Transcript Lecture 15 Notes

Chapter 6 Problems



6.19, 6.21, 6.24
6-29, 6-31, 6-39, 6.41
6-42, 6-48,
6-19.

A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can
99% of Ca2+ be precipitated by sulfate without precipitating
Ag+? What will be the concentration of Ca2+ when Ag2SO4
begins to precipitate?
6-19.

A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can
99% of Ca2+ be precipitated by sulfate without precipitating
Ag+? What will be the concentration of Ca2+ when Ag2SO4
begins to precipitate?
Ca2+ at equilibrium
 0.0500M * 0.01  0.000500M
2+ + SO 22CaSO44  Ca2+
44
=
-5
2=
-5
2+][SO 2-]
Ksp
[0.000500][SO
sp 2.4 x 10 = [Ca
4 ]
4
[SO42-] = 0.048 M
Ag2SO4  2Ag+ + SO42Q = [Ag+]2[SO42-]
Ksp = 1.5 x 10-5
Q = [0.03]2[0.000500]
Q>K
Q = 4.3 x 10-5
6-19.

A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can
99% of Ca2+ be precipitated by sulfate without precipitating
Ag+? What will be the concentration of Ca2+ when Ag2SO4
begins to precipitate?
Ag2SO4  2Ag+ + SO42K/[Ag2+]2 = [SO42-]
Ksp = 1.5 x 10-5
1.5 x 10-5/[0.0300]2 = [SO42-]
1.67 x 10-2= [SO42-]
Find Ca2+
CaSO4  Ca2+ + SO42[Ca2+] = 0.0014 M
Ksp= 2.4 x 10-5 = [Ca2+][1.67 x 10-2]
About 2.8 % remains in solution
6.21

If a solution containing 0.10 M Cl-, Br-, I- and Cr2O42- is
treated with Ag+, in what order will the anions precipitate?
AgCl  Ag+ + ClAgBr  Ag+ + BrAgI  Ag+ + IAg2 CrO4  2Ag+ + CrO4-
K
x 10-10=[Ag][0.1]
=[Ag][Cl]
1.8
x 1.8
10-9=[Ag]
sp =
-13=[Ag][Br]
-12
K
=
5.0
x
10
5.0
sp x 10 =[Ag]=[Ag][0.1]
=[Ag][I]
K
8.3-16x=[Ag]
10-17=[Ag][0.1]
sp =
8.3
x 10
-6=[Ag]
K
1.2
x 10-12 =[Ag]2[0.1]
[Cl]
3.5
x 10
sp =
SOLVE for Ag+ required at equilibrium
6-24.

The cumulative formation constant for SnCl2(aq) in 1.0 M
NaNO3 is b2=12. Find the concentration of SnCl2 for a
solution in which the concentration of Sn2+ and Cl- are
both somehow fixed at 0.20 M.
Sn2+ (aq) + 2Cl- (aq)  SnCl2 (aq)
b2=12
[SnCl2 (aq )]
[SnCl2 (aq )]

b2=12 
2
 2
[0.20][0.20]2
[Sn ][Cl ]
[SnCl2 (aq )]  0.096M
Complex Formation
complex ions (also called coordination ions)
Lewis Acids and Bases
acid => electron pair acceptor (metal)
base => electron pair donor (ligand)
Effects of Complex Ion
Formation on Solubility
Consider the addition of I- to a solution
of Pb+2 ions
Pb2+ + I-
PbI+
[ PbI  ]
2
K1 

1
.
0
x
10
[ Pb2 ][ I  ]
PbI+ + I-
PbI2
K2 = 1.4 x 101
PbI2 + I-
PbI3-
K3 =5.9
PbI3+ I-
PbI42-
K4 = 3.6
Effects of Complex Ion
Formation on Solubility
Consider the addition of I- to a solution
of Pb+2 ions
Pb2+
+
I-
<=>
PbI+
[ PbI  ]
2
K1 

1
.
0
x
10
[ Pb2 ][ I  ]
PbI+ + I- <=> PbI2
K2 = 1.4 x 101
Pb2+ + 2I- <=> PbI2
K’ =?
Overall constants are designated with b
This one is b2
Effects of Complex Ion
Formation on Solubility
Consider the addition of I- to a solution
of Pb+2 ions
Pb2+ + I-
PbI+
[ PbI  ]
2
K1 

1
.
0
x
10
[ Pb2 ][ I  ]
PbI+ + I-
PbI2
K2 = 1.4 x 101
PbI2 + I-
PbI3-
K3 =5.9
PbI3+ I-
PbI42-
K4 = 3.6
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?
Use of double arrows! Said to reach
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 x 10-14 = [1 x 10-3][OH-] temperature
be 25oC unless
1 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
+
H 2 O + H2 O
Kb
Ka

NH4+ + OH-
[ NH 4 ][OH  ]
Kb 
[ NH 3 ]

NH3 + H3
O+
OH- + H3O+
[ NH 3 ][ H 3O ]
Ka 

[ NH 4 ]
[ NH 3 ][ H 3O ] [ NH 4 ][OH ]
K


[ NH 3 ]
[ NH34 ]
w
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-
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
2
2
 2
K sp  [ Hg ][ IO3 ]  1.3 1018
K sp  [ x][2 x]2  1.3 1018
[ 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
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Why?
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Complex Ion?
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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 IO3NO3NO3
NO3NO
3
NO3
-
2+
-
NO3NO3-
Add KNO3
K+
-
NO
Electrostatic
attraction
3
K+
NO3- NO3
K+
K+
-
K+
K+
The Explanation

Consider Hg22+ and the IO3NO3NO3
NO3-
NO3NO
3
K+
NO3-
K+
2+
-
NO3
NO3-
K+
-
NO3- NO3
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