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Chapter 6 Problems

6.6, 6.9, 6.15, 6.16,

6.19, 6.21, 6.24
Comments on

“Lab Report & Pop Rocks”
6.6

Flip
From the equations
HOCl H+ + OClHOCl + OBr-  HOBr + OClHOBr + OCl-  HOCl + OBr-
K = 3.0 x 10-8
K= 15
K’ = 1/K= 1/15
Find K for
HOBr  H+ + OBr—
K=
? x 10-8/15
K= 3.0
K= 0.2 x 10-8
6.9

The formation of Tetrafluorethlene from
its elements is highly exothermic.

2 F2 (g) + 2 C (s)  F2C=CF2 (g)


(a) if a mixture of F2, graphite, and C2F4 is at
equilibrium in a closed container, will the
reaction go to the right or to the left if F2 is
added?
(b) Rare bacteria … eat C2F4 and make teflon
for cell walls. Will the reaction go to the right or
to the left if these bacteria are added?
6.9

The formation of Tetrafluorethlene from
its elements is highly exothermic.

2 F2 (g) + 2 C (s)  F2C=CF2 (g)



(C) will the reaction go to the right or to the left
if graphite is added?
(d) will reaction go left or right if container is
crushed to one-eighths of original volume?
(e) Does “Q” get larger or smaller if vessel is
Heated?
6-15.

What concentration of Fe(CN)64- is in
equilibrium with 1.0 uM Ag+ and Ag4Fe(CN)6
(s).
Ag4Fe(CN)6  4Ag+ + Fe(CN)64-
Ksp = [Ag+]4 [Fe(CN)64-]
8.5 x 10-45 = [1.0 x 10-6]4 [Fe(CN)64-]
[Fe(CN)64-] = 8.5 x 10-21 M = 8.5 zM
6-16.
Cu4(OH)6(SO4)  4 Cu2+ + 6OH- + SO42I’d first set up an ICE table:
I
C
E
Cu4(OH)6(SO4)
Some
-x
Some –x

4 Cu2+
+ 4x
+ 4x
+ 6 OH
1.0 x 10-6 M
Fixed at 1.0 x 10-6 M
Fixed at 1.0 x 10-6 M
+ SO42+x
+x
Ksp = [Cu2+]4 [OH-]6 [SO42-] = 2.3 x 10-69
Ksp = [4x]4 [1.0 x 10-6]6 [x] = 2.3 x 10-69
X = 9.75 x 10-8 M
Cu2+ = 4x = 3.90 x 10-7 M
Chapter 6
Chemical
Equilibrium
Chemical Equilibrium

Equilibrium Constant




Solubility product (Ksp)
Common Ion Effect
Separation by precipitation
Complex formation
Separation by
Precipitation
Separation by
Precipitation
Complete separation can mean a lot …
we should define complete.
Complete means that the concentration of
the less soluble material has decreased
to 1 X 10-6M or lower before the more
soluble material begins to precipitate
Separation by
Precipitation
EXAMPLE:
Can Fe+3 and Mg+2 be separated
quantitatively as hydroxides from a solution that is
0.10 M in each cation? If the separation is possible,
what range of OH- concentrations is permissible.
Add OH-
Mg2+
Mg2+
Fe3+
Fe3+
Fe3+
3+
Fe
2+
Mg2+
Mg
Mg2+
Mg2+
Fe3+
3+
Fe
3+
Mg2+
2+
Fe
Mg
Fe3+
Fe3+
Fe3+
Mg2+
2+
Mg2+ Mg
Fe3+ Fe3+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+ Mg
Mg2+
Fe3+
@ equilibrium
What is the [OH-] ^when
this happens
Mg2+
2+
Fe(OH)3(s)
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume [Fe+3]eq = 1.0 X 10-6M when “completely”
precipitated.
What will be the [OH-] @ equilibrium required
to reduce the [Fe+3] to [Fe+3] = 1.0 X 10-6M ?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
EXAMPLE: Separate Iron and
Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
(1.0 X 10-6M)*[OH-]3 = 2 X 10-39
[OH  ]3  2 1033
[OH  ]  1.3 1011
Dealing with Mg2+
Find [OH-] to start precipitating Mg2+
• Conceptually –
•Will assume a minimal amount
of Mg2+ will precipitate and
determine the respective
concentration of OH-
Evaluate Q
• If
•Q>K
•Q<K
•Q=K
“Left”
“Right”
“Equilibrium”
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+ Mg
2+
Mg2+
Mg2+
Fe3+
@ equilibrium
-11
[OH-]^ = 1.3 x 10
Is this [OH-] (that is in
solution) great enough
to start precipitating
Mg2+?
Fe(OH)3(s)
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+
Mg2+ Mg
2+
Mg2+
Mg2+
Fe3+
@ equilibrium
-11
[OH-]^ = 1.3 x 10
Is this [OH-] (that is in
solution) great enough
to start precipitating
Mg2+?
Fe(OH)3(s)
EXAMPLE: Separate Iron and
Magnesium?
What [OH-] is required to begin the
precipitation of Mg(OH)2?
Really, Really close to 0.1 M
+2
[Mg ] = 0.10 M
[Mg2+]eq = 0.09999999999999999 M
Ksp = (0.10 M)[OH-]2 = 7.1 X 10-12
[OH-] = 8.4 X 10-6M
EXAMPLE: Separate Iron and
Magnesium?
@ equilibrium
[OH-] to ‘completely’ remove Fe3+
^ -11
= 1.3 X 10 M
[OH-] to start removing Mg2+
= 8.4 X 10-6M
“All” of the Iron will be precipitated b/f any
of the magnesium starts to precipitate!!
EXAMPLE: Separate Iron and
Magnesium?
Q vs. K
Mg(OH)2(s)

Mg2+ + 2OH-
Ksp = [Mg2+][OH-]2 = 7.1 X 10-12
Q = [0.10 M ][1.3 x 10-11 ]2 = 1.69 x 10-23
Q<K
Reaction will proceed to “Right”
Dealing with Mg2+
Find [OH-] to start precipitating Mg2+
• Conceptually –
•Will assume a minimal amount
of Mg2+ will precipitate and
determine the respective
concentration of OH-
NO PPT
Evaluate Q
• If
•Q>K
•Q<K
•Q=K
“Left”
“Right”
“Equilibrium”
“Real Example”

Consider a 1 liter solution that contains
0.3 M Ca2+ and 0.5 M Ba2+.

Can you separate the ions by adding






Sodium
Sodium
Sodium
Sodium
Sodium
Sodium
Carbonate?
Chromate ?
Fluoride?
Hydroxide?
Iodate?
Oxylate?
An example

Consider Lead Iodide
PbI2 (s)
Pb2+ + 2I-
Ksp = 7.9 x 10-9
What should happen if I- is added to a
solution?
Should the solubility go up or down?
Complex Ion Formation
Complex Formation
complex ions (also called coordination ions)
Lewis Acids and Bases
acid => electron pair acceptor (metal)
base => electron pair donor (ligand)
I-
Pb2+
I-
Pb2+
I-
Pb2+
I-
Pb2+
I-
II-
I-
I-
I-
I-
IPb2+
I-
I-
I-
Pb2+
Pb2+
-
-
2+
I-
I-
II-
I-
I-
II-
Pb2+
I-
2+
Pb
-
-
I- Pb-2+
2+
Pb
I
I
2+
-Pb - 2+
2+
I-
I-
I-
II-
I-
I-
I-
I-
I-
I-
IPb2+
I-
I-
I-
I-
II-
II-
2+
Pb
-
-
I- Pb-2+
2+
Pb
I
I
2+
-Pb - 2+
2+
II-
I-
I-
I-
I-
I-
II-
Pb2+
I-
I-
I-
I-
I-
II-
II-
2+
Pb
-
-
I- Pb-2+
2+
Pb
I
I
2+
-Pb - 2+
2+
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
Protic Acids and Bases
Section 6-7
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?
Protic Acids and Bases Salts

Consider Ammonium chloride

Can ‘generally be thought of as the product
of an acid-base reaction.
NH4+Cl- (s)
NH4+ + Cl-
From general chemistry – single positive and single negative
charges are STRONG ELECTROLYTES – they dissolve completely
into ions in dilute aqueous solution
Protic Acids and Bases
Conjugate Acids and Bases in the B-L concept
CH3COOH + H2O  CH3COO- + H3O+
acid
+
base
<=> conjugate base + conjugate acid
conjugate base => what remains after a B-L acid donates its proton
conjugate acid => what is formed when a B-L base accepts a proton
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-] = 1.01 X 10-14
KW=(X)(X) = 1.01 X 10-14
(X) = 1.00 X 10-7
Example
Concentration of OHif [H+] is 1.0 x 10-3 M @ 25 oC?

“From now on,
Kw =
assume the
to
1 x 10-14 = [1 x 10-3][OH-] temperature
be 25oC unless
1 x 10-11 = [OH-]
otherwise
stated.”
[H+][OH-]
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.
6-9 Strengths of Acids and Bases
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)
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
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
base.
NH3 + H2O
NH4 + H2O
+
H2O + H2O
Ka
Kb

NH4+ + OH-
[ NH 4 ][OH  ]
Ka 
[ NH 3 ]

NH3 + H3
O+
OH- + H3O+
[ NH 3 ][ H 3O ]
Kb 

[ 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
1st Insurance Problem
Challenge on page 120