Chapter 6 Problems

6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
Outline

Equilibrium of Acids and Bases

Bronsted-Lowry Acids/Bases






Define strong
Define weak
pH of pure water at 25oC
Define Ka and Kb
Relationship b/w Ka and Kb
Chapter 8 – Activity

Relationship with K
Acids and Bases & Equilibrium
Section 6-7
Strong Bronsted-Lowry Acid

A strong Bronsted-Lowry Acid is one
that donates all of its acidic protons to
water molecules in aqueous solution.
(Water is base – electron donor or the
proton acceptor).

HCl as example
Strong Bronsted-Lowry Base


Accepts protons from water molecules
to form an amount of hydroxide ion,
OH-, equivalent to the amount of base
added.
Example: NH2- (the amide ion)
Question


Can you think of a salt that when dissolved
in water is not an acid nor a base?
Can you think of a salt that when dissolved
in water IS an acid or base?
Weak Bronsted-Lowry acid

One that DOES not donate all of its
acidic protons to water molecules in
aqueous solution.

Example?

Equilibrium
Weak Bronsted-Lowry base


Does NOT accept an amount of protons
equivalent to the amount of base
added, so the hydroxide ion in a weak
base solution is not equivalent to the
concentration of base added.
example:
NH3
Common Classes of Weak
Acids and Bases
Weak Acids


carboxylic acids
ammonium ions
Weak Bases


amines
carboxylate anion
Equilibrium and Water
Question:
Calculate the Concentration of
H+ and OH- in Pure water at 250C.
EXAMPLE: Calculate the Concentration
of H+ and OH- in Pure water at 250C.
Initial
Change
Equilibrium
H2O
H+ + OH-
liquid
-
-
-x
+x
+x
+x
+x
Liquid-x
Kw = [H+][OH-] = ?
KW=(X)(X) = ?
EXAMPLE: Calculate the Concentration
of H+ and OH- in Pure water at 250C.
Initial
Change
Equilibrium
H2O
H+ + OH-
liquid
-
-
-x
+x
+x
+x
+x
Liquid-x
Kw = [H+][OH-] = 1.01 X 10-14
KW=(X)(X) = 1.01 X 10-14
(X) = 1.00 X 10-7
Example
What is the concentration of OH- in a
solution of water that is 1.0 x 10-3 M in
[H+] (@ 25 oC)?
“From now on,
+
Kw = [H ][OH ]
assume the

to
1.0 x 10-14 = [1 x 10-3][OH-]temperature
be 25oC unless
1.0 x 10-11 = [OH-]
otherwise
stated.”
pH
~ -3 -----> ~ +16
pH + pOH = - log Kw = pKw = 14.00
Is there such a thing as Pure
Water?


In most labs the answer is NO
Why?
CO2 + H2O

HCO3- + H+
A century ago, Kohlrausch and his students
found it required to 42 consecutive
distillations to reduce the conductivity to a
limiting value.
Weak Acids and Bases
HA
Ka
H+ + A-
[ H  ][ A ]
Ka 
[ HA]
Ka’s ARE THE SAME
HA + H2O(l)
H3O+ + A-
[ H 3O  ][ A ]
Ka 
[ HA]
Weak Acids and Bases
B + H2O
Kb
BH+ + OH

[ BH ][OH ]
Kb 
[ B]
Relation Between Ka and Kb
Relation between Ka and Kb

Consider Ammonia and its conjugate
acid.
NH3 + H2O
NH4+ + H2O
H2O + H2O
Kb
Ka

NH4+ + OH-
[ NH 4 ][OH  ]
Kb 
[ NH 3 ]
NH3 + H3O+
[ NH 3 ][H 3O ]
Ka 

[ NH 4 ]
OH- + H3O+
[ NH 3 ][H 3O  ] [ NH 4 ][OH ]
K


[ NH 3 ]
w [ NH 43]
K  [H O ][OH ]
Example
The Ka for acetic acid is 1.75 x 10-5. Find Kb
for its conjugate base.
Kw = Ka x Kb
Kw
Kb 
Ka
1.0 1014
10
Kb 

5
.
7

10
1.75105
Example
Calculate the hydroxide ion concentration in a
0.0100 M sodium hypochlorite solution.
OCl- + H2O  HOCl + OH[ HOCl][OH  ]
Kb 
[OCl  ]

The acid dissociation constant = 3.0 x 10-8
1st Insurance Problem
Challenge on page 120
Chapter 8
Activity
Write out the equilibrium constant for
the following expression
Fe3+ + SCN-
D Fe(SCN)2+
[ Fe( SCN ) 2 ]
K
[ Fe3 ][SCN  ]
Q: What happens to K when we add, say KNO3 ?
A: Nothing should happen based on our K, our K is
independent of K+ & NO3-
Keq
K decreases when an inert salt is added!!! Why?
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?
Ionic strength is dependent on the
number of ions in solution and their
charge.
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?

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
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
Anyway … we generally don’t need to calculate g – can get it from a table
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-
K sp  AHg 2 A
2
2
IO3
Ksp=1.3x10-18
 [ 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 ‘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
2
IO3