Transcript Lecture 17 Notes

Chapter 6 Problems

6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
6-29

Distinguish between Lewis Acids/Bases
& Bronsted-Lowry acids and bases.
Give an example.
6-31.

Why is the pH of water usually < 7?
How can you prevent this from
happening?
6-39.

The equilibrium constant for autoprotolysis
of water is 1.0 x 10-14 at 25oC. What is the
value of K for 4 H2O  4H+ + 4OH


K = [H+]4 [OH]4
K = (1x10-7)4(1x10-7)4
K = 1 x 10-56
6-41
Use Le
Chatelier’s
principle
and
K
in
w
7E-12
Table6E-12
6-1 to decide whether the
5E-12
autoprotolysis
of water is exothermic or
4E-12
endothermic
at
3E-12
K




25o2E-12
C
oC
1001E-12
0
o
300 C
0
100
200
Temp
300
400
6-42

Make a list of strong acid and strong
bases. Memorize this list.
6-48

Which is the stronger acid?
Dichloracetic acid
Ka = 8 x 10-2
Chloroacetic acid
Ka = 1.36 x 10-3
Stronger Base?
Hydrazine
Kb = 1.1 x 10-6
Urea
Kb = 1.5 x 10-14
Chapter 8
Activity
Homework
Chapter 8 - Activity

8.2, 8.3, 8.6, 8.9, 8.10, 8.12
8-1 Effect of Ionic Strength on
Solubility of Salts

Consider a saturated solution of Hg2(IO3)2
in ‘pure water’. Calculate the concentration
of mercurous ions.
Hg2(IO3)2(s) D Hg22+ + 2IO3I
C
E
some
-x
some-x
+x
+x
+2x
+2x
Ksp=1.3x10-18

K sp  [ Hg22 ][IO3 ]2  1.3 1018
2
18
Ksp  [ x][2x]  1.310
[ x]  6.9 107
A seemingly strange effect is observed when a salt such as KNO3 is
added. As more KNO3 is added to the solution, more solid dissolves until
[Hg22+] increases to 1.0 x 10-6 M. Why?
Increased solubility

Why?

LeChatelier’s Principle?


Complex Ion?



NO – not a product nor reactant
No
Hg22+ and IO3- do not form complexes with K+
or NO3-.
How else?
The Explanation

Consider Hg22+ and the IO3Electrostatic attraction
2+
-
The Explanation

Consider Hg22+ and the IO3Electrostatic attraction
2+ -
Hg2(IO3)2(s)
The Precipitate!!
The Explanation

Consider Hg22+ and the IO3Electrostatic attraction
2+
K+ NO3-
Add KNO3
-
The Explanation

Consider Hg22+ and the IO3NO3NO3
NO3NO
3
NO3
-
K+
-
NO3-
2+
Add KNO3
NO3NO3NO3-
K+
K+
-
NO3
-
K+
K+
K+
K+
-
K+
K+
K+
K+ K+
The Explanation

Consider Hg22+ and the IO3NO3NO3
NO3NO
3
NO3
-
K+
-
NO3-
2+
NO3NO3NO3-
K+
K+
-
NO3-
K+
K+
-
K+
K+
K+
K+ K+ K+
K+
Hg22+ and IO3- can’t get
CLOSE ENOUGH to form Crystal lattice
Or at least it is a lot “Harder” to form crystal lattice
The potassium hydrogen
tartrate example
OH
O
K+-O
OH
O
OH
potassium hydrogen tartrate
Alright, what do we mean by
Ionic strength?

Consider Hg22+ and the IO3NO3NO3
NO3NO
3
NO3
-
K+
-
NO3-
2+
Add KNO3
NO3NO3NO3-
K+
K+
-
NO3
-
K+
K+
K+
K+
-
K+
K+
K+
K+ K+
Low
Ionic
Strength
High
Higher
Ionic
Ionic
Strength
Strength
Alright, what do we mean by
Ionic strength?


measure of the total
concentration of ions in solution.
Ionic strength is a
Ionic strength is dependent on the number of ions
in solution and their charge.

Not dependent on the chemical nature of the ions
Ionic strength (m) = ½ (c1z12+ c2z22 + …)
Or Ionic strength (m) = ½ S cizi2
Examples

Calculate the ionic strength of (a) 0.1 M solution of
KNO3 and (b) a 0.1 M solution of Na2SO4 (c) a
mixture containing 0.1 M KNO3 and 0.1 M Na2SO4.
(m) = ½ (c1z12+ c2z22 + …)
Alright, that’s great but how does
it affect the equilibrium constant?
A+BC+D

Activity = Ac = [C]gc

AND
A A
[C ] g [ D] g
K

b
A A
[ A] g [ B] g
c
C
a
A
d
D
b
B
c
c
C
a a
A
d
d
D
b
B
Relationship between activity
coefficient and ionic strength
Debye-Huckel Equation
0.51z x m
2
 log g x 
1  3.3 x m
m = ionic strength of solution
g = activity coefficient
Z = Charge on the species x
 = effective diameter of ion
(nm)
2 comments
(1) What happens to g when m approaches zero?
(2) Most singly charged ions have an effective radius of about 0.3 nm
We generally don’t need to calculate g – values are tabulated
Concept Test
List at least three properties of activity coefficients
• Dimensionless
• Depends on size of the ions (ex. Hg22+ and IO3-)
• Depends on the Ionic Strength of the Solution (K+ &
NO3-)
• Depends on the charge of the ions (ex. Hg22+ and IO3-)
• In dilute solutions, where ionic strength is minimal, the
activity coefficient -> 1, and has little effect on
equilibrium constant
Activity coefficients are
related to the hydrated
radius of atoms in
molecules
Relationship between m and g
Back to our original problem

Consider a saturated solution of Hg2(IO3)2
in ‘pure water’. Calculate the concentration
of mercurous ions.
Hg2(IO3)2(s) D Hg22+ + 2IO3-
Ksp=1.3x10-18
1
1
K sp  AHg 2 A
2
2
IO3
 [ Hg ]g Hg 2 [ IO ] g
2
2
2
At low ionic strengths g -> 1
 2
3
2
IO3
Back to our original problem

Consider a saturated solution of Hg2(IO3)2
in ‘pure water’. Calculate the concentration
of mercurous ions.
Hg2(IO3)2(s) D Hg22+ + 2IO3-
K sp  AHg 2 A
2
2
IO3
Ksp=1.3x10-18
 [ Hg ]g Hg 2 [ IO ] g
2
2
2
 2
3
In 0.1 M KNO3 - how much Hg22+ will be dissolved?
2
IO3
Back to our original problem

Consider a saturated solution of Hg2(IO3)2
in 0.1 M KNO3. Calculate the
concentration of mercurous ions.
Hg2(IO3)2(s) D Hg22+ + 2IO3-
Ksp=1.3x10-18
K
Kspsp  AAHg
A


[
[
Hg
Hg
]
]
0
g
.
355
[
IO
IO
]
g
0
.
775
22 A
2 [
Hg
Hg
22
IO
IO33
22
K sp  AHg 2 A
18
1.3x10
2
2
IO3
22
22
 22 2
33
IO3
2
 [ x]0.355[2 x] 0.775
2
 0.213122(4x )
3
2
IO3
2
6
1.15 10  x

Consider a saturated solution of
Hg2(IO3)2 Calculate the concentration of
mercurous ions in:
-7
 In Pure water = 6.9 x 10
-6
 in 0.1 M KNO = 1.1 x 10
3
5
pH revisited
Definition of pH



pH = -log AH
or
pH = - log [H+] gH
pH of pure water
H2O (l)  H+ (aq) + OH- (aq)
Kw = AH AOH
Kw = [H+] gH [OH-] gOH
Kw =1.0 x 10-14
pH of pure water
H2O (l)  H+ (aq) + OH- (aq)
I
C
E
+x
+x
Kw = AH AOH
+x
+x
1
Kw = [H+] gH [OH-] gOH
Kw = x2
1
x = 1.0 x 10-7 M
Kw =1.0 x 10-14
pH of pure water
x = 1.0 x 10-7 M
Therefore [H+] = 1.0 x 10-7 M
1
pH = -log AH
= -log [H+] g
H
= - log [1.0 x 10-7]
= 7.00
pH of pure water containing
salt

Calculate the pH of pure water containing
0.10 M KCl at 25oC.