Transcript ADDITIONAL AQUEOUS EQUILIBRIA CHAPTER 17

Additional Aqueous
Equilibria
CHAPTER 16
I. Buffers
A. Definitions
Buffer- solutions that resist changes in pH
when acid or base are added to it.
- they are composed of a weak acid and
its conjugate base or a weak base and
its conjugate acid.
acetic acid / sodium acetate
carbonic acid / carbonate
B. Explanation of Buffers
1. Shifting of Equilibrium
For some weak acid and its conjugate base:
HA ⇌ H+ + Aa. What happens if you add excess acid (H+)?
b. What happens if you add excess base (OH-)?
** Best buffering capacity when approximately
equal amounts of both acid and conjugate base.
Why strong acids and bases not good buffers?
2. Titration Curve For Buffer
Acetic acid / acetate (CH3COOH / CH3COO-)
3. Mathematical Relationship
a. Henderson-Hasselbalch Equation
For: HA H+ + A[ H + ][ A- ]
Ka =
[HA]

[ A- ]
pH = pK a + log
[HA]
When [HA] = [A-], it is true that pH = pKa
Best buffering region = ±1 unit of pKa
C. Buffer Problem Solving
Similar to other equilibrium problems, except
we will use the Henderson Hasselbalch Equation
base

pH  pKa  log
 acid 
Always utilizes assumption that we can neglect small
amount of dissociation of weak acid or base!!!!!!!!
Different from Tro text, all our buffer problem solving
will be done using the Henderson-Hasselbalch!!!!
1. Determining the pH of a Buffer Solution
What is the pH of a buffered solution that is
0.015 M acetic acid and 0.045 M sodium
acetate?
For acetic acid, Ka = 1.8 x 10-5 ?
2. Preparation of a Buffer Solution
a. Which acid-base pair should be used to prepare
a buffer pH 5.0?
Acetic acid / acetate Ka = 1.8 x 10-5
Ammonium / ammonia Ka = 5.6 x 10-10
b. A buffer solution pH 5.0 is to be prepared using
acetic acid (Ka = 1.8 x 10-5) and sodium acetate .
How many grams of sodium acetate (NaC2H3O2)
should you add to 250.0 mL of a 0.0500 M acetic
acid (HC2H3O2) solution?
3. Addition of Acid or Base To A Buffer
A buffer is prepared by adding 0.30 mol of lactic
acid and 0.40 mol of sodium lactate to sufficient
water to make 2.00 L of buffer solution. The Ka of
lactic acid is 1.4 x 10-4. (Assume Ka small enough
to neglect dissociation of weak acid).
a) Calculate the pH of the buffer.
b) Calculate the pH of the buffer after the addition
of 0.050 moles of NaOH.
When Adding Strong Acid or Base To Buffer:
1. Addition of strong acid to buffer → converts a
stoichiometric amount of the base to the conjugate acid.
2. Addition of strong base to buffer → converts a
stoichiometric amount of the acid to the conjugate base.
(Adding acid creates more acid, adding base creates more base)
CONCEPT QUESTION
A 1.00 liter buffer solution is 0.10 M in HF and
0.050 M in NaF. Which action will destroy the
buffer? Why?
a. addition of 0.050 mol HCl
b. addition of 0.050 mol NaOH
c. addition of 0.050 mol NaF
d. addition of 0.05 mol HF
D. Buffer Capacity
1. Definition- amount (moles) of acid or
base required to change the pH of a
given volume of buffer by ±1 pH unit.
A measure of the ability of a buffer to resist
changes in pH.
Addition of more acid or base will cause the
buffer to fail, meaning inability to resist
changes in pH.
II. Acid-Base Titrations
A. Titration Process
1. Discuss Process (acid + base  salt + water)
2. Equivalence Point: point in titration when
stoichiometric amounts of titrant have been
added to neutralize acid or base titrated.
3. Endpoint: point where titration is actually
ended because of indicator color change.
(Endpoint is used to visually determine equivalence pt.)
B. Explanation of Acid-Base Indicators
1. Weak organic acids which undergo color change
when there is a shift from acid  base at complete
neutralization.
HIn (aq) + H2O (l) ⇌ H3O+ (aq) + In-(aq)
acid
one color
base
another color
2. Different indicators change colors at different pH’s
Bromocresol green: pH 3.8 to 5.4
Phenolphthalein:
pH 8.2 to 10
C. Titration of Strong Acid with Strong Base
Example: HCl + NaOH  H2O + NaCl
Strong Acid-Strong Base Titration Problem
A 50.00 mL sample of 0.100 M HCl is titrated with a
0.100 M NaOH solution.
a.
Determine the pH of the HCl solution before the
addition of any NaOH solution.
a.
Determine the pH after the addition of 25.00 mL
of the NaOH solution.
b.
Determine the pH after the addition of 50.00 mL of
the NaOH solution.
D. Titration of Weak Acid with Strong Base
Example: HC2H3O2 + NaOH  C2H3O2- Na+ + H2O
Weak Acid-Strong Base Titration Problem
A 50.00 mL sample of 0.100 M acetic acid (HC2H3O2, Ka =
1.8 x 10-5) is titrated with a 0.100 M NaOH solution.
a.
Determine the pH of the solution before the addition
of the NaOH solution.
b.
Determine the pH after the addition of 50.00 mL of
the NaOH solution.
E. Titration of Weak Base with Strong Acid
Example: NH3 + HCl  NH4+ Cl1. Is the equivalence point less than or
greater than pH 7? Explain you answer.
2. Draw a titration curve for the titration.
(titrate NH3 using HCl as the titrant)
F. Titration of Polyprotic Acid with Strong Base
Example: H3PO4
III. Solubility Equilibria
A. Equilibrium Expressions for Sparingly
Soluble Salts
1. Ksp :
Solubility Product Constant
(Write like a normal Kc and recall it does not
include solids in the equilibrium expression).
Meaning: idea of extent of solubility of salt
Larger Ksp  Greater solubility of salt
Write the chemical equations and Ksp
expressions for the slightly soluble salts:
AgCl
CaCO3
Ag3PO4
Mg(OH)2
2. Solubility: amount of substance dissolved
based on grams / volume.
3. Molar Solubility: maximum amount of
substance dissolved based on moles / liter.
Molar solubility is not the same as Ksp !!!!!
The molar solubility of BaF2 is 6.27 x 10-3 M. The
Ksp of BaF2 is 9.8 x 10-7. What is the concentration
of F- ions in a BaF2 saturated solution?
B. Problem Solving
1. Calculating Ksp from Solubility Data
One liter of water is able to dissolve 2.15 x 10-3 mol
of PbF2 at 250C.
What is the molar solubility of PbF2?
What is the Ksp for PbF2?
2. Calculating Molar Solubility From Ksp
Determine the molar solubility of barium
phosphate, Ba3(PO4)2, given a Ksp value of
3.4 x 10-23.
What are the equilibrium concentrations of each
ion in solution?
3. Common Ion Effect (Factor Affecting Solubility)
Ag2CO3 has a Ksp of 8.1 x 10-12 at 250C. The molar
solubility of Ag2CO3 in water is 1.3 x 10-4 M. What is
the molar solubility of Ag2CO3 in a 0.10 M NaCO3
solution. (NaCO3 is very soluble).
a. Qualitative Explanation – Le Chatelier’s Principle
Ag2CO3 (s)  2 Ag+ (aq) + CO32- (aq)
What is the effect of adding excess CO32- ?
What is the effect on the molar solubility of Ag2CO3 ?
b. Terminology Used:
Common Ion: the ion which is common
to both salts being considered.
Common Ion Effect: the solubility of a
compound is always lowered due to
the addition of the common ion.
Now solving problem:
4. Predicting If Precipitation Occurs
Comparison of known Ksp to the calculated ion
product (Q) for a given mixture can serve to indicate
whether or not a precipitate will be formed.
Q
Ion Product > Ksp
ppt. forms
Ion Product = Ksp
ppt. forms
Ion Product < Ksp
no ppt. forms
Will a precipitate form at equilibrium when
50.00 mL of 0.00100 M BaCl2 is added to
50.00 mL of 0.000100 M Na2SO4 ? The
solubility product constant for barium sulfate
is 1.1 x 10-10.