Chapter 17
Additional Aspects of Aqueous
Equilibria
The Common Ion Effect
• The presence of a “common ion” will affect
the ionization of a weak acid or base and the
solubility of a slightly soluble salt.
• For example:
• Calculate the pH of 0.200 M acetic acid. (Ka = 1.8 x 10-5)
The Common Ion Effect
• Now calculate the pH of acetic acid in a
0.500M soution of sodium acetate:
Buffers
• Buffers are solutions of a
weak conjugate acidbase pair.
• They are particularly
resistant to pH changes,
even when strong acid or
base is added.
• Buffers contain a
component that will
react with H+ ion and a
component that will
react with OH- ion
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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.
OH- + HF ↔ H2O + F© 2009, Prentice-Hall, Inc.
Buffers
Similarly, if acid
is added, the F−
reacts with it to
form HF and
water.
H3O+ + F- ↔ HF + H2O
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Buffer Calculations
Find the pH of a buffer which is 0.500 M in acetic
acid and 0.350 M in sodium acetate.
HAc + H2O
H3O+ + Ac−
[H3O+] [Ac−]
Ka =
[HAc]
NOTE that this is essentially a common ion calculation!
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Another Way……
If we rearrange the Ka expression:
[H3O+] [Ac−]
Ka =
[HAc]
Ka =
−]
[A
[H3O+]
[HA]
And then take the negative log of both side, we get
−log Ka = −log [H3O+] + −log
[A−]
[HA]
base
pKa
acid
pH
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Another way….
• So
pKa = pH − log
[base]
[acid]
• Rearranging, this becomes
pH = pKa + log
[base]
[acid]
• This is the Henderson–Hasselbalch equation.
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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.
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pH Range and buffer capacity
• The pH range is the range of pH values over
which a buffer system works effectively.
• When preparing a buffer, it is best to choose
an acid with a pKa close to the desired pH.
• The buffer capacity is the amount of acid or
base that can be neutralized before there is a
significant change in pH. The buffer capacity
depends on the concentrations of the
component of the buffer.
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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.
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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. (think ICE table!!)
2. Find the new [ H3O+ ] to determine
the new pH of the solution.
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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.
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Titration
• In this technique a known
concentration of base (or
acid) is slowly added to a
solution of acid (or base).
• 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.
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Titration of a Strong Acid with a Strong
Base
• Before starting the titration, the pH of the acid is
determined by [acid] = [H+].
• Before equivalence, the pH is determined by
calculating the moles of base added,, determining
the moles of H+ remaining AND accounting for the
NEW total volume. From the start of the titration to
near the equivalence point, the pH slowly increases.
• Just before (and after) the equivalence point, the pH
increases rapidly.
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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.
• As more base is added,
the increase in pH again
levels off.
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Titration of a Weak Acid with a Strong Base
• Before titration, the pH is determined by using
the Ka to find the [H+].
• Before equivalence, the pH is determined by
calculating the moles of base added, determining
the moles of weak acid remaining and
determining the number of moles of conjugate
base that has formed AND accounting for the
NEW total volume. Use the Ka expression to find
the [H+].
OH- + HW ↔ W- + H2O
Titration of a Weak Acid with a Strong Base
•At equivalence, all of the acid has reacted
and the conjugate base has formed. The
base now reacts with water:
W- + H2O ↔ HW + OH•The equilibrium expression for this
reaction is a Kb!!!
•Find Kb = Kw/Ka and solve for the [OH-]
and [HW] by assuming that the original
moles of acid = moles of W- and
accounting for the NEW volume.
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.
• After equivalence, the
pH depends on the
excess OH- ion
added…..exactly like a
strong acid titration!!
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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.
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Titration of a Weak Base with a Strong Acid
• The pH at the
equivalence point in
these titrations is < 7.
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Titrations of Polyprotic Acids
When one
titrates a
polyprotic acid
with a base there
is an equivalence
point for each
dissociation.
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Solubility Products
Consider the equilibrium that exists in a saturated
solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
The equilibrium constant expression for this
equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the
solubility product.
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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).
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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)
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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.
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Factors Affecting Solubility
• Complex Ions
– Metal ions can act
as Lewis acids and
form complex
ions with Lewis
bases in the
solvent
– The formation of
these complex
ions increases the
solubility of these
salts.
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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.
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Selective Precipitation of Ions
One can use
differences in
solubilities of salts
to separate ions
in a mixture.
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