AP Chapter 17
Ionic Equilibria of Weak Electrolytes
pH and pOH scales
acidic
basic
pH + pOH = 14.00
+
H
=
+
H3O
pH = -log[H+]
• The pH scale is a logarithmic scale.
• This means that in order to change the pH
by one unit there must be a tenfold change
in the [H+].
• Find the pH of a solution with [H+] = 0.010M
• Find the pH of a solution with [H+] = 0.10M
Example 17.1 Page 521
• What is the pH of a solution of HCl with a
concentration of 1.2 x 10-3M?
pH = 2.92
Example 17.2 Page 521
• Calculate the pH of:
– 0.10 M solution of HNO3 pH = 1.00
– 0.10 M solution of CH3CO2H (1.3% ionized)
pH = 2.89
Example 17.4 Page 522
• Calculate the [H3O+] of a solution with a pH of 9.0
[H3O+] = 1.0 x 10-9
Example 17.6 Page 523
• Calculate the pH and pOH of a 0.0125 M
KOH solution.
pOH = 1.903
pH = 12.097
Calculate the [OH-] of a solution with a pH of 5.56.
[OH ]
= 3.63 x
-9
10
What are the ion concentrations
in a 0.10M HCl solution?
0.10M HCl
0.10M H+
0.10M Cl-
Strong electrolytes
dissociate completely
What are the ion concentrations
in a 0.15 M K2SO4 solution?
0.15M K2SO4
0.30M K+
0.15M SO42-
At equilibrium, a solution of acetic acid, [ CH3CO2H ] = 0.0788M
and [H3O+] = [CH3CO2-] = 0.0012M. What is the Ka of acetic acid?
Example 17.8 page 527
Example 17.9 page 527
The pH of a 0.0516 M solution of nitrous acid, HNO2,
is 2.34. What is the Ka?
Additional Ka values
Table contains more Ka values than are pictured here
Calculate the [H3O+], [CH3CO2-], and [ CH3CO2H ] in a 0.100 M
solution of acetic acid. What is the Ka = 1.8 x 10-5.
Example 17.10 page 528
What else can you determine in
the previous example?
• pH, pOH, [OH-]
• percent ionization Calculate this value.
Percent ionization = 1.3%
What is the percent ionization of a 0.25 M solution of
trimethylamine, (CH3)3N, a weak base with a Kb = 7.4 x 10-5
1.7%
What is the pH of the trimethylamine solution?
[OH-] = 4.3 x 10-3
pOH = -log[4.3 x 10-3] = 2.37
pH = 14 – 2.37 = 11.63
[OH-] = 4.3 x 10-3
[H+] = 1 x 10-14 ÷ 4.3 x 10-3 = 2.3 x 10-12
pH = -log[2.3 x 10-12] = 11.63
Appendix G has additional values
Diprotic and Triprotic Acids
• A diprotic acid ionizes in two steps
because it has two ionizable hydrogens.
• A triprotic acid ionizes in three steps
because it has three ionizable hydrogens.
The Stepwise Dissociation of Phosphoric Acid.
A triprotic acid.
H3PO4 (aq) + H2O(l)
H2PO4-(aq) + H3O+(aq)
H2PO4-(aq) + H2O(l)
HPO42-(aq) + H3O+(aq)
HPO42-(aq) + H2O(l)
PO43-(aq) + H3O+(aq)
H3PO4 (aq) + 3 H2O(l)
PO43-(aq) + 3 H3O+(aq)
Each ionization step occurs to a lesser
extent than one preceding it.
Ka = 7.5 x 10-3
H3PO4 (aq) + H2O(l)
Ka = 6.3 x 10-8
H2PO4-(aq) + H2O(l)
2Ka = 3.6 x 10-13 HPO4 (aq) + H2O(l)
Ka = ?
H3PO4 (aq) + 3 H2O(l)
H2PO4-(aq) + H3O+(aq)
HPO42-(aq) + H3O+(aq)
PO43-(aq) + H3O+(aq)
PO43-(aq) + 3 H3O+(aq)
0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate [H3O+].
Example 17.19 page 548
0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate [H3O+].
We can work the problem using either Ka or Kb
Example 17.19 page 548
Compare this answer to [H3O+] from problem 17.10 pp. 528.
How does this illustrates LeChatlier’s Principle.
Example 17.10 page 528
Calculate the [H3O+], [CH3CO2-], and [ CH3CO2H ] in a 0.100 M
solution of acetic acid. What is the Ka = 1.8 x 10-5.
CH3CO2H + H2O ↔ CH3CO2- + H3O+
[H3O+] = 1.3 x 10-3
0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in
1.0L of solution. Calculate [H3O+].
[H3O+] = 3.6 x 10-6
10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3
are mixed. Calculate the [OH-].
Example 17.20 page 549
10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3
are mixed. Calculate the [OH-].
Common Ion Problems.
What is the “common ion” in each example?
Example 17.20 page 549
• 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are
mixed. Calculate the [OH-].
Example 17.19 page 548
• 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate [H3O+].
Buffers are also examples of the common ion effect
since they are mixtures where both substances
produce the same ion.
Both of the solution mixtures above are buffers.
Buffers
• A buffer solution is an aqueous solution
consisting of a mixture of a weak acid and
its conjugate base or a weak base and its
conjugate acid.
• Buffer solutions have the property that the
pH of the solution changes very little when a
small amount of acid or base is added to it.
Demonstration:
Buffered vs. Non-buffered solutions
Why are these solutions buffers?
Example 17.20 page 549
• 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are
mixed. Calculate the [OH-].
Example 17.19 page 548
• 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate [H3O+].
How buffers work
• Weak acids and weak bases tend to
remain in high concentrations when added
to water because by definition they do not
ionize much in water since they are weak.
• However, they are very likely to react with
any added strong base or strong acid.
Why are weak acids/bases used to
create buffers?
Remember this problem
• Calculate the pH of:
– 0.10 M solution of HNO3 pH = 1.00
– 0.10 M solution of CH3CO2H (1.3% ionized)
pH = 2.89
• What would the [CH3CO2H] have to be for it to
have the same pH as the HNO3 assuming the
1.3% ionization factor does not change?
0.10M = [CH3CO2H](0.013)
[CH3CO2H] = 7.7 M
Why are weak acids/bases used to
create buffers?
• That’s right 7.7 M [CH3CO2H] vs. 0.10M
have the same pH.
• In other words the [CH3CO2H] is 77 times
greater [HNO3].
• It takes much more base to change the pH
of a weak acid solution because there is a
large reservoir of undissociated weak acid.
Weak Acid (HA) and its conjugate
base (A-) buffer
Adding a strong base to the buffer
• If a strong base is added to a buffer, the weak acid will give up its H+ in
order to transform the base (OH-) into water (H2O) and the conjugate
base:
• HA + OH- → A- + H2O.
• Since the added OH- is consumed by this reaction, the pH will change
only slightly.
Adding a strong acid to the buffer
• If a strong acid is added to a buffer, the weak base will react with the H+
from the strong acid to form the weak acid, HA:
• H+ + A- → HA
• The H+ gets absorbed by the A- instead of reacting with water to form
H3O+ (H+), so the pH changes only slightly.
An effective buffer requires relatively equal
amounts of weak acid and conjugate base.
Basic Buffers
• Note that the same ideas hold true for weak
bases, (B), and their conjugate acids (BH+).
For the most effective buffers Ka = [H3O+]
HA + H2O
↔A

+
O
+ H3
-
[H 3O ][A ]
Ka 
[HA]
Consider acetic acid
You want to make the most “effective” buffer you
can using acetic acid. What would the [H3O+] be?
Calculate the pH of this buffer.
I want to make a buffer with a pH of
3.14. Which acid should I use?
I want to make a buffer with a pH of
3.14. Which acid should I use?
Ka = [H3O+]
[H3O+] = antilog (-pH)
Ka = antilog (-3.14) = 7.2 x 10-4
HF
I want to make a buffer with a pH of
3.75. Which acid should I use?
I want to make a buffer with a pH of
3.75. Which acid should I use?
Ka = [H3O+]
[H3O+] = antilog (-pH)
Ka = antilog (-3.14) = 7.2 x 10-4
Formic Acid
(HCO2H)
For the most effective buffers Kb = [OH-]
Consider ammonia
For the most effective buffers Kb = [OH-]
B + H2 O ↔
+
BH

+
OH

[OH ][BH ]
Kb 
[B]
I want to make a buffer with a pH of
10.64. Which base should I use?
I want to make a buffer with a pH of
10.64. Which base should I use?
Kb = [OH-]
Kb = antilog (-pOH)
Kb = antilog (-3.36) = 4.4 x 10-4
Methylamine
(CH3NH2)
Effective Buffers
• The previous formulas only apply when the
concentrations of weak acid and conjugate base
or weak base and conjugate acid are equal.
• However it is important to note that you do not
have to have equal concentrations to have a
buffer.
Henderson – Hasselbach Equation
-
[A ]
pH  pK a  log
[HA]
Example 17.19 page 548
0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are
dissolved in 1.0L of solution. Calculate [H3O+].
Henderson – Hasselbach Equation

[BH ]
pOH  pKb  log
[B]
Example 17.20 page 549
10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3
are mixed. Calculate the [OH-].
I want to make a buffer using acetic acid.
What chemicals should I get from the stock room?
• Describe how you would make the buffer?
• What would the pH of the buffer be?
I have no acetate salts. Can I still make
the buffer using acetic acid? Explain.
I want to make a buffer using ammonia.
What chemicals should I get from the stock room?
• Describe how you would make the buffer?
• What would the pH of the buffer be?
I have no ammonium salts. Can I still
make a buffer using ammonia? Explain.
Making Effective Buffers
1. Equal moles of a weak acid and a salt of
its conjugate base.
2. Equal moles of a weak base and a salt of
its conjugate acid.
3. A weak acid and half as many moles of
strong base.
4. A weak base and half as many moles of
strong acid.
5. “Use a pH meter”
pH meter
Example 17.21 page 551
Calculate the pH of a buffer that is 0.10 M acetic
acid and 0.10 M sodium acetate.
pH = 4.74
Buffer Capacity
• Buffer capacity: the amount of an acid or base
that can be added to a volume of a buffer
solution before its pH changes significantly.
• Buffer capacity depends on the amount (moles)
of the conjugate pair used to make the buffer.
• Buffer solutions have essentially lost their
buffering capabilities when one component of
the conjugate pair is about 10% or less of the
other.
Page 555
Buffers
• Which buffer has the greater capacity?
• Which buffer is more effective?
Buffer A:
– 100mL of 0.1M CH3CO2H with 100 mL of 0.1M NaCH3CO2.
Buffer B:
– 100mL of 1.0M CH3CO2H with 100 mL of 1.0M NaCH3CO2.
Calculate the [OH-] of a 0.050 M
solution of NaCH3CO2
[OH-] = 5.3 x 10-6
Calculate the pH of a 0.050 M
solution of NaCH3CO2
pH = 8.72
Calculate the percent reaction of
a 0.050 M solution of NaCH3CO2
0.011%
Calculate the pH of a 0.10 M
solution of AlCl3 (Ka = 1.4 x 10-5)
pH = 2.93
Titration
titrant
analyte
Titration Curves
• A titration curve is a plot of the pH against the
volume of acid or base added in a titration.
• The equivalence point or endpoint for a titration is
the point at which exactly enough of the titrant has
been added to completely react with the analyte.
• In other words, at the equivalence point, the number of
moles of titrant added corresponds exactly to the
number of moles of substance being titrated, the
analyte, according to the reaction stoichiometry.
Table 17.8 Page 568
Page 568
Page 569
How do we choose an appropriate
indicator for a titration?
Indicators
• An acid – base indicator is a substance that
indicates how acidic or basic a solution is using
certain color changes.
• Indicators are normally weak acids that have a
different color than their conjugate base.
HA + H2O ↔ H3O+ + AIndicators can also be weak bases that have a
different color than their conjugate acid.
Indicators
• Remember that the equivalence point of a
titration is where you have mixed the two
substances in exact stoichiometric proportions.
• You obviously need to choose an indicator
which changes color as close as possible to that
equivalence point.
• The indicator should have a pKa value near the
pH of the titration's endpoint.
• That varies from titration to titration.
pKa and pH range of
Acid-Base Indicators
Which indicator(s) would be
appropriate for the titrations below?
Indicators
Now that we have chosen an appropriate
indicator for a titration. Let’s consider in
more detail what happens during the
titration of different types of acids.
(a)
(c)
(b)
(d)
What species are present at half way to
the equivalence point?
What species are present at half way to
the equivalence point?
NaOH + HCl → NaCl + H2O
•At half way to the
equivalence point
the main species
present are H+
(hydronium), Cland Na+.
•The solution is
neutral.
What species are present at half way to
the equivalence point?
NaOH + CH3CO2H → NaCH3CO2 + H2O
•At half way to the
equivalence point the
main species present
are CH3CO2H, Na+,
and CH3CO2- and to a
lesser extent H+
(hydronium).
•The solution is
neutral.
The titration curve is produced when a 10.0 ml
sample of HCl is titrated with 0.100M NaOH. What is
the concentration of the HCl solution.
0.25M HCl
The titration curve is produced when a 10.0 ml
sample of CH3CO2H is titrated with 0.100M NaOH.
What is the concentration of the CH3CO2H solution.
0.25M CH3CO2H
Determine the Ka of the weak acid using the titration curve.
Determine the Ka of the weak acid using the titration curve.
Diprotic Acid
• When titrating a diprotic acid with a strong
base it is essentially like doing two
titrations at once.
Titration curve of a diprotic acid
HSO4- → H+ + SO42-
H2SO4 → H+ + HSO4-
The titration curve shown above is for a diprotic acid
such as H2SO4. This proves that polyprotic acids lose
their protons in a stepwise manner.
How would you determine the Ka
for each acid (H2SO4 and HSO4-)?
How would you determine the Ka
for each acid (H2SO4 and HSO4-)?
Why do you have to learn all this very
challenging information?
This stuff is hard. Why do you
have to learn it?
Why? Because I had to learn this
crap and now its payback time!