Download AP Chem Unit 6 Chpt 15 acidbase

Unit 6 - Chpt 15 - Acid/Base Equilibria
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Common Ion Effect
Buffers / Buffer Capacity
Titration / pH curves
Acid / Base Indicators
• HW set1: Chpt 15 - pg. 736-742 # 17, 19, 21, 23,
25, 34, 38, 40, 44
Use Appendix 5 for Ka Kb
values - Due Mon. Mar 3
• HW set2: Chpt 15 #49, 50, 52, 59, 103
Common Ion Effect
• Shift in equilibrium position that occurs because of the
addition of an ion already involved in the equilibrium
reaction.
• An application of Le Châtelier’s principle.
HCN(aq) + H2O(l)
H3O+(aq) + CN-(aq)
• Addition of NaCN will shift the equilibrium to the left
because of the addition of CN-, which is already involved in
the equilibrium reaction.
• A solution of HCN and NaCN is less acidic than a solution
of HCN alone.
Ka Problem with common ion
Calculate the pH of a 0.50 M aqueous solution of the
weak acid HF (Ka = 7.2 x 10–4) and 0.10 M NaF (a
strong electrolyte).
Initial
Change
Equilibrium
HF(aq) +
0.50 M
–x
0.50–x
H2O
H3O    A  
K =
HA 
pH = 2.44
H3O+(aq) +
~0
+x
x
F–(aq)
0.10M
+x
0.10M + x
[H3O+]= 3.6 x 10–3
What was pH before adding NaF?
Buffered Solutions
• Buffered Solution – resists a
change in pH.
• They are weak acids or bases
containing a common ion.
• After addition of strong acid or
base, deal with stoichiometry first,
then the equilibrium.
Buffer Problems flow chart
Buffers - How do they work?
Buffers - How do they work? (2)
Henderson–Hasselbalch Equation
 A  
pH = pK a + log
HA 
• For a particular buffering system
(conjugate acid–base pair), all
solutions that have the same ratio
[A–] / [HA] will have the same pH.
Henderson–Hasselbalch Derivation
Ka = [H+] [A-] / [HA]
[H+] = Ka [HA] / [A-]
can use this formula for calculating [H+] in buffer solutions
Take -log of everything
pH = pKa - log ( [HA] / [A-] )
pH = pKa + log ([A-] / [HA] ) base / acid
Buffer Problem
What is the pH of a buffer solution
that is 0.45 M acetic acid
(HC2H3O2) and 0.85 M sodium
acetate (NaC2H3O2)? The Ka for
acetic acid is 1.8 × 10–5.
pH = 5.02
Buffers graphic
Buffers - Recap
• Buffers contain relatively large
amounts of weak acid and
corresponding conjugate base.
• Added H+ reacts to completion with
the conjugate base.
• Added OH reacts to completion
with the weak acid.
Buffers - Recap (2)
• The pH in the buffered solution is
determined by the ratio of the
concentrations of the weak acid and
weak base. As long as this ratio remains
virtually constant, the pH will remain
virtually constant. This will be the case
as long as the concentrations of the
buffering materials (HA and A– or B and
BH+) are large compared with amounts
of H+ or OH– added.
Buffer Capacity
• The amount of protons or hydroxide
ions the buffer can absorb without a
significant change in pH.
• Determined by the magnitudes of [HA]
and [A–].
• A buffer with large capacity contains
large concentrations of the buffering
components.
Buffer Capacity
• Optimal buffering occurs when [HA] is equal
to [A–].
• It is for this condition that the ratio [A–] / [HA]
is most resistant to change when H+ or OH– is
added to the buffered solution.
• pKa of the weak acid to be used in the
buffer should be as close as possible to
the desired pH.
pKa = -log Ka
Titration Curves
• Plotting the pH of the
solution being analyzed
as a function of the
amount of titrant added.
• Equivalence
(Stoichiometric) Point –
point in the titration
when enough titrant has
been added to react
exactly with the
substance in solution
The pH Curve for the Titration of 50.0 mL
being titrated.
of 0.200 M HNO3 with 0.100 M NaOH
Weak acid / strong base problem
Step 1: A stoichiometry problem
(reaction is assumed to run to
completion) then determine remaining
species.
Step 2: An equilibrium problem
(determine position of weak acid
equilibrium and calculate pH).
Concept check - example 1
Consider a solution made by mixing 0.10
mol of HCN (Ka = 6.2 x 10–10) with 0.040
mol NaOH in 1.0 L of aqueous solution.
What are the major species immediately
upon mixing (that is, before a reaction)?
HCN, Na+, OH–, H2O
Concept Check 1 (cont)
The possibilities for the dominant reaction are:
1.
2.
3.
4.
5.
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
HCN(aq) + H2O(l)
H3O+(aq) + CN–(aq)
HCN(aq) + OH–(aq)
CN–(aq) + H2O(l)
Na+(aq) + OH–(aq)
NaOH
Na+(aq) + H2O(l)
NaOH + H+(aq)
Concept check 1 (cont)
• How do we decide which reaction
controls the pH?
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
HCN(aq) + H2O(l)
H3O+(aq) + CN–(aq)
HCN(aq) + OH–(aq)
CN–(aq) + H2O(l)
• Check out Ka’s and Kb’s
Concept check 2
Calculate the pH of a solution made
by mixing 0.20 mol HC2H3O2 (Ka =
1.8 x 10–5) with 0.030 mol NaOH in
1.0 L of aqueous solution.
Concept check 2 (cont)
• What are the major species in solution?
Na+, OH–, HC2H3O2, H2O
• Why isn’t NaOH a major species?
• Why aren’t H+ and C2H3O2– major
species?
Concept check 2 (cont)
• What are the possibilities for the dominant
reaction?
1. H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
2. HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2–(aq)
3. HC2H3O2(aq) + OH–(aq)
C2H3O2–(aq) + H2O(l)
4. Na+(aq) + OH–(aq)
NaOH
5. Na+(aq) + H2O(l)
NaOH + H+(aq)
• Which of these reactions really occur?
Concept check 2 (cont)
• Which reaction controls the pH?
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2–(aq)
HC2H3O2(aq) + OH–(aq)
C2H3O2–(aq) + H2O(l)
• How do you know?
• Ka’s and Kb’s
Concept check 2 (cont)
HC2H3O2(aq) +
Initial
0.170 M
Change
–x
Equilibrium
0.170 – x
H2O
H3O+
~0
+x
x
+ C2H3O2-(aq)
0.030 M
+x
0.030 + x
Ka = 1.8 x 10–5
pH = 3.99
Can also use pH = pKa + log [A-]/[HA]
Concept check 3
Calculate the pH of a solution at the
equivalence point when 100.0 mL of a
0.100 M solution of acetic acid
(HC2H3O2), which has a Ka value of 1.8
x 10–5, is titrated with a 0.10 M NaOH
solution.
pH = 8.72
Acetic Acid / Strong Base
Titration Curve
• The pH Curve for
the Titration of
50.0 mL of 0.100
M HC2H3O2 with
0.100 M NaOH
Weak acid / Strong base
general titration curve
• The pH Curves
for the Titrations
of 50.0-mL
Samples of 0.10
M Acids with
Various Ka
Values with 0.10
M NaOH
Weak Base / Strong Acid Titration
Curve
• The pH Curve for
the Titration of
100.0mL of
0.050 M NH3
with 0.10 M HCl
Molarity as mmol/mL
Molarity M = moles/Liter
(each divide/1000)
because the concentrations tend to be low for titration
problems and we titrate small volumes (mL)
or
Molarity M = mmol/mL
mmol = Molarity x mL
Titration Table for weak base with
strong acid using mmol
Indicators
• Typically a weak acid or weak base
• Marks the end point of a titration by
changing color.
• The equivalence point is not necessarily
the same as the end point (but they are
ideally as close as possible).
Phenolphthalein Indicator
The Acid and
Base Forms of
the Indicator
Phenolphthalein
(clear in acid,
pink in basic
solution)