Chapter 15
Acid-Base Equilibria
Buffer - A Definition
• Solutions that contain components that
resist changes in pH
– Imperative to living things which require
specific pH ranges (ex: blood = 7.4 even
during creation of lactic acid…)
Solutions of Acids/Bases Containing
a Common Ion
• Weak acid (HA) is present in solution along
with a salt (NaA --> Na+ and A-)
– Salt will break up (strong electrolyte)
– HA will determine the pH (ex: HF and NaF)
HF(aq) <-> H+(aq) + F-(aq)
NaF -> Na+ + F-
• The common ion (in this case F-) will shift
equilibrium LEFT (Le Chatelier’s principle)
• Fewer H+ ions present, higher/more basic pH
Example
• Which way will the equilibrium shift if NH4Cl
is added to a 1.0 M NH3 solution? How will
this affect the pH?
NH4Cl -> NH4+ + ClNH3 + H2O <-> NH4+ + OH-
• The creation of NH4+ ions in the first reaction
will shift the second reaction to the LEFT.
This will lower the [OH-] and increase the [H+]
and lowering pH.
Example #2
• Calculate the pH and the percent
dissociation of the acid
– 0.200 M solution of HC2H3O2 (Ka = 1.8 X 10-5)
• Answer: pH = 2.7, 0.95%
– 0.200 M HC2H3O2 in the presence of 0.500 M
NaC2H3O2
• Answer: pH = 5.14, 3.6 X 10-3%
Change From Chapter 14 Problems
• The initial concentration of the products will not
be zero in the presence of the salt
**ALWAYS write the major species in
solution!!
• Example: The equilibrium [H+] in a 1.0 M HF
solution is 2.7 X 10-2 M, and the percent
dissociation of HF is 2.7%. Calculate the [H+]
and the percent dissociation of HF in a solution
containing 1.0 M HF (Ka = 7.2 X 10-4) and 1.0 M
NaF.
– Answer: [H+] = 7.2 X 10-4, 0.072% dissociation
Buffered Solutions
• A buffered solution is one that resists a change
in its pH with addition of OH- ions or protons
– BLOOD (HCO3- and H2CO3)
– Weak acid and its salt (ex HF and NaF)
– Weak base and its salt (ex NH3 and NH4Cl)
• Same calculation approach as chapter 14
Example #1
• A buffered solution contains a 0.50 M
acetic acid (HC2H3O2, Ka = 1.8 X 10-5)
and 0.50 M sodium acetate
(NaC2H3O2). Calculate the pH of this
solution.
– Answer: 4.74
Flow
• When dealing with a strong acid or base,
rxn stoichiometry comes first, then use
ICE table.
Example #2
• Calculate the change in pH that occurs when
0.010 mol solid NaOH is added to 1.0 L of the
buffered solution described in the last example.
• Compare this pH change with that which occurs
when 0.010 mol solid NaOH is added to 1.0 L
water.
Example #3
• Calculate the pH of a solution that contains
0.250 M formic acid, HCOOH (Ka = 1.8 X 10-4),
and 0.100 M sodium formate, HCOONa
• Calculate the pH of the buffer after the addition
of 10.0 mL of 6.00 M NaOH to the original
buffered solution volume of 500.0 mL
Adding Acid to a Solution (HA
and A-)
• If acid (H+) is added instead of OH-, the
reaction that will occur is
H+ + A- --> HA
– The H+ ions are replaced by HA, which
won’t change the pH very much.
But WHY?
• When a weak acid (HA) dissociates, HA --> H+
+ A- the corresponding equilibrium expression is
Ka = [H+][A-]/[HA]. In order to determine pH, the
equation should be rearranged:
• [H+] = Ka [HA]/[A-]
• A buffered solution may contain weak acid
(HA) and it’s conjugate base (A-). When a
base (OH-) is added, the reaction that occurs
is
OH- + HA --> A- + H2O
– The added OH- ions react with HA to make Aions. With less HA and more A-, the equilibrium
expression from the last slide is affected. The [H+]
is reduced which will change the pH. The reason
why it doesn’t affect the pH much is because [HA]
and [A-] are very large compared to the amount of
OH- added. The change in [HA]/[A-] will be small
and won’t affect the pH much.
Henderson-Hasselbalch Equation
• [H+] = Ka [HA]/[A-] will be very helpful in
determining [H+] and pH in buffered
solutions…can be manipulated to make:
pH = pKa + log ([A-]/[HA]) or
pH = pKa + log ([base]/[acid])
• NOTE: In a particular buffering system, all
solutions that have the same ratio of [A-] to [HA]
will have the same pH.
• ICE table assumption for x is generally
accepted when using this equation
Example
• Calculate the pH of a solution containing
0.75 M lactic acid (Ka = 1.4 X 10-4) and 0.25
M sodium lactate. Lactic acid (HC3H5O3) is a
common constituent of biologic systems. For
example, it is found in milk and is present in
human muscle tissue during exertion.
– Answer: 3.38
Example #2
• A solution is prepared by adding 31.56 g NaCN
and 22.30 g HCN to 600.0 mL of water (Ka for
HCN = 6.2 X 10-10).
– What is the pH of this solution?
• Answer: 9.100
– What is the pH after the addition of 50.0 mL of 3.00
M HCl?
• Answer: 8.91
– What is the pH after a further addition of 80.0 mL of
4.00 M NaOH?
• Answer: 9.30
Example #3
• The Ka of propionic acid, HC3H5O2, is
1.34 X 10-5 (pKa = 4.87). What is the
pH when [HC3H5O2] = [C3H5O2-]?
– Answer: 4.87
Buffering Capacity
• Definition: The amount of protons or hydroxide
ions a buffer can absorb without a significant
change in pH.
• Large buffering capacity means it contains a
large amount of buffering components and it
can absorb a lot of protons/hydroxide ions and
show little pH change
• NOTE: The pH of a buffered solution is
determined by [A-]/[HA]. The capacity is
determined by how large [HA] and [A-] are
Example
• Calculate the pH of a 0.500 L solution that
contains 0.15 M HCOOH (Ka = 1.8 X 10-4) and
0.20 M HCOONa.
• Then calculate the pH of the solution after the
addition of 10.0 mL of 12.0 M NaOH
– Answer: 3.87, 12.95
• When the strong base was added, the pH
changed drastically. When a buffer is used, the
goal is for this NOT to happen.
Optimal Buffering
• We want to avoid large changes in the
[A-]/[HA] ratio
– Best buffered solution will have [HA]=[A-],
or [A-]/[HA] = 1
Example
• We wish to buffer a solution at pH = 10.07.
Which one of the following bases (and
conjugate acid salts) would be most useful?
– NH3 (Kb = 1.8 X 10-5)
– C6H5NH2 (Kb = 4.2 X 10-10)
– N2H4 (Kb = 9.6 X 10-7)
Titration Term Review
• Titrant in buret = solution of known
concentration
• Equivalence point = stoichiometric point =
moles of H+ equal moles of OH• Endpoint = color changes due to pH and
depends on indicator used
• pH curve = titration curve is a plot of the pH
of the solution being analyzed as a function of
the amount of titrant added
Titrations and pH Curves
• Strong Acid - Strong Base (ch. 4)
• Weak Acid with Strong Base (ch. 14)
• Weak Base with Strong Acid
• Mole is too large of a unit when working with
mililiters, so generally a millimole (mmol) is
used…1000 mmol = 1 mol and mmol/mL = M
Strong Acid - Strong Base Titration
• pH changes gradually until
the titration is close to the
equilvalence point where a
dramatic change occurs
• pH = 7.00 at equivalence
point
• Curve points right/left based
on beginning solution
• Polyprotic acids have
multiple curves
Weak Acid with Strong Base
Titration
•
•
Essentially a set of buffer problems
**Even though it is a weak acid, it reacts
essentially to completion with the strong
base’s hydroxide ion
1. Stoichiometry Problem: OH- reacts with weak
acid. [ ] of acid remaining and conjugate base
formed are determined
2. Equilibrium Problem: Position of the weak acid
equilibrium is determined, and pH can be
calculated
Example
•
Hydrogen cyanide gas (HCN), a powerful
respiratory inhibitor, is highly toxic. It is a
very weak acid (Ka = 6.2 X 10-10) when
dissolved in water. If a 50.0 mL sample of
0.100 M HCN is titrated with 0.100 M NaOH,
calculate the pH of the solution
a. After 8.00 mL of 0.100 M NaOH has been added.
b. At the halfway point of the titration.
c. At the equivalence point of the titration.
Important Notes
• pH at the equivalence point of a titration
of a weak acid with a strong base is
always greater than 7.00 (basic)
• pH is determined by the amount of
excess OH- present
• Curve looks different before and the
same after the equivalence point
• The AMOUNT, not strength of acid
determines the equivalence point. The
STRENGTH affects the pH at the
equivalence point, however. This pH
affects the titration curve.
Titrations vs. Ka
• Titration curves can be used to determine
equilibrium constant values
• EX: 2.00 mmol of a monoprotic weak acid is
dissolved in 100.0 mL of water and is titrated
with 0.0500 M NaOH. After 20.0 mL NaOH
has been added, the pH is 6.00. What is the
Ka value for the acid?
– Answer: 1.0 X 10-6
Weak Base - Strong Acid Titration
• **Always think about the major species in
solution, then use stoichiometry, then choose
the dominant equilibrium and find pH
• The pH at the equivalence point will be less
than 7.00 (acid)
Example
• Calculate the pH at each of the following points
in the titration of 50.00 mL of a 0.01000 M
sodium phenolate (NaOC6H5) solution with
1.000 M HCl solution (Ka for HOC6H5 = 1.05 X
10-10)…
- Initially
- Answers:
-Midpoint
10.99
- Equivalence Point
9.99
5.99
Determining the Equivalence Point
1. pH meter can be used and a plot of
the titration curve can be made.
2. Acid-base indicator can be used to
see the endpoint (NOT SAME AS
EQUIVALENCE POINT, however
various indicators can be used so this
error won’t be a big deal).
Acid-Base Indicators
• Indicators are represented by HIn. As the H+
ions react with OH- ions from the basic titrant
(are removed from the HIn), In- ions remain.
These In- ions cause the color to change based
on the indicator present.
• Indicators can be chosen where the endpoint
and equivalence point are very close.
– Determine the pH at the equivalence point
– Use pH range chart on page 732