Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 17
Additional Aspects of
Aqueous Equilibria
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Overview
Application of equilibrium:
• common ion effect
• buffers (application of common ion
effect to acid-base situations)
• titrations, especially acid-base
• solubility product (Ksp)
• complex ions
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
• Consider a solution of acetic acid:
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO−(aq)
• If acetate ion is added to the solution,
Le Châtelier says the equilibrium will
shift to the left.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
“The extent of ionization of a weak
electrolyte is decreased by adding to the
solution a strong electrolyte that has an ion
in common with the weak electrolyte.”
Use the same principle to solve common ion
problems – include the initial concentration
of the common ion in the RICE table.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.1 (p. 720)
What is the pH of a solution made by
adding 0.30 mol of acetic acid
(HC2H3O2) and 0.30 mol of sodium
acetate (NaC2H3O2) to enough water to
make 1.0 L of solution?
(4.74)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.1 (p. 720)
Dissociation of sodium acetate:
NaCH3COO(aq)
totally dissociates

Na+(aq)
no effect on pH
+ CH3COO-(aq)
+0.30 M
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.1 (p. 720)
What is the pH of a 0.30 M solution of
acetic acid? (i.e. without the added
sodium acetate)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.1 (p. 720)
Calculate the pH of a solution
containing 0.085 M nitrous acid (HNO2;
Ka = 4.5 x 10-4) and 0.10 M potassium
nitrite (KNO2).
(3.42)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.2 (p. 722)
Calculate the fluoride ion concentration and pH
of a solution that is 0.20 M in HF and 0.10 M in
HCl. (Note: common ion is H3O+)
Ka for HF is 6.8  10−4.
[H3O+] [F−]
Ka =
= 6.8  10-4
[HF]
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect

HCl(aq) + H2O(l)
totally dissociates
H3O+(aq) + Cl- (aq)
0.10 M
no effect
H3O+(aq) + F−(aq)
HF(aq) + H2O(l)
Because HCl, a strong acid, is also present,
the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M
[H3O+], M
[F−], M
Initially
0.20
0.10
0
Change
−x
+x
+x
0.20 − x  0.20
0.10 + x  0.10
x
At Equilibrium
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
6.8 
10−4
(0.10) (x)
=
(0.20)
(0.20) (6.8  10−4)
=x
(0.10)
1.4  10−3 = x
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
• Therefore, [F−] = x = 1.4  10−3
[H3O+] = 0.10 + x = 0.10 + 1.4  10−3 = 0.10 M
• So,
pH = −log (0.10)
pH = 1.00
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice 17.2
Calculate the formate ion concentration
and pH of a solution that is 0.050 M in
formic acid (HCOOH; Ka = 1.8 x 10-4)
and 0.10 M in HNO3.
([HCOO-] = 9.0 x 10-5 M; pH = 1.00)
(for HW)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffers
• Buffers are solutions
of a weak conjugate
acid-base pair.
• They are particularly
resistant to pH
changes, even when
strong acid or base is
added.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffers
• Consists of a mixture of a weak acid
and its conjugate base
Contains both an acidic species (to
neutralize OH-) and a basic species (to
neutralize H+)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffers
• The most effective buffers contain equal
amounts of the weak acid and its
conjugate base
• Add the salt of the conjugate base to
make [HX] and [X-] equal
 moves the equilibrium to the left
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffers
If a small amount of OH- is added to an equimolar
solution of HF in NaF, for example, the HF reacts
with the OH− to make F− and water.
The [HX]/[X-] ratio remains more or less constant 
no significant change in pH.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffers
Similarly, if acid is added, the F− reacts with it to form
HF and water.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffer Calculations
Consider the equilibrium constant
expression for the dissociation of a
generic acid, HA:
H3O+ (H+) + A−
HA + H2O
Ka =
[H+] [A−]
[HA]
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffer Calculations
Rearranging slightly, this becomes
Ka =
−]
[A
[H+]
[HA]
Taking the negative log of both sides, we get
−log Ka = −log [H+] + −log
pKa
pH
[A−]
[HA]
base
acid
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffer Calculations
• So
[base]
pKa = pH − log
[acid]
• Rearranging, this becomes
[base]
pH = pKa + log
[acid]
• This is the Henderson–Hasselbalch equation.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffer Capacity and pH Range
• Buffer capacity = the amount of acid or
base that can be neutralized before
there is a significant change in pH
• depends on [buffer components]
[conjugate acid-base pair]
 buffer capacity
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
pH Range
• The pH range is the range of pH values
over which a buffer system works
effectively.
• It is best to choose an acid with a pKa
close to the desired pH.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.3 (p. 725)
What is the pH of a buffer that is 0.12 M in lactic acid,
CH3CH(OH)COOH, and 0.10 M in sodium lactate?
Ka for lactic acid is 1.4  10−4.
(3.77)
Note: You may solve this problem using either the
standard equilibrium method, with RICE table, or the
Henderson-Hasselbalch Equation.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.3
pH = ?
[HC2H3O2] = 0.12 M
[C2H3O2-] = 0.10 M
Ka = 1.4 x 10-4
from NaC3H5O3
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Henderson–Hasselbalch Equation
[base]
pH = pKa + log
[acid]
pH = −log (1.4 
10−4)
(0.10)
+ log (0.12)
pH = 3.85 + (−0.08)
pH = 3.77
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.4 (p. 726)
How many moles of NH4Cl must be
added to 2.0 L of 0.10 M NH3 to form a
buffer whose pH is 9.00?
(Assume that the addition of NH4Cl
does not change the volume of the
solution.)
(0.36 mol)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.4
Calculate the concentration of sodium
benzoate that must be present in a
0.20 M solution of benzoic acid
(HC7H5O2) to produce a pH of 4.00.
(0.13 M)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
When Strong Acids or Bases Are
Added to a Buffer…
…it is safe to assume that all of the strong acid
or base is consumed in the reaction.
1st RICE table
or stoich calculation
2nd RICE table
or H-H calculation
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Addition of Strong Acid or Base to a
Buffer
1. Determine how the neutralization
reaction affects the amounts of
the weak acid and its conjugate
base in solution.
2. Use the Henderson–Hasselbalch
equation to determine the new
pH of the solution (or use a RICE
table)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers –
Sample Exercise 17.5 (p. 728)
A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol
NaC2H3O2 to enough water to make 1.00 L of solution. The pH
of the buffer is 4.74. (We have done this calculation before.)
a) Calculate the pH of this solution after 0.020 mol of NaOH is
added.
(4.80)
b) Calculate the pH of a 0.020 M NaOH solution for
comparison. (no buffer).
(12.30)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
Using the H-H equation:
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8  10−5) = 4.74
i.e. Ka = [H+][C2H3O2−] =>
[HC2H3O2]
[H+] = (1.8 x 10-5)(0.300)
(0.300)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
Stoich:
The 0.020 mol NaOH will react with 0.020 mol of the
acetic acid:
HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l)
HC2H3O2
OH−
C 2 H 3 O2 −
Before reaction
0.300 mol
0.020 mol
0.300 mol
After reaction
0.280 mol
0.000 mol
0.320 mol
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to
calculate the new pH:
(0.320)
pH = 4.74 + log
(0.280)
pH = 4.74 + 0.06
pH = 4.80
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
Or use a second RICE table
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.5 (cont’d)
and, for comparison,
b) calculate the pH that would result if 0.020
mol of NaOH were added to 1.00 L of pure
water (neglect any volume changes). Note:
strong base, \ simple calculation.
(12.30)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.5
Determine
a) the pH of the original buffer described
in Sample Exercise 17.5 after the
addition of 0.020 mol HCl, and
(4.68)
b) the pH of the solution that would result
from the addition of 0.020 mol HCl to
1.00 L of pure water.
(1.70)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration
In this technique a
known concentration of
base (or acid) is slowly
added to a solution of
acid (or base).
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration
A pH meter or
indicators are used to
determine when the
solution has reached
the equivalence point,
at which the
stoichiometric amount
of acid equals that of
base.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration Curve
= a plot of pH versus volume during a
titration
• Titration labs – early April
– Choose appropriate indicators
– Standardize an NaOH solution
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a
Strong Base
Four regions of the titration curve:
1. Initial pH – given by the strong acid solution
(pH < 7)
2. Between initial pH and equivalence point
(pH still < 7)
3. At the equivalence point: pH = 7.00
4. After the equivalence point – given by the amount of
excess base (pH > 7)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a
Strong Base
From the start of the
titration to near the
equivalence point,
the pH goes up
slowly.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a
Strong Base
Just before (and
after) the equivalence
point, the pH
increases rapidly.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a
Strong Base
At the equivalence
point, moles acid =
moles base, and the
solution contains
only water and the
salt from the cation
of the base and the
anion of the acid.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a
Strong Base
As more base is
added, the increase
in pH again levels
off.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Seven steps to solving titration
problems:
Analyze the problem, then:
1. Determine # moles of each reactant.
2. Determine # moles of all species after reaction.
3. Calculate new volume after reaction.
4. Determine molarities of all species (combine Steps 2 & 3).
5. Equilibrium calculation – RICE table. *
6. Equilibrium calculation, substituting results from Step 5 into
Ka. *
7. [H+]  pH. May mean [OH-]  pOH  14.00 – pOH = pH.
*not required for strong acid-strong base titrations
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.6 (p. 731)
Calculate the pH when the following
quantities of 0.100 M NaOH solution
have been added to 50.0 mL of 0.100 M
HCl solution:
a) 49.0 mL
( 3.00)
b) 51.0 mL
(11.00)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.6
Calculate the pH when the following
quantities of 0.10 M HNO3 have been
added to 25.0 mL of 0.10 M KOH
solution:
a) 24.9 mL
(10.30)
B) 25.1 mL
( 3.70)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
How to perform and analyze a titration
curve
e.g. strong base added to strong acid
• Equivalence point is at pH = 7.00
• Choose an indicator that changes color near
pH = 7.00, e.g. phenolphthalein (colorless in
acid, fuschia in base)
• End point = the volume of the titrant at the
point when the indicator changes color
• Titration error = difference between
equivalence point and end point
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Strong Base-Strong Acid Titration
Note that the pH begins
high then decreases
as acid is added.
After the equivalence
point, the pH is given
by the strong acid
in excess.
 pH < 7
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Weak Acid with a
Strong Base
• Unlike in the previous
case, the conjugate
base of the acid
affects the pH when it
is formed.
• At the equivalence
point the pH is >7.
• Phenolphthalein is
commonly used as an
indicator in these
titrations.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Weak Acid-Strong Base Titration
Four regions of the titration curve:
1. Initial pH: from equilibrium
calculation of weak acid
2. Buffer: neutralization  weak acid
and its conjugate base:
pH does not change very much
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Weak Acid – Strong Base Titration
3. equivalence point: pH depends on
[conjugate base] and equilibrium
calculation: pH > 7
4. after the equivalence point:
pH depends on strong base: pH > 7
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Weak Acid with a
Strong Base
At each point below the equivalence point, the
pH of the solution during titration is determined
from the amounts of the acid and its conjugate
base present at that particular time.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.7 (p. 735)
Calculate the pH of the solution formed
when 45.0 mL of 0.100 M NaOH
solution is added to 50.0 mL of 0.100 M
HC2H3O2. (Ka = 1.8 x 10-5)
(2.0 x 10-6 M)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.7
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.7
a) Calculate the pH in the solution formed
by adding 10.0 mL of 0.050 M NaOH
to 40.0 mL of 0.0250 M benzoic acid
(HC7H5O2, Ka = 6.3 x 10-5).
(4.20)
b) Calculate the pH in the solution formed
by adding 10.0 mL of 0.100 M HCl to
20.0 mL of 0.100 M NH3. (9.26)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.8 (p. 735)
Calculate the pH at the equivalence
point in the titration of 50.0 mL of
0.100 M HC2H3O2 with 0.100 M NaOH.
(8.72)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.8
Calculate the pH at the equivalence point
when
a) 40.0 mL of 0.025 M benzoic acid
(HC7H5O2, Ka = 6.3 x 10-5) is titrated
with 0.050 M NaOH
(8.21)
b) 40.0 mL of 0.100 M NH3 is titrated with
0.100 HCl
(5.28)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Weak Acid with a
Strong Base
With weaker acids,
the initial pH is
higher and pH
changes near the
equivalence point
are more subtle.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Strong acid/Weak acid Titrations
• pH begins at 7, gradual
as base is added
• pH dramatically near
equivalence point
• pH at equivalence point
= 7.00
• Shape of curve after
equivalence point is due
to [base]
• Initial pH is steeper
than for strong acid
• pH levels off (buffer
effect)
• pH at equivalence point
> 7.00
• Shape of curve after
equivalence point is due
to [base]
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titration of a Weak Base with a
Strong Acid
• The pH at the
equivalence point in
these titrations is < 7.
• Methyl red is the
indicator of choice.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Titrations of Polyprotic Acids
When one
titrates a
polyprotic acid
with a base
there is an
equivalence
point for each
dissociation.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Solubility Products
• Ksp = equilibrium constant for the equilibrium
between an ionic solid solute and its
saturated aqueous solution.
• Called the solubility-constant or the solubility
product
• Ksp = [cation]m[anion]n
where m, n = stoichiometric coefficients
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Solubility Products
Consider the equilibrium that exists in a
saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Solubility Products
The equilibrium constant expression for this
equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called
the solubility product.
(BaSO4 falls out of K because it is a solid.)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.9 (p. 738)
Write the expression for the solubilityproduct constant for CaF2, and look up
the corresponding Ksp value in Appendix
D in your textbook.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.9
Give the solubility-product constant
expressions and the values of the
solubility-product constants (from
Appendix D) for the following
compounds:
a) barium carbonate
b) silver sulfate
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Solubility Products
• Ksp is not the same as solubility.
• Solubility = the maximum amount of solute
that can be dissolved in a standard volume of
solvent
– often expressed as grams of solid that will
dissolve per liter of solution (g/L).
• Molar solubility = the number of moles of
solute that dissolve to form a liter of saturated
solution.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Solubility Products
• We can use solubility to find Ksp and vice versa.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Solubility  Ksp
1. Solubility (g/vol)  molar solubility (M)
2. Molar solubility  molar [ions] at
equilibrium.
3. Use equilibrium [ions] in the Ksp
expression.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Ksp  Solubility
1. Write Ksp expression.
2. Let x = molar solubility of the salt.
3. Use stoichiometry of the reaction to find the
exponents in the Ksp expression.
4.
Substitute into Ksp and solve for x.
e.g. Ksp Ag2SO4 = 1.5 x 10-5 = [Ag+]2[SO42-] = [2x]2[x]
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.10 (p. 739)
Solid silver chromate is added to pure water at 25oC. Some of
the solid remains undissolved at the bottom of the flask. The
mixture is stirred for several days to ensure that equilibrium is
achieved between the undissolved Ag2CrO4(s) and the solution.
Analysis of the equilibrated solution shows that its silver ion
concentration is 1.3 x 10-4 M. Assuming that Ag2CrO4
dissociates completely in water and that there are no other
important equilibria involving the Ag+ or CrO42- ions in the
solution, calculate Ksp for this compound.
(1.1 x 10-12)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.10
A saturated solution of Mg(OH)2 in contact
with undissolved solid is prepared at 25oC.
The pH of the solution is found to be 10.17.
Assuming that Mg(OH)2 dissociates
completely in water and that there are no
other simultaneous equilibria involving the
Mg2+ or OH- ions in the solution, calculate Ksp
for this compound.
(1.6 x 10-12)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.11 (p. 740)
The Ksp for CaF2 is 3.9 x 10-11 at 25oC. Assuming
that CaF2 dissociates completely upon dissolving and
that there are no other important equilibria affecting
its solubility, calculate the solubility of CaF2 in grams
per liter.
(1.6 x 10-2 g CaF2/L soln)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.11 (cont’d)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.11
The Ksp for LaF3 is 2 x 10-19. What is
the solubility of LaF3 in water in moles
per liter?
(9 x 10-6 mol/L)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors That Affect Solubility
1. The presence of a common ion.
2. The pH of the solution.
3. The presence or absence of
complexing agents.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
1. The Common-Ion Effect
Solubility  when a common ion is added
(Le Châtelier’s principle):
e.g. CaF2: CaF2(s) D Ca2+(aq) + 2F–(aq)
• Add more F– (i.e., + NaF), the equilibrium
shifts to offset the increase.
 CaF2(s) is formed and precipitation
occurs.
• As NaF is added to the system, the
solubility of CaF2 
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.12 (p. 742)
Calculate the molar solubility of CaF2 at
25oC in a solution that is
a) 0.010 M Ca(NO3)2
(3.1 x 10-5 mol CaF2/L
in 0.010 M Ca(NO3)2)
b) 0.010 M in NaF
(3.9 x 10-7 mol CaF2/L
in 0.010 M NaF)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.12 (p. 742)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.12
The value for Ksp for manganese (II)
hydroxide, Mn(OH)2, is 1.6 x 10-13.
Calculate the molar solubility of
Mn(OH)2 in a solution that contains
0.020 M NaOH.
(4.0 x 10-10 M)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
2. pH
– If a substance has a
basic anion, it will be
more soluble in an
acidic solution.
– Substances with
acidic cations are
more soluble in
basic solutions.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
Add more info from
packet here??
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.13 (p. 745)
Which of the following substances will be
more soluble in acidic solution than in basic
solution:
a) Ni(OH)2(s)
b) CaCO3(s)
c) BaF2(s)
d) AgCl(s)
(a-c)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.13
Write the net ionic equation for the
reaction of the following copper (II)
compounds with acid:
a) CuS
b) Cu(N3)2
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
MOMA Demo
•
Mg(OH)2 D Mg2+(aq) + 2OH-(aq)
•
Solubility of Mg(OH)2 = 9 x10-4 g/100. mL
•
What is its molar solubility?
•
What is the pH of a saturated Mg(OH)2 solution?
•
What happens when HCl is added?
Mg(OH)2 + 2 HCl  MgCl2(aq) + H2O,
i.e. OH- ions are removed from the solution, thus allowing more Mg(OH)2
to dissolve. As the Mg(OH)2 dissolves, it immediately reacts with the
HCl. After all the added HCl has been used up, the pH gradually
increases as more Mg(OH)2 is added, up to its maximum solubility.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Complex Ions
• Add more from packet??
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Formation of Complex Ions
e.g. AgCl has a very low solubility: Ksp = 1.8 x 10-10
Ag+(aq) + 2 NH3(aq) D Ag(NH3)2+(aq)
Kf = the formation constant, for the complex ion
Overall reaction: AgCl(s) + 2 NH3(aq) Ag(NH3)2+(aq) + Cl–(aq)
As the small amount of AgCl dissolves, the Ag+ is taken up by the NH3
allowing more Ag+ to form.
 By Le Châtelier’s principle, the forward reaction (the dissolving of AgCl)
is favored.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
3. Complex Ions
– Metal ions can act as Lewis acids and form
complex ions with Lewis bases in the solvent.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
• Complex Ions
– The formation
of these
complex ions
increases the
solubility of
these salts.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Complex Ions
• Common ions that form complexes:
NH3
CNOHSCNhalogens
Clue in Reactions Question = An excess of
concentrated …. is added to ….
Ch 5 & 13 The Ultimate Equations Handbook
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.14 (p. 748)
Calculate the concentration of Ag+ present in
solution at equilibrium when concentrated
ammonia is added to a 0.010 M solution of
AgNO3 to give an equilibrium concentration of
[NH3] = 0.20 M. Neglect the small volume
change that occurs when NH3 is added.
([Ag+] = 1.5 x 10-8 M)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.14 (cont’d)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.14
Calculate [Cr3+] in equilibrium with
Cr(OH)4- when 0.010 mol of Cr(NO3)3 is
dissolved in a liter of solution buffered at
pH 10.0.
([Cr3+] = 1 x 10-16 M)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
• Amphoterism
– Substances that are capable of acting either as an
acid or a base are amphoteric.
– Compare with amphiprotic, which relates more
generally to any species that can either gain or
lose a proton.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
• Amphoterism
– Amphoteric metal
oxides and hydroxides
are soluble in strong
acid or base, because
they can act either as
acids or bases.
– Examples of such
cations are Al3+, Cr3+,
Zn2+, and Sn2+.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Amphoterism
The hydroxides generally form complex ions with several
hydroxide ligands attached to the metal:
Al(OH)3(s) + OH–(aq)  Al(OH)4–(aq)
Hydrated metal ions act as weak acids.
As strong base is added, protons are removed:
• Al(H2O)63+(aq) + OH–(aq)  Al(H2O)5(OH)2+(aq) + H2O(l)
• Al(H2O)5(OH)2+(aq) + OH–(aq)  Al(H2O)4(OH)2+(aq) + H2O(l)
• etc.
• Addition of an acid reverses these reactions
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Precipitation and Separation of Ions
• Consider the following:
BaSO4(s) D Ba2+(aq) + SO42–(aq)
At any instant in time, Q = [Ba2+][ SO42– ]
- If Q < Ksp, more solid can dissolve until Q = Ksp.
- If Q = Ksp, the system is at equilibrium and the
solution is saturated.
- If Q > Ksp, the salt will precipitate until Q = Ksp.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.15 (p.751)
Will a precipitate form when 0.10 L of
8.0 x 10-3 M Pb(NO3)2 is added to 0.40
L of 5.0 x 10-3 M Na2SO4?
(yes)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.15
Will a precipitate form when 0.050 L of
2.0 x 10-2 M NaF is mixed with 0.010 L
of 1.0 x 10-2 M Ca(NO3)2?
(yes)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Selective Precipitation of Ions
One can use
differences in
solubilities of
salts to separate
ions in a
mixture.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Exercise 17.16 (p. 751)
A solution contains 1.0 x 10-2 M Ag+ and 2.0 x 10-2 M
Pb2+. When Cl- is added to the solution, both AgCl
(Ksp = 1.8 x 10-10) and PbCl2 (Ksp = 1.7 x 10-5)
precipitate from the solution.
What concentration of Cl- is necessary to begin the
precipitation of each salt?
Which salt precipitates first?
(> 2.9 x 10-2 M for PbCl2; > 1.8 x 10-8 M for AgCl,
AgCl precipitates first)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Practice Exercise 17.16
A solution consists of 0.050 M Mg2+ and 0.020 M Cu2+.
• Which ion will precipitate first as OH- is added to the
solution?
• What concentration of OH- is necessary to begin the
precipitation of each cation?
• (Ksp = 1.8 x 10-11 for Mg(OH)2 and Ksp = 4.8 x 10-20 for
Cu(OH)2)
• (Cu(OH)2 precipitates first, when [OH-] > 1.5 x 10-9 M;
Mg(OH)2 precipitates when [OH-] > 1.9 x 10-5 M)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Sample Integrative Exercise 17 (p. 755)
A sample of 1.25 L of HCl gas at 21oC and
0.950 atm is bubbled through 0.500 L of
0.150 M NH3 solution. Calculate the pH of
the resulting solution, assuming that all of the
HCl dissolves and that the volume of the
solution remains 0.500 L.
(pH = 8.97)
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.