
Analysis of Silver Group
Ag+ Pb2+ Hg22+
Ag+ Pb2+ Hg22+
AgCl PbCl2 Hg2Cl2
AgCl PbCl2 Hg2Cl2
AgCl(s)  Ag+(aq) + Cl-(aq)
Saturated solution has
[Ag+] = [Cl-] = 1.67  10-5 M
Use this to calculate Kc
Kc = [Ag+] [Cl-]
= (1.67  10-5)(1.67  10-5)
= 2.79  10-10
Lead(II) Chloride
PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)
Ksp = 1.9  10-5 = [Pb2+][Cl–]2
AgCl(s) 
+ Cl-(aq)
Kc = [Ag+] [Cl-] = 2.79  10-10
Because this is the product of “solubilities”,
we call it:
Ksp = solubility product constant
Ag+(aq)
Relating Solubility & Ksp
Problem:
The solubility of lead (II) iodide is found to be
0.00130M. What is the Ksp for PbI2?
Chapter 18 — Advanced Aqueous Equilibria
Relating Solubility & Ksp
15
Relating Solubility & Ksp
Problem:
Problem:
The solubility of lead (II) iodide is found to be
0.00130M. What is the Ksp for PbI2?
The solubility of lead (II) iodide is found to be
0.00130M. What is the Ksp for PbI2?
Recall that lead (II) iodide dissociates via:
From the reaction stoichiometry,

 Pb2  (aq)  2I (aq)
PbI2 (s) 

K sp
K sp  [Pb2 ][I ]2
sp

 Pb2  (aq)  2I (aq)
PbI2 (s) 

K
[Pb2+] = 0.00130M & [I] = 2  0.00130M = 0.00260M
Relating Solubility & Ksp
Relating Solubility & Ksp
Problem:
The solubility of lead (II) iodide is found to be
0.00130M. What is the Ksp for PbI2?
Entering these values into the Ksp expression,
Ksp for MgF2 = 5.2 1011. Calculate the solubility in:
K sp  [Mg2 ][F ]2
(a) moles/L & (b) in g/L
K sp  [Pb2 ][I ]2
K sp   0.00130    2  0.00130 
2
K sp  4   0.00130   8.79  109
3
For PbI2, Ksp = 4  (solubility)3
Relating Solubility & Ksp
Relating Solubility & Ksp
Ksp for MgF2 = 5.2 1011. Calculate the solubility in:
K sp  [Mg2 ][F ]2
(a) moles/L & (b) in g/L
(a) The solubility of the salt is governed by equilibrium
so let’s set up an ICE table:
Ksp for MgF2 = 5.2 1011. Calculate the solubility in:
K sp  [Mg2 ][F ]2
(a) moles/L & (b) in g/L
(a) Entering the values into the Ksp expression:
K sp  Mg2  I 
2
K sp   x  2x   4x 3
2
Initial
Change
Equilibrium
[MgF2(s)]
[Mg2+]
[F]
-
0
+x
x
0
+ 2x
2x
x
3
K sp
4

3
5.2  10 11
 2.4  10 4
4
The solubility of MgF2 = 2.4  104 mols/L
Chapter 18 — Advanced Aqueous Equilibria
Relating Solubility & Ksp
Ksp for MgF2 = 5.2 1011. Calculate the solubility in:
K sp  [Mg2 ][F ]2
(a) moles/L & (b) in g/L
(b) the solubility of the salt in g/L is found using the
formula weight:
16
Relating Solubility & Ksp
What is the maximum [Cl] in solution with 0.010 M
Hg22+ without forming Hg2Cl2 (s)?
Hg2Cl2 (s)
K sp
Hg22 (aq)  2Cl (aq)
 1.1  1018
2.4  104 mols MgF2 62.3g MgF2
MgF2

 0.015g
L
L
mol
Relating Solubility & Ksp
What is the maximum [Cl] in solution with 0.010 M
Hg22+ without forming Hg2Cl2 (s)?
Hg2Cl2 (s)
K sp
Hg22 (aq)  2Cl (aq)
 1.1  1018
Precipitation will initiate when the product of the
concentrations exceeds the Ksp.
Relating Solubility & Ksp
What is the maximum [Cl] in solution with 0.010 M
Hg22+ without forming Hg2Cl2 (s)?
The maximum chloride concentration can be found
from the Ksp expression.
Hg2Cl2 (s)
K sp
Hg22 (aq)  2Cl (aq)
 1.4  10 18 = Hg22  Cl 
2
K sp
1.4  10 18
Cl  

 1.2  10 8
0.010
Hg22 
Solubility & the Common Ion Effect
Adding an ion “common” to an equilibrium
causes the equilibrium to shift towards
reactants according to Le Chatelier’s principle.
Common Ion Effect
PbCl2 (s)
Pb2 (aq)  2Cl (aq)
K sp  1.7  105
Chapter 18 — Advanced Aqueous Equilibria
Solubility & the Common Ion Effect
What is the the solubility of BaSO4(s) in
(a) pure water and (b) in 0.010 M Ba(NO3)2?
Ksp for BaSO4 = 1.11010
17
Solubility & the Common Ion Effect
What is the the solubility of BaSO4(s) in
(a) pure water and (b) in 0.010 M Ba(NO3)2?
Ksp for BaSO4 = 1.11010
(a)
Ba2 (aq)  SO24 (aq)
BaSO4 (s)
K sp  Ba2  SO24   x 2
x  K sp  1.0  10 5 M
Solubility & the Common Ion Effect
Solubility & the Common Ion Effect
What is the the solubility of BaSO4(s) in
(a) pure water and (b) in 0.010 M Ba(NO3)2?
Ksp for BaSO4 = 1.11010
What is the the solubility of BaSO4(s) in
(a) pure water and (b) in 0.010 M Ba(NO3)2?
Ksp for BaSO4 = 1.11010
(b)
(b)
[BaSO4(s)]
[Ba2+]
[SO42]
Initial
-
0.010
0
Change
Equilibrium
-
+x
0.010 + x
+x
x
[BaSO4(s)]
[Ba2+]
Initial
-
0.010
[SO42]
0
Change
Equilibrium
-
+x
0.010 + x
+x
x
K sp  Ba2  SO22 
K sp  0.010  x  x 
Solubility & the Common Ion Effect
What is the the solubility of BaSO4(s) in
(a) pure water and (b) in 0.010 M Ba(NO3)2?
Ksp for BaSO4 = 1.11010
Solubility & the Common Ion Effect
What is the the solubility of BaSO4(s) in
(a) pure water and (b) in 0.010 M Ba(NO3)2?
Ksp for BaSO4 = 1.11010
(b)
K sp  Ba2  SO22 
K sp  0.010  x  x 
(b)
K sp  Ba2  SO22 
K sp  0.010  x  x 
Since x << 0.010,
K sp  0.010 x 
Since x << 0.010,
K sp  0.010 x 
K sp
1.1 10 10

0.010
0.010
x  1.1 10 8 M
x
K sp
1.1 10 10

0.010
0.010
x  1.1 10 8 M
x
Solubility is
significantly
decreased by
the presence
of a common
ion.
Chapter 18 — Advanced Aqueous Equilibria
Effect of Basic Salts on Solubility
18
Effect of Basic Salts on Solubility
Some anions of precipitates are the conjugate
bases of weak acids.
In solution these anions can hydrolyze to form the
weak acid.

X (aq)  H2O(l)

HX(aq)  OH (aq)
Add a base:
X (aq)  H2O(l)
Addition of additional base can shift equilibrium to
the right thus reducing the concentration of X.
This in turn would increase the solubility of the
ppt.
Effect of Basic Salts on Solubility
Decrease the
concentration of
HX
Effect of Basic Salts on Solubility
Eq. Shifts to the
right
X (aq)  H2O(l)
HX(aq)  OH (aq)
HX(aq)  OH (aq)
Eq. Shifts to the
right
X (aq)  H2O(l)
HX(aq)  OH (aq)
The
concentration of
X drops
Effect of Basic Salts on Solubility
Ksp & the Reaction Quotient
Relating Q to Ksp
Eq. Shifts to the
right
MX(s)
M (aq)  X (aq)
The solubility of MX increases!
Q = Ksp
The solution is saturated
Q < Ksp
The solution is not saturated
Q = Ksp
The solution is over saturated,
precipitation will occur
Chapter 18 — Advanced Aqueous Equilibria
Ksp & the Reaction Quotient
Suppose you have a solution that is
1.5 106 M Mg2+.
Enough NaOH(s) is added to give a
[OH] = 1.0 106 M.
Will a precipitate form?
19
Ksp & the Reaction Quotient
Suppose you have a solution that is
1.5 106 M Mg2+.
Enough NaOH(s) is added to give a
[OH] = 1.0 106 M.
Will a precipitate form?
Solution:
Calculate “Q” and compare to Ksp.
Ksp & the Reaction Quotient
Ksp & the Reaction Quotient
1.5 106 M Mg2+
1.5 106 M Mg2+
Solution:
&
[OH] = 1.0 104 M.
Calculate “Q” and compare to Ksp.
Mg(OH)2 (s)
Mg2 (aq)  2OH (aq)
Q  Mg2  OH 
Solution:
&
Calculate “Q” and compare to Ksp.
Mg(OH)2 (s)
[OH] = 1.0 104 M.
 K sp  5.6  10
2
Ksp & the Reaction Quotient
What concentration of hydroxide ion will precipitate a
1.5 106 M Mg2+ solution.
Mg(OH)2 (s)
Q  1.5  10
Mg2 (aq)  2OH (aq)
Q  1.5  10 14
Calculate “Q” and compare to Ksp.
14
[OH] = 1.0 104 M.
Q  1.5  10 6  1.0  10 4 
2
Ksp & the Reaction Quotient
1.5 106 M Mg2+
Solution:
&
Mg2 (aq)  2OH (aq)
12
K sp  1.5  106   x 
2
Since Q < Ksp, no precipitation will occur.
Chapter 18 — Advanced Aqueous Equilibria
Ksp & the Reaction Quotient
What concentration of hydroxide ion will precipitate a
1.5 106 M Mg2+ solution.
x  OH  
K sp
5.6  1012

6
1.5  10
1.5  10 6

3
OH   1.9  10
20
Equilibria & Complex Ions
Metal ions exist in solution as complex ions.
A complex ion involves a metal ion bound to
molecules or ions called “ligands”.
Ligands are Lewis bases that form “coordinate
covalent bonds” with the metal.
Examples are:
Ni(H2O)6 2
or
Cu(NH3 )2
Equilibria & Complex Ions
Equilibria & Complex Ions
The equilibrium constants for complex ion are
very product favored:
Cu2+ (aq) + 4 H2O(l)
[Cu(NH3 )4 ]2+
K f  2.1 1013
Kf is known as the formation equilibrium
constant.
Equilibria & Complex Ions
The extent of dissociation of a complex ions is
given by the “dissociation constant”, KD
[Cu(NH3 )4 ]2+
KD 
Dissolving Precipitates by forming
Complex Ions
The presence of a ligand dramatically affect
the solubility of a precipitate:
Cu2+ (aq) + 4 H2O(l)
1
1

 4.8  10 14
K f 2.1 1013
AgCl(s)  2 NH3 (aq)
 Ag(NH3 )2   Cl (aq)
Chapter 18 — Advanced Aqueous Equilibria
Solubility of Complex Ions
Solubility of Complex Ions
The presence of a ligand dramatically affect
the solubility of a precipitate:
AgCl(s)  2NH3 (aq)
 Ag(NH3 )2 +
21
 Cl (aq)
The presence of a ligand dramatically affect
the solubility of a precipitate:
 Ag(NH3 )2 +
AgCl(s)  2NH3 (aq)
 Cl (aq)

 Ag+ (aq)  Cl (aq)
AgCl(s) 

K sp
f

  Ag(NH3 )2 
Ag (aq)  2NH3 (aq) 
K
Solubility of Complex Ions
Solubility of Complex Ions
The presence of a ligand dramatically affect
the solubility of a precipitate:
AgCl(s)  2NH3 (aq)
 Ag(NH3 )2 +
+
 Cl (aq)

 Ag+ (aq)  Cl (aq)
AgCl(s) 

K sp
Kf

  Ag(NH3 )2 +
Ag (aq)  2NH3 (aq) 
What is the solubility of AgCl(s) in grams per liter in a
0.010M NH3(aq) solution:
AgCl(s)  2NH3 (aq)
 Ag(NH3 )2 +
 Cl (aq)
 Ag(NH3 )2   Cl 

K net  
 2.9  10 3
NH3 2
+
K net  K f  K sp  1.6  107   1.8  10 10 
 Ag(NH3 )2 +  Cl 

K net  
 2.9  10 3
NH3 2
Solubility of Complex Ions
Solubility of Complex Ions
What is the solubility of AgCl(s) in grams per liter in a
0.010M NH3(aq) solution:
AgCl(s)  2NH3 (aq)
K net
 Ag(NH3 )2 +
Change
Equilibrium
[[Ag(NH3)2]+]
0
0
- 2x
0.010 – 2x
+x
x
+x
x
Initial
Change
Equilibrium
[NH3]
[[Ag(NH3)2]+]
[Cl]
0.010
0
0
- 2x
0.010 – 2x
+x
x
+x
x
[Cl]
[NH3]
0.010
 Cl (aq)
 Ag(NH3 )2 +  Cl 


 2.9  10 3
NH3 2
Initial
What is the solubility of AgCl(s) in grams per liter in a
0.010M NH3(aq) solution:
K net 
x2
 2.9  103
0.010  2x
Chapter 18 — Advanced Aqueous Equilibria
Solubility of Complex Ions
What is the solubility of AgCl(s) in grams per liter in a
0.010M NH3(aq) solution:
K net 
x2
 2.9  103
0.010  2x
3
x  5.8  10  2.9  10
2
5
0
22
Solubility of Complex Ions
What is the solubility of AgCl(s) in grams per liter in a
0.010M NH3(aq) solution:
AgCl(s)  2NH3 (aq)
1L
 Ag(NH3 )2 +
 Cl (aq)
0.0032 mol Cl 1 mol AgCl 143.32g AgCl


 0.046 g
1L
1 mol Cl
1 mol AgCl
Solving using the quadratic equation:
X = 0.0032M = [Cl]
Separating Metal Ions Cu2+, Ag+,
Pb2+
Ksp Values
AgCl
1.8 x 10-10
PbCl2
1.7 x 10-5
PbCrO4 1.8 x 10-14
In pure water, the solubility of AgCl is only
0.0019 g/L