A.P. Chemistry
Chapter 15
Applications of Aqueous Equuilibria
15.1 Solutions of Acids or Bases Containing a Common Ion
Common ion- the presence of an ion which appears in both the acid
(or base) and a salt in the solution
Common ion effect- the shift in equilibrium position that occurs
because of the addition of anion already involved in the equilibrium
reaction
For example: if a solution of NaF and HF are mixed together, F- would
be the common ion. If we look at the dissociation of each
compoundNaF  Na+(aq) + F-(aq)
HF(aq)  H+(aq) + F-(aq)
According to LeChatelier’s Principle, an increase in concentration of an
ion will shift the equilibrium away from the ion added.
HF(aq)  H+(aq) + F-(aq)
Equilibrium shifts away
added F- ions from
From added component;
NaF
Fewer H+ ions present
This makes the solution less acidic than a solution of HF alone.
Example: 41 g of sodium acetate, CH3COONa
have been added to 500.0 mL of a 0.50 M
CH3COOH solution. Assume no change in
volume. Determine the pH of the solution.
(Ka for CH3COOH is 1.8 x 10-5)(5.05)
15.2 Buffered Solutions
Buffered Solution- a solution that resists a change in pH
when either hydroxide ions or protons (hydrogen ions)
are added. A buffered solution is composed of a weak
acid and its salt (or a weak base and its salt)
They are simply solutions of weak acids or bases
containing a common ion. The pH calculations on
buffered solutions require exactly the same procedures
introduced in ch. 14.
When a strong acid or base is added to a buffered
solution, it is best to deal with the stoichiometry of the
resulting reaction first. After the stoichiometric
calculations are completed, then consider the
equilibrium calculations.
How does a buffer work? Suppose in a buffered solution
we have large amounts of a weak acid HA and its
conjugate base AIf OH- ions are added to this system (the reaction will
proceed in the forward direction)
HA + OH-  A- + H2O
The net result is that OH- are not allowed to accumulate
but are replaced by A- ions. So pH changes only slightly.
If H+ ions are added to this system (the reaction will
proceed in the reverse direction)
H+ + A-  HA
The net result is that H+ ions are not allowed to
accumulate but are replaced by HA molecules, so the
pH changes only slightly.
Thus the stability of the pH under these conditions
can be understood by examining the equilibrium
expression for the dissociation of HA:
Ka = [H+][A-]
[HA]
Rearrange to
[H+] = Ka [HA]
[A-]
In other words, the equilibrium concentration of H+
and thus pH, is determined by the ratio [HA]/[A-].
If the amounts of HA and A- are very large to start
with compared to the amount of H+ (or OH-) ions
added, the change in [HA]/[A-] will be small.
What if the buffered solution is made from a weak base and the salt of
the conjugate acid?
If H+ ions are added to this system (the reaction will proceed in the
forward direction)
B + H+  BH+
If OH- ions are added to this system(the reaction will proceed in the
reverse direction)
BH+ + OH-  B + H2O
Henderson-Hasselbalch equation- using the above relationship, if we
were to take the log form of the Ka expression we would get an
equation which is useful for calculating the pH of solutions when
the [HA]/[A-] is known.
pH = pKa + log( [A-]/[HA]) = pKa + log([base]/[acid])
For a particular buffering system (acid conjugate base pair) all solutions
that have the same ratio
[A-]/[HA] will have the same pH.
Example: Determine which of the following solutions form
buffers and identify each solution as acidic (pH < 7.00),
basic (pH > 7.00), or neutral (pH = 7.00)
• 50.0 mL of 1.0 M HCl & 50.0 mL of NaCl
(not buffer; pH < 7.00)
• 100.0 mL of 1.0 M HF & 100.0 mL of 1.0 M NaF
(acidic buffer; pH < 7.00)
• 100.0 mL of 1.0 M KOH & 50.0 mL of 2.0 M HBr
(not buffer; pH = 7.00)
• 30.0 mL of 1.0 M NH3 & 15.0 mL of 1.0 M HCl
(basic buffer; pH > 7.00)
• 30.0 mL of 1.0 M NH3 & 45.0 mL of 1.0 M HCl
(not buffer; pH < 7.00)
Example: Calculate the pH of a buffer solution
that is 0.25 M HF and 0.50 M NaF. (3.45)
Example: What ratio of [F-] to [HF] would you
use to make a buffer of pH = 2.85?
15.3 Buffer capacity
Buffer capacity-represents the amount of
protons (hydrogen ions) or hydroxide ions the
buffer can absorb without a significant change
in pH. The capacity of a buffered solution is
determined by the magnitude of [HA] and [A-]
Example: Adding Strong Acid to a Buffer
Suppose 3.0 mL of 2.0 M HCl is added to exactly
100. mL of the buffer described in the above
example; what is the new pH of the buffer
after the HCl is neutralized?
15.4 Titrations and pH curves (Examples on in-class
worksheet)
pH curve (titration curve)- the progress of an acid-base
titration monitored by plotting the pH of the solution
being analyzed as a function of the amount of titrant
added.
Millimole- since most titrations involve very small
quantities, the mole is inconveniently large. Since most
volumes will be measured in milliliters, it would also be
convenient to measure in a quantity equal to a
thousandth of a mole.
Molarity = mol solute = mol solute/1000 = mmol
L solution
L solution/1000 mL
solution
Strong acid-strong base titration- since both the acid and
the base are strong, each will dissociate completely.
Therefore it is important to determine the number of
moles of H+ and OH- produced. Three situations could
occur:
If you have equal moles of H+ and OH-, they will cancel
out each other and you will be left with a neutral
solution.
If you have more H+ ions than OH- ions, the extra H+ will
make the solution acidic (the other H+ ions will
combine with the OH- ions to make water)
If you have more OH- ions than H+ ions, the extra OH- will
make the solution basic (or alkaline)(the other OH- ions
will combine with the H+ ions to make water).
Stoichiometric Point (or equivalence point)- the point in
the titration where an amount of base has been added
to exactly react with all the acid originally present.
Weak acid- strong base titration- this type of titration
actually has two components: the stoichiometry of the
reaction and the equilibrium which will occur. Each
should be dealt with separately.
The stoichiometry- the reaction of hydroxide ion with the
weak acid is assumed to run to completion and the
concentrations of the acid remaining and the conjugate
base formed are determined.
The equilibrium- the position of the weak acid
equilibrium is determined and the pH is calculated. The
pH of the equivalence point of a titration of a weak
acid and a strong base is always greater than 7.
Weak base- strong acid titration-(see above)
The pH at the equivalence point of a titration of
a weak base with a strong acid is always less
than 7.
(Make sure you know the general shapes of
acid-base titration curves.)
Example: Calculate the pH when the 10.0 mL of
a 0.15 M HNO3 is added to 50.0 mL of 0.10 M
NaOH.
15.5 Acid-Base Indicators
Acid-base indicator: marks the end point of a
titration by changing color. It is important to
note that the end point is defined by the
change in color of the indicator. The
equivalence point is defined by the reaction
stoichiometry.
Example: Choose an indicator for the titration of
50.0 mL of a 0.10 M HI solution with 0.10 M
NH3.
Second Half of the Chapter!!
15.6 Solubility Equilibria and the Solubility Product
Solubility product constant (Ksp)- expresses the
equilibrium position that occurs for the solubility
of solids. It also expresses at what point solution
would be considered to be saturated.
For the following reaction: AB(s)  nA+(aq) +
mB-(aq)
The equilibrium expression would be
Ksp =
[A+]n[B-]m
Example: The solubility of magnesium fluoride in
water is 7.3 x 10-3 g per 100. mL of solution.
What is the solubility product constant?
Example:
The Ksp of Zn(OH)2 at 25oC is 1.8 x 10-14.
Determine the molar solubility of zinc hydroxide in
pure water at 25oC.
Determine the molar solubility of zinc hydroxide if
0.10 mol of Zn(NO3)2 are added to 1.0 L of a
saturated solution of zinc hydroxide at 25oC.
(assume no volume change)
Determine the molar solubility of zinc hydroxide if
the solution has an adjusted pH of 2.0 at 25oC.
15.7 Precipitation and Qualitative Analysis
Ion product (Q) – just like the expression for Ksp
for a given solid except that initial
concentrations are used instead of equilibrium
concentrations.
If Q > Ksp, precipitation occurs, and will
continue until the concentrations are reduced
to the point that they satisfy Ksp.
Is Q < Ksp, no precipitation will occur.
Selective precipitation: uses a reagent whose
anion forms a precipitate with only one or a
few of the metal ions in a mixture.
Example:10.0 mL of 0.010 M Pb(NO3)2 are mixed
with 10.0 mL of 0.01 M NaI solution.
Determine whether a precipitate will occur
and justify your answer.
Example: Determine all ion concentrations at
equilibrium when 10.0 mL of a 0.050 M
Pb(NO3)2 solution are mixed with 20.0 mL of a
0.010 M Na2CrO4.
Example: What is the solubility of PbCl2 in 0.50
M NaCl solution?
15.8 Equilibria Involving Complex ions (read this section)
Take note: The complex ions that appear most often on
the AP exam are: Ag(NH3)2+, Ag(CN)2-, Al(OH)4-, AlF63-,
Cu(NH3)42+, Fe(SCN)2+, Fe(SCN)63-, Ni(NH3)x2+, Ni(OH)42-,
Zn(OH)42-, Zn(NH3)62+, Zn(NH3)42+, CoCl42See Eqx Writing on Complex Ion.
Example: Write equations for the formation reaction of
each of the following complex ions:
Cu(NH3)42+
Cu(CN)42Ag(CN)2-