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7/3/2012
1
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Chapter 17
Additional Aspects of
Aqueous Equilibria
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
John D. Bookstaver
St. Charles Community College
Cottleville, MO
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Chapter 17 Problems
• Problems 11, 15, 17, 19, 21, 27, 33, 35,
47, 49, 51, 55, 61
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
• Consider a solution of acetic acid:
• If acetate ion is added to the solution,
Le Châtelier says the equilibrium will
shift to the left.
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO−(aq)
7/3/2012
2
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.”
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
Calculate the fluoride ion concentration and pH
of a solution that is 0.20 M in HF and 0.10 M in
HCl.
Ka for HF is 6.8 10−4.
[H3O+] [F−]
[HF] Ka = = 6.8 10-4
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
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
At Equilibrium 0.20 − x 0.20 0.10 + x 0.10 x
HF(aq) + H2O(l) H3O+(aq) + F−(aq)
7/3/2012
3
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
The Common-Ion Effect
= x
1.4 10−3 = x
(0.10) (x)
(0.20) 6.8 10−4 =
(0.20) (6.8 10−4)
(0.10)
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.
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.
7/3/2012
4
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffers
If a small amount of hydroxide is added to an
equimolar solution of HF in NaF, for example, the HF
reacts with the OH− to make F− and water.
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+] [A−]
[HA] Ka =
HA + H2O H3O+ + A−
7/3/2012
5
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffer Calculations
Rearranging slightly, this becomes
[A−]
[HA] Ka = [H3O
+]
Taking the negative log of both side, we get
[A−]
[HA] −log Ka = −log [H3O
+] + −log
pKa
pH acid
base
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Buffer Calculations
• So
pKa = pH − log [base]
[acid]
• Rearranging, this becomes
pH = pKa + log [base]
[acid]
• This is the Henderson–Hasselbalch equation.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Henderson–Hasselbalch Equation
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.
7/3/2012
6
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Henderson–Hasselbalch Equation
pH = pKa + log [base]
[acid]
pH = −log (1.4 10−4) + log (0.10)
(0.12)
pH
pH = 3.77
pH = 3.85 + (−0.08)
pH
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.
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.
7/3/2012
7
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.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
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.
Calculate the pH of this solution after
0.020 mol of NaOH is added.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8 10−5) = 4.74
7/3/2012
8
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
The 0.020 mol NaOH will react with 0.020 mol of the
acetic acid:
HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)
HC2H3O2 C2H3O2− OH−
Before reaction 0.300 mol 0.300 mol 0.020 mol
After reaction 0.280 mol 0.320 mol 0.000 mol
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to
calculate the new pH:
pH = 4.74 + log (0.320)
(0.200)
pH = 4.74 + 0.06 pH
pH = 4.80
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).
7/3/2012
9
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 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.
7/3/2012
10
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.
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.
7/3/2012
11
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.
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.
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.
7/3/2012
12
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.
The Solubility Product
Constant, Ksp
Many ionic compounds are only slightly soluble
in water: ex. Ag salts, sulfides
Equations are written to represent the equilibrium
between the compound and the ions present in a
saturated aqueous solution
AgCl(s) Ag+(aq) + Cl–(aq)
c c
[Ag ][Cl ]... [AgCl] [Ag ][Cl ]
[AgCl]K but K
EOS
Ksp = [Ag+][Cl–]
Aqueous
Equilibria
SAMPLE EXERCISE Writing Solubility-Product (Ksp) Expressions
Write the expression for the solubility-product constant for CaF2, and look up the corresponding Ksp
value in Appendix D.
In Appendix D we see that this Ksp has a value of 3.9 10–11.
Solution
Analyze and Plan: We are asked to write an equilibrium-constant expression for the
process by which CaF2 dissolves in water. We apply the same rules for writing any
equilibrium-constant expression, making sure to exclude the solid reactant from the
expression. We assume that the compound dissociates completely into its component
ions.
Solve: Following the italicized rule stated previously, the expression for
Ksp is
7/3/2012
13
Aqueous
Equilibria
Ksp and Molar Solubility
The solubility product constant is related to
the solubility of an ionic solute
Ksp = [Ag+][Cl–]; solubility given by [Ag+]
From stoichiometry, the ion ratio is 1:1, so
[Ag+] = [Cl–], both of which are unknown
(x)
EOS
Ksp = x2 and [Ag+] = (Ksp)1/2
Ag+ Cl– Ag+ Cl– +
Aqueous
Equilibria
SAMPLE EXERCISE Calculating Ksp from Solubility
Solid silver chromate is added to pure water at 25°C. 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 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. Solution
Analyze: We are given the equilibrium concentration of Ag+ in a saturated solution of Ag2CrO4. From
this, we are asked to determine the value of the solubility-product constant for the dissolution of
Ag2CrO4.
Plan: The equilibrium equation and the expression for Ksp are
To calculate Ksp, we need the equilibrium concentrations of Ag+ and CrO42–. We know that at
equilibrium [Ag+] = 1.3 10–4 M. All the Ag+ and CrO42– ions in the solution come from the Ag2CrO4
that dissolves. Thus, we can use [Ag+] to calculate [CrO42–].
Solve: From the chemical formula of silver chromate, we know that there must be 2 Ag+ ions in
solution for each CrO42– ion in solution. Consequently, the concentration of CrO4
2– is half the
concentration of Ag+.
We can now calculate the value of Ksp.
Aqueous
Equilibria
SAMPLE EXERCISE Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9 10–11 at 25°C. 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. Solution
Analyze: We are given Ksp for CaF2 and are asked to determine solubility. Recall that the solubility of
a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, Ksp, is
an equilibrium constant.
Plan: We can approach this problem by using our standard techniques for solving equilibrium
problems. We write the chemical equation for the dissolution process and set up a table of the initial
and equilibrium concentrations. We then use the equilibrium-constant expression. In this case we
know Ksp, and so we solve for the concentrations of the ions in solution.
Solve: Assume initially that none of the salt has dissolved, and then allow x moles/liter of CaF2 to
dissociate completely when equilibrium is achieved.
7/3/2012
14
Aqueous
Equilibria
SAMPLE EXERCISE continued
The stoichiometry of the equilibrium dictates that 2x moles/liter of F– are produced for each x
moles/liter of CaF2 that dissolve. We now use the expression for Ksp and substitute the equilibrium
concentrations to solve for the value of x:
(member that to calculate the cube root of a number, you can use the yx function on your
calculator, with ) Thus, the molar solubility of CaF2 is 2.1 10–4 mol/L. The mass of CaF2 that
dissolves in water to form a liter of solution is
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.
7/3/2012
15
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Ksp’s (25 oC)
EOS
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Solubility Products
• Ksp is not the same as solubility.
• Solubility is generally expressed as the mass
of solute dissolved in 1 L (g/L) or 100 mL
(g/mL) of solution, or in mol/L (M).
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
• The Common-Ion Effect
– If one of the ions in a solution equilibrium
is already dissolved in the solution, the
equilibrium will shift to the left and the
solubility of the salt will decrease.
BaSO4(s) Ba2+(aq) + SO42−(aq)
7/3/2012
16
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
• 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
• 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.
7/3/2012
17
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+, Zn2+,
and Sn2+.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Will a Precipitate Form?
• In a solution,
– If Q = Ksp, the system is at equilibrium
and the solution is saturated.
– If Q < Ksp, more solid can dissolve
until Q = Ksp.
– If Q > Ksp, the salt will precipitate until
Q = Ksp.
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Applying the Criteria for Precipitation of a Slightly Soluble
Solute. Three drops of 0.20 M KI are added to 100.0 mL of 0.010
M Pb(NO3)2. Will a precipitate of lead iodide form?
(1 drop 0.05 mL)
PbI2(s) → Pb2+(aq) + 2 I-(aq) Ksp= 7.110-9
Determine the amount of I- in the solution:
= 310-5 mol I-
nI- = 3 drops 1 drop
0.05 mL
1000 mL
1 L
1 L
0.20 mol KI
1 mol KI
1 mol I-
EXAMPLE
7/3/2012
18
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
[I-] = 0.1000 L
310-5 mol I- = 310-4 M I-
Determine the concentration of I- in the solution:
Apply the Precipitation Criteria:
Q = [Pb2+][I-]2 = (0.010)(310-4)2
= 910-10 < Ksp = 7.110-9
EXAMPLE
Aqueous
Equilibria
© 2009, Prentice-Hall, Inc.
Selective Precipitation of Ions
One can use
differences in
solubilities of
salts to separate
ions in a
mixture.