CHEMISTRY REVIEW
ROUND 3
Kinetics, Equilibrium, Acids
and Bases and Buffers
KINETICS
• The study of reaction rates.
• Spontaneous reactions are reactions that
will happen - but we can’t tell how fast.
• Diamond will spontaneously turn to
graphite – eventually.
• Reaction mechanism- the steps by which a
reaction takes place.
REACTION RATE
• Rate = Conc. of A at t2 -Conc. of A at t1
t2- t1
• Rate = D[A]
Dt
• Change in concentration per unit time
• For this reaction
• N2 + 3H2
2NH3
• As the reaction progresses the concentration H2
goes down
C
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[H2]
Time
• As the reaction progresses the concentration N2 goes
down 1/3 as fast
C
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[N2]
[H2]
Time
• As the reaction progresses the concentration NH3
goes up.
C
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[N2]
[H2]
[NH3]
Time
RATE LAWS
• Reactions are reversible.
• As products accumulate they can begin to
turn back into reactants.
• Early on the rate will depend on only the
amount of reactants present.
• We want to measure the reactants as soon
as they are mixed.
• This is called the Initial rate method.
Rate Laws
• Two key points
• The concentration of the products do not
appear in the rate law because this is an
initial rate.
• The order must be determined
experimentally,
• can’t be obtained from the equation
2 NO2
2 NO + O2
• You will find that the rate will only depend
on the concentration of the reactants.
n
• Rate = k[NO2]
• This is called a rate law expression.
• k is called the rate constant.
• n is the order of the reactant -usually a
positive integer.
TYPES OF RATE LAWS
• Differential Rate law - describes how rate
depends on concentration.
• Integrated Rate Law - Describes how
concentration depends on time.
• For each type of differential rate law there
is an integrated rate law and vice versa.
• Rate laws can help us better understand
reaction mechanisms.
DETERMINING RATE LAWS
• The first step is to determine the form
of the rate law (especially its order).
• Must be determined from
experimental data.
• For this reaction
2 N2O5 (aq)
4NO2 (aq) + O2(g)
The reverse reaction won’t play a role
THE METHOD OF INITIAL RATES
• This method requires that a reaction be run
several times.
• The initial concentrations of the reactants
are varied.
• The reaction rate is measured bust after the
reactants are mixed.
• Eliminates the effect of the reverse
reaction.
AN EXAMPLE
• For the reaction
+
BrO3 + 5 Br + 6H
3Br2 + 3 H2O
• The general form of the Rate Law is
n
m
+
p
Rate = k[BrO3 ] [Br ] [H ]
• We use experimental data to
determine the values of n,m,and p
Initial concentrations (M)
Rate (M/s)
BrO30.10
0.20
0.20
0.10
Br0.10
0.10
0.20
0.10
H+
0.10
0.10
0.10
0.20
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Now we have to see how the rate
changes with concentration
INTEGRATED RATE LAW
• Expresses the reaction concentration
as a function of time.
• Form of the equation depends on the
order of the rate law (differential).
n
• Changes Rate = D[A]
Dt
• We will only work with n=0, 1, and 2
FIRST ORDER
• For the reaction 2N2O5
4NO2 + O2
• We found the Rate = k[N2O5]1
• If concentration doubles rate doubles.
• If we integrate this equation with
respect to time we get the Integrated
Rate Law
• ln[N2O5] = - kt + ln[N2O5]0
• ln is the natural log
• [N2O5]0 is the initial concentration.
FIRST ORDER
• General form Rate = D[A] / Dt = k[A]
• ln[A] = - kt + ln[A]0
• In the form y = mx + b
• y = ln[A]
m = -k
•x=t
b = ln[A]0
• A graph of ln[A] vs time is a straight
line.
FIRST ORDER
• By getting the straight line you can
prove it is first order
• Often expressed in a ratio
FIRST ORDER
• By getting the straight line you can prove it is first
order
• Often expressed in a ratio
  A 0 
ln
 = kt
  A 
HALF LIFE
• The time required to reach
half the original
concentration.
• If the reaction is first order
• [A] = [A]0/2 when t = t1/2
HALF LIFE
• The time required to reach half the
original concentration.
• If the reaction is first order
• [A] = [A]0/2 when t = t1/2
•ln(2) = kt1/2

 A 0

ln
  0
 A 2

 = kt
12


HALF LIFE
• t1/2 = 0.693/k
• The time to reach half the original
concentration does not depend
on the starting concentration.
• An easy way to find k
SECOND ORDER
• Rate = -D[A] / Dt = k[A]2
• integrated rate law
• 1/[A] = kt + 1/[A]0
• y= 1/[A]
m=k
• x= t
b = 1/[A]0
• A straight line if 1/[A] vs t is graphed
• Knowing k and [A]0 you can calculate
[A] at any time t
SECOND ORDER HALF LIFE
• [A] = [A]0 /2 at t = t1/2
1
[ A ]0
2
[ A] 0
1
[A] 0
2
1
= kt1 2 +
[A] 0
1
= kt 1 2
[ A] 0
= kt 1 2
t1 2 =
1
k[A] 0
ZERO ORDER RATE LAW
• Rate = k[A]0 = k
• Rate does not change with
concentration.
• Integrated [A] = -kt + [A]0
• When [A] = [A]0 /2 t = t1/2
• t1/2 = [A]0 /2k
ZERO ORDER RATE LAW
• Most often when reaction
happens on a surface because
the surface area stays constant.
• Also applies to enzyme chemistry.
C
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i
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Time
C
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c
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k = DA]/Dt
DA]
Dt
Time
MORE COMPLICATED
REACTIONS
+
• BrO3 + 5 Br + 6H
3Br2 + 3 H2O
• For this reaction we found the rate law to
be
-
-
+2
• Rate = k[BrO3 ][Br ][H ]
• To investigate this reaction rate we need to
control the conditions
SUMMARY OF RATE LAWS
REACTION MECHANISMS
• The series of steps that actually occur
in a chemical reaction.
• Kinetics can tell us something about
the mechanism
• A balanced equation does not tell us
how the reactants become products.
REACTION MECHANISMS
• 2NO2 + F2
2NO2F
• Rate = k[NO2][F2]
• The proposed mechanism is
• NO2 + F2
NO2F + F
(slow)
• F + NO2
NO2F
(fast)
• F is called an intermediate It is
formed then consumed in the
reaction
REACTION MECHANISMS
• Each of the two reactions is called an
elementary step .
• The rate for a reaction can be written from
its molecularity .
• Molecularity is the number of pieces that
must come together.
• Unimolecular step involves one
molecule - Rate is rirst order.
• Bimolecular step - requires two
molecules - Rate is second order
• Termolecular step- requires three
molecules - Rate is third order
• Termolecular steps are almost never
heard of because the chances of
three molecules coming into contact
at the same time are miniscule.
•A
• A+A
• 2A
• A+B
• A+A+B
• 2A+B
• A+B+C
products
products
products
products
Products
Products
Products
Rate = k[A]
Rate= k[A]2
Rate= k[A]2
Rate= k[A][B]
Rate= k[A]2[B]
Rate= k[A]2[B]
Rate=k[A][B][C]
COLLISION THEORY
• Molecules must collide to react.
• Concentration affects rates because
collisions are more likely.
• Must collide hard enough.
• Temperature and rate are related.
• Only a small number of collisions
produce reactions.
P
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Reactants
E
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Products
Reaction Coordinate
P
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Activation
Energy Ea
Reactants
E
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Products
Reaction Coordinate
P
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t
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Activated
complex
Reactants
E
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g
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Products
Reaction Coordinate
P
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Reactants
E
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y
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Products
Reaction Coordinate
DE
TERMS
• Activation energy - the minimum
energy needed to make a
reaction happen.
• Activated Complex or Transition
State - The arrangement of atoms
at the top of the energy barrier.
ARRHENIUS
• Said the at reaction rate should
increase with temperature.
• At high temperature more molecules
have the energy required to get over
the barrier.
• The number of collisions with the
necessary energy increases
exponentially.
ARRHENIUS EQUATION
•k = Ae-Ea/RT
• A is called the frequency factor
• ln k = -(Ea/R)(1/T) + ln A
• Another line !!!!
• ln k vs t is a straight line
CATALYSTS
• Speed up a reaction without being used up
in the reaction.
• Enzymes are biological catalysts.
• Homogenous Catalysts are in the same
phase as the reactants.
• Heterogeneous Catalysts are in a different
phase as the reactants.
HOW CATALYSTS WORK
• Catalysts allow reactions to proceed by a
different mechanism - a new pathway.
• New pathway has a lower activation
energy.
• More molecules will have this activation
energy.
• Do not change DE
HETEROGENOUS CATALYSTS
H H
• Hydrogen bonds to surface of
metal.
H H
• Break H-H bonds
H H
H
H
Pt surface
HETEROGENOUS CATALYSTS
H
H
H
C
C
H
H H
H
H
Pt surface
HETEROGENOUS CATALYSTS
• The double bond breaks and bonds to the catalyst.
H
H
H
C
H
C
H
H
Pt surface
H H
HETEROGENOUS CATALYSTS
• The hydrogen atoms bond with the carbon
H
H
H
C
H
C
H
H
Pt surface
H H
HETEROGENOUS CATALYSTS
H
H
H
H
C
C
H
H
H
Pt surface
H
HOMOGENOUS CATALYSTS
• Chlorofluorocarbons catalyze the
decomposition of ozone.
• Enzymes regulating the body
processes. (Protein catalysts)
CATALYSTS AND RATE
• Catalysts will speed up a reaction but
only to a certain point.
• Past a certain point adding more
reactants won’t change the rate.
• Zero Order
CATALYSTS AND RATE.
R
a
t
e
• Rate increases until the active sites of catalyst
are filled.
• Then rate is independent of concentration
Concentration of reactants
REACTIONS ARE
REVERSIBLE
•A +B
C + D ( forward)
•C + D
A + B (reverse)
• Initially there is only A and B so only
the forward reaction is possible
• As C and D build up, the reverse
reaction speeds up while the forward
reaction slows down.
• Eventually the rates are equal
Reaction Rate
Forward Reaction
Equilibrium
Reverse reaction
Time
A
Experiment
Initial [NO] (mol L- Initial [O2] (mol L-1)
1
1
2
3
)
0.10
0.20
0.20
0.10
0.10
0.40
Initial Rate of
Formation of NO2 (mol
L-1s-1)
2.5x10–4
5.0x10–4
8.0x10–3
36.
The initial-rate data in the table above were
obtained for the reaction represented below. What is the
experimental rate law for the reaction?
2 NO(g) + O2(g) ↔ NO2(g)
(A)
Rate = k[NO][O2]
(B)
Rate = k[NO][O2]2
(C)
Rate = k[NO]2[O2]
(D)
Rate = k[NO]2[O2]2
(E)
Rate = k([NO]/[O2])
B
D
C
C
D
63.
The graph above shows
the results of a study of the
reaction of X with a large
excess of Y to yield Z. The
concentrations of X and Y were
measured over a period of time.
According to the results, which
of the following can be
concluded about the rate law for
the reaction under the
conditions studied?
(A)
It is zero order in [X].
(B)
It is first order in [X].
(C)
It is second order in [X].
(D)
It is first order in [Y].
(E)
The overall order of the
reaction is 2.
B
EQUILIBRIUM
WHAT IS EQUAL AT
EQUILIBRIUM?
• Rates are equal
• Concentrations are not.
• Rates are determined by
concentrations and activation
energy.
• The concentrations do not change at
equilibrium.
• or if the reaction is verrrry slooooow.
LAW OF MASS ACTION
• For any reaction
• jA + kB
lC + mD
• K = [C]l[D]m
[A]j[B]k
PRODUCTSpower
REACTANTSpower
• K is called the equilibrium constant.
•
is how we indicate a reversible
reaction
PLAYING WITH K
• If we write the reaction in reverse.
• lC + mD
jA + kB
• Then the new equilibrium constant is
j
k
• K’ = [A] [B] = 1/K
l
m
[C] [D]
PLAYING WITH K
• If we multiply the equation by a constant
• njA + nkB
nlC + nmD
• Then the equilibrium constant is
•K’ =
nj nk
[A] [B]
=
nl
nm
[C] [D]
([A] j[B]k)n
l
m
n
([C] [D] )
=
n
K
THE UNITS FOR K
• Are determined by the various
powers and units of
concentrations.
• They depend on the reaction.
EQUILIBRIUM
CONSTANT
One for each Temperature
CALCULATE K
• N2 + 3H2
3NH3
• Initial
At Equilibrium
• [N2]0 =1.000 M [N2] = 0.921M
• [H2]0 =1.000 M
• [NH3]0 =0 M
[H2] = 0.763M
[NH3] = 0.157M
CALCULATE K
• N2 + 3H2
• Initial
• [N2]0 = 0 M
• [H2]0 = 0 M
3NH3
At Equilibrium
[N2] = 0.399 M
[H2] = 1.197 M
• [NH3]0 = 1.000 M
[NH3] = 0.157M
• K is the same no matter what the
amount of starting materials
EQUILIBRIUM AND
PRESSURE
• Some reactions are gaseous
• PV = nRT
• P = (n/V)RT
• P = MRT
• M is a concentration in moles/Liter
• M = P/RT
EQUILIBRIUM AND
PRESSURE
• 2SO2(g) + O2(g)
•
Kp =
2SO3(g)
(PSO3)2
(PSO2)2 (PO2)
•
K=
[SO3]2
[SO2]2 [O2]
•
•
•
K=
K=
EQUILIBRIUM AND
PRESSURE
(PSO3/RT)2
(PSO2/RT)2(PO2/RT)
(PSO3)2 (1/RT)2
(PSO2)2(PO2) (1/RT)3
K = Kp
(1/RT)2
(1/RT)3
= Kp RT
GENERAL EQUATION
• jA + kB
lC + mD
(l+m)-(j+k)
=
Dn
• Kp = K (RT)
K (RT)
• Dn=(l+m)-(j+k)=Change in moles of gas
HOMOGENEOUS
EQUILIBRIA
• So far every example dealt with
reactants and products where all
were in the same phase.
• We can use K in terms of either
concentration or pressure.
• Units depend on reaction.
HETEROGENEOUS
EQUILIBRIA
• If the reaction involves pure solids or pure
liquids the concentration of the solid or the
liquid doesn’t change.
• As long s they are not used up they are not
used up we can leave them out of the
equilibrium expression.
• For example
FOR EXAMPLE
• H2(g) + I2(s)
2
• K = [HI]
2HI(g)
[H2][I2]
• But the concentration of I2 does not
change.
SOLVING EQUILIBRIUM
PROBLEMS
Type 1
THE REACTION QUOTIENT
• Tells you the directing the reaction will go to reach
equilibrium
• Calculated the same as the equilibrium constant, but
for a system not at equilibrium
coefficient
coefficient
[Reactants]
• Q = [Products]
• Compare value to equilibrium constant
WHAT Q TELLS US
• If K>Q
Not enough products
Shift to right
• If K<Q
Too many products
Shift to left
• If Q=K system is at equilibrium
EXAMPLE
• for the reaction
• 2NOCl(g)
2NO(g) + Cl2(g)
-5
• K = 1.55 x 10 M at 35ºC
• In an experiment 0.10 mol NOCl, 0.0010
mol NO(g) and 0.00010 mol Cl2 are mixed
in 2.0 L flask.
• Which direction will the reaction proceed
to reach equilibrium?
SOLVING EQUILIBRIUM
PROBLEMS
• Given the starting concentrations and
one equilibrium concentration.
• Use stoichiometry to figure out other
concentrations and K.
• Learn to create a table of initial and
final conditions.
• Consider the following reaction at
600ºC
• 2SO2(g) + O2(g)
2SO3(g)
• In a certain experiment 2.00 mol of
SO2, 1.50 mol of O2 and 3.00 mol of
SO3 were placed in a 1.00 L flask. At
equilibrium 3.50 mol were found to be
present.
• Calculate the equilibrium
concentrations of O2 and SO2, K and
KP
• Consider the same reaction at 600ºC
• In a different experiment 0.500 mol
SO2 and 0.350 mol SO3 were placed
in a 1.000 L container. When the
system reaches equilibrium 0.045 mol
of O2 are present.
• Calculate the final concentrations of
SO2 and SO3 and K
WHAT IF YOU’RE NOT GIVEN
EQUILIBRIUM CONCENTRATION?
• The size of K will determine what
approach to take.
• First let’s look at the case of a small
value of K ( <0.01).
• Allows us to make simplifying
assumptions.
EXAMPLE
• H2(g) + I2(g)
2HI(g)
-3
• K = 7.1 x 10 at 25ºC
• Calculate the equilibrium
concentrations if a 5.00 L container
initially contains 15.9 g of H2 294 g I2 .
• [H2]0 = (15.7g/2.02)/5.00 L = 1.56 M
• [I2]0 = (294g/253.8)/5.00L = 0.232 M
• [HI]0 = 0
• Q= 0<K so more product will
be formed.
• Assumption since K is small
reaction is reactant favored
at equilibrium.
• Set up table of initial, final and
change in concentrations.
initial
change
final
H2(g)
1.56 M
I2(g)
0.232 M
• Using to stoichiometry we can find
• Change in H2 = -X
• Change in HI = -X
• Change in HI = +2X
HI(g)
0M
H2(g)
initial
1.56 M
change -X
final
I2(g)
0.232 M
-X
HI(g)
0M
+2X
• Now we can determine the final concentrations by
adding.
H2(g)
initial
1.56 M
change -X
final
1.56-X
I2(g)
0.232 M
-X
0.232-X
HI(g)
0M
+2X
2X
• Now plug these values into the equilibrium expression
• K=
2
(2X)
(1.56-X)(0.232-X)
-3
= 7.1 x 10
2
• K = (2X)
WHY WE CHOSE X
-3
= 7.1 x 10
(1.56-X)(0.232-X)
• Since X is going to be small, we can
ignore it in relation to 0.464 and 1.328
• So we can rewrite the equation
-3
2
• 7.1 x 10 = (2X)
(1.56)(0.232)
• Makes the algebra easy
CHECKING THE
ASSUMPTION
• The rule of thumb is that if the value of
X is less than 5% of all the other
concentrations, our assumption was
valid.
• If not we would have had to use the
quadratic equation
• More on this later.
• Our assumption was valid.
PRACTICE
• For the reaction Cl2 + O2
2ClO(g) K =
156
• In an experiment 0.100 mol ClO, 1.00 mol
O2 and 0.0100 mol Cl2 are mixed in a 4.00 L
flask.
• If the reaction is not at equilibrium, which
way will it shift?
• Calculate the equilibrium concentrations.
ANOTHER PROBLEM
WITH SMALL K
K< .01
• For the reaction
2NOCl
-5
FOR EXAMPLE
2NO +Cl2
• K= 1.6 x 10
• If 1.20 mol NOCl, 0.45 mol of NO and 0.87
mol Cl2 are mixed in a 1 L container
• What are the equilibrium concentrations
2
2
• Q = [NO] [Cl2] = (0.45) (0.87) = 0.15 M
2
2
[NOCl]
(1.20)
Initial
2NOCl
2NO
Cl2
1.20
0.45
0.87
+2X
+X
Change -2X
Final
1.20-2X 0.45+2X
0.87+X
 K = (0.45+2X)2(0.87+X)
= 1.6 x 10-5
(1.20-2X)2

Figure out X
Initial
2NOCl
2NO
Cl2
1.20
0.45
0.87
+2X
+X
Change -2X
Final
1.20-2X
0.45+2X
Make assumptions
 K = (0.45)2(0.87) = 1.6 x 10-5
(1.20-2X)2
 X= 8.2 x 10-3

0.87+X
Initial
2NOCl
2NO
Cl2
1.20
0.45
0.87
+2X
+X
Change -2X
Final
1.20-2X
0.45+2X
Check assumptions
 2(.0082)/1.20 = 1.2 % OKAY!!!

0.87+X
PRACTICE PROBLEM
• For the reaction
2ClO(g)
Cl2 (g) + O2 (g)
• K = 6.4 x 10-3
• In an experiment 0.100 mol ClO(g),
1.00 mol O2 and 1.00 x 10-2 mol Cl2 are
mixed in a 4.00 L container.
• What are the equilibrium
concentrations.
PROBLEMS INVOLVING
PRESSURE
• Solved exactly the same, with same rules
for choosing X depending on KP
• For the reaction N2O4(g)
2NO2(g) KP
= .131 atm. What are the equilibrium
pressures if a flask initially contains 1.000
atm N2O4?
LE CHATELIER’S
PRINCIPLE
• If a stress is applied to a system at
equilibrium, the position of the
equilibrium will shift to reduce the
stress.
• 3 Types of stress
CHANGE AMOUNTS OF
REACTANTS AND/OR PRODUCTS
• Adding product makes K<Q
• Removing reactant makes K<Q
• Adding reactant makes K>Q
• Removing product makes K>Q
• Determine the effect on Q, will tell
you the direction of shift
CHANGE IN PRESSURE
• By changing volume
• System will move in the direction that
has the least moles of gas.
• Because partial pressures (and
concentrations) change a new
equilibrium must be reached.
• System tries to minimize the moles of
gas.
CHANGE IN PRESSURE
• By adding an inert gas
• Partial pressures of reactants and
product are not changed
• No effect on equilibrium position
CHANGE IN
TEMPERATURE
• Affects the rates of both the forward
and reverse reactions.
• Doesn’t just change the equilibrium
position, changes the equilibrium
constant.
• The direction of the shift depends on
whether it is exo- or endothermic
EXOTHERMIC
• DH<0
• Releases heat
• Think of heat as a product
• Raising temperature push toward
reactants.
• Shifts to left.
ENDOTHERMIC
• DH>0
• Produces heat
• Think of heat as a reactant
• Raising temperature push toward
products.
• Shifts to right.
2 SO3(g)  2 SO2(g) + O2(g)
41.
After the equilibrium represented above is established,
some pure O2(g) is injected into the reaction vessel at constant
temperature. After equilibrium is re-established, which of the
following has a lower value compared to its value at the original
equilibrium?
(A)
Keq for the reaction
(B)
The total pressure in the reaction vessel
(C)
The amount of SO3(g) in the reaction vessel
(D)
The amount of O2(g) in the reaction vessel
(E)
The amount of SO2(g) in the reaction vessel
E
C
A
A
B
B
ARRHENIUS DEFINITION
• Acids produce hydrogen ions in aqueous
solution.
• Bases produce hydroxide ions when
dissolved in water.
• Limits to aqueous solutions.
• Only one kind of base.
• NH3 ammonia could not be an Arrhenius
base.
BRONSTED-LOWRY
DEFINITIONS
+
• And acid is an proton (H ) donor and a base is a
proton acceptor.
• Acids and bases always come in pairs.
• HCl is an acid..
• When it dissolves in water it gives its proton to
water.
• HCl(g) + H2O(l)
H3O
+ + Cl-
• Water is a base makes hydronium ion
• General equation
• HA(aq) + H2O(l)
• Acid + Base
PAIRS
+
H3O (aq) + A (aq)
Conjugate acid +
Conjugate base
• This is an equilibrium.
+
• Competition for H between H2O
and A
• The stronger base controls direction.
+
• If H2O is a stronger base it takes the H
• Equilibrium moves to right.
ACID DISSOCIATION
CONSTANT KA
• The equilibrium constant for the general
equation.
• HA(aq) + H2O(l)
+
• Ka = [H3O ][A ]
+
[HA]
+
H3O (aq) + A (aq)
+
• H3O is often written H ignoring the water
in equation (it is implied).
ACID DISSOCIATION
CONSTANT KA
+
• HA(aq)
-
H (aq) + A (aq)
+
-
• Ka = [H ][A ]
[HA]
• We can write the expression for any acid.
• Strong acids dissociate completely.
• Equilibrium far to right.
• Conjugate base must be weak.
BACK TO PAIRS
• Strong acids
• Ka is large
• [H+] is equal to
[HA]
• A- is a weaker
base than water
• Weak acids
• Ka is small
• [H+] <<< [HA]
• A- is a stronger
base than water
TYPES OF ACIDS
• Polyprotic Acids- more than 1 acidic
hydrogen (diprotic, triprotic).
• Oxyacids - Proton is attached to the
oxygen of an ion.
• Organic acids contain the Carboxyl group COOH with the H attached to O
• Generally very weak.
AMPHOTERIC
• Behave as both an acid and a base.
• Water autoionizes
+
+
+
• KW= [H3O ][OH ]=[H ][OH ]
-14
• At 25ºC KW = 1.0 x10
• 2H2O(l)
H3O (aq) + OH (aq)
• In EVERY aqueous solution.
+
-7
• Neutral solution [H ] = [OH ]= 1.0 x10
+
• Acidic solution [H ] > [OH ]
+
• Basic solution [H ] < [OH ]
PH
+
• pH= -log[H ]
+
• Used because [H ] is usually very small
+
• As pH decreases, [H ] increases
exponentially
• Sig figs only the digits after the decimal
place of a pH are significant
+
-8
• [H ] = 1.0 x 10
pH= 8.00 2 sig figs
• pOH= -log[OH ]
• pKa = -log K
+
-
RELATIONSHIPS
• KW = [H ][OH ]
+
+
= -log[H ]+ -log[OH ]
• -log KW = -log([H ][OH ])
• -log KW
• pKW = pH + pOH
• KW = 1.0 x10
-14
• 14.00 = pH + pOH
+
-
• [H ],[OH ],pH and pOH
Given any one of these we can find the
other three.
[H+]
100 10-1
0
1
Acidic
14
13
10-3 10-5 10-7 10-9 10-11 10-13 10-14
pH
3
11
5
9
7
9
Neutral
7
5
11
13
14
Basic
3
1
0
pOH
10-14 10-13 10-11 10-9Basic
10-7 10-5 10-3 10-1 100
[OH-]
CALCULATING PH OF
SOLUTIONS
• Always write down the major ions in
solution.
• Remember these are equilibria.
• Remember the chemistry.
• Don’t try to memorize there is no one
way to do this.
STRONG ACIDS
• HBr, HI, HCl, HNO3, H2SO4, HClO4
• ALWAYS WRITE THE MAJOR SPECIES
• Completely dissociated
+
• [H ] = [HA]
• [OH ] is going to be small because of
equilibrium
-14
+
• 10
= [H ][OH ]
-7
+
• If [HA]< 10 water contributes H
WEAK ACIDS
• Ka will be small.
• ALWAYS WRITE THE MAJOR SPECIES.
• It will be an equilibrium problem from
the start.
EXAMPLE
• Calculate the pH of 2.0 M acetic acid
-5
HC2H3O2 with a Ka 1.8 x10
+
• Calculate pOH, [OH ], [H ]
A MIXTURE OF WEAK
ACIDS
• The process is the same.
• Determine the major species.
• The stronger will predominate.
• Bigger Ka if concentrations are comparable
• Calculate the pH of a mixture 1.20 M HF (Ka =
-4
-10
7.2 x 10 ) and 3.4 M HOC6H5 (Ka = 1.6 x 10
)
PERCENT DISSOCIATION
•=
amount dissociated
x 100
initial concentration
• For a weak acid percent dissociation
increases as acid becomes more dilute.
• Calculate the % dissociation of 1.00 M
and .00100 M Acetic acid (Ka = 1.8 x 10-5)
• As [HA]0 decreases [H+] decreases but %
dissociation increases.
• Le Chatelier
THE OTHER WAY
• What is the Ka of a weak acid that is 8.1 %
dissociated as 0.100 M solution?
BASES
-
• The OH is a strong base.
• Hydroxides of the alkali metals are strong
bases because they dissociate completely
when dissolved.
• The hydroxides of alkaline earths Ca(OH)2
etc. are strong dibasic bases, but they
don’t dissolve well in water.
• Used as antacids because [OH ] can’t
build up.
BASES WITHOUT OH
• Bases are proton acceptors.
• NH3 + H2O
+ + OH-
NH4
• It is the lone pair on nitrogen that accepts the
proton.
• Many weak bases contain N
• B(aq) + H2O(l)
• Kb = [BH+][OH- ]
[ B]
BH
+(aq) + OH- (aq)
STRENGTH OF BASES
• Hydroxides are strong.
• Others are weak.
• Smaller Kb weaker base.
• Calculate the pH of a solution of 4.0 M pyridine
-9
(Kb = 1.7 x 10 )
N:
POLYPROTIC ACIDS
• Always dissociate stepwise.
+
• The first H comes of much easier than the
second.
• Ka for the first step is much bigger than Ka
for the second.
• Denoted Ka1, Ka2, Ka3
POLYPROTIC ACID
• H2CO3
• HCO3
+
H + HCO3
-7
Ka1= 4.3 x 10
+
-2
H + CO3
-10
Ka = 4.3 x 10
2
• Base in first step is acid in second.
• In calculations we can normally
ignore the second dissociation.
CALCULATE THE
CONCENTRATION
• Of all the ions in a solution of 1.00 M Arsenic
acid H3AsO4
Ka1 = 5.0 x 10-3
Ka2 = 8.0 x 10-8
Ka3 = 6.0 x 10-10
SULFURIC ACID IS
SPECIAL
• In first step it is a strong acid.
-2
• Ka2 = 1.2 x 10
• Calculate the concentrations in a 2.0 M
solution of H2SO4
-3
• Calculate the concentrations in a 2.0 x 10
M solution of H2SO4
SALTS AS ACIDS AN
BASES
• Salts are ionic compounds.
• Salts of the cation of strong bases and the
anion of strong acids are neutral.
• for example NaCl, KNO3
• There is no equilibrium for strong acids and
bases.
• We ignore the reverse reaction.
BASIC SALTS
• If the anion of a salt is the conjugate base
of a weak acid - basic solution.
• In an aqueous solution of NaF
+ HF + OH
• The major species are Na , F , and H2O
• F + H 2O
• Kb =[HF][OH ]
[F ]
+
-
• but Ka = [H ][F ]
[HF]
-
• Ka x Kb = [HF][OH ]
[F ]
BASIC SALTS
x
+ -
[H ][F ]
[HF]
BASIC SALTS
+
• Ka x Kb = [HF][OH ] x [H ][F ]
[F ]
[HF]
+
• Ka x Kb =[OH ] [H ]
• K a x Kb = K W
KA TELLS US KB
• The anion of a weak acid is a weak base.
• Calculate the pH of a solution of 1.00 M
-10
NaCN. Ka of HCN is 6.2 x 10
-
-
• The CN ion competes with OH for the H
+
ACIDIC SALTS
• A salt with the cation of a weak base and
the anion of a strong acid will be basic.
• The same development as bases leads to
• Ka x K b = K W
• Calculate the pH of a solution of 0.40 M
NH4Cl (the Kb of NH3 1.8 x 10-5).
• Other acidic salts are those of highly
charged metal ions.
• More on this later.
ANION OF WEAK ACID,
CATION OF WEAK BASE
• Ka > Kb
acidic
• Ka < Kb
basic
• Ka = Kb
Neutral
STRUCTURE AND ACID BASE
PROPERTIES
• Any molecule with an H in it is a potential
acid.
• The stronger the X-H bond the less acidic
(compare bond dissociation energies).
• The more polar the X-H bond the stronger
the acid (use electronegativities).
• The more polar H-O-X bond -stronger acid.
STRENGTH OF OXYACIDS
• The more oxygen hooked to the central
atom, the more acidic the hydrogen.
• HClO4 > HClO3 > HClO2 > HClO
• Remember that the H is attached to an
oxygen atom.
• The oxygens are electronegative
• Pull electrons away from hydrogen
Strength of oxyacids
Electron Density
Cl
O
H
Strength of oxyacids
Electron Density
O
Cl
O
H
Strength of oxyacids
Electron Density
O
Cl
O
O
H
STRENGTH OF OXYACIDS
Electron Density
O
O
O
Cl
O
H
ACID-BASE PROPERTIES OF OXIDES
• Non-metal oxides dissolved in water can
make acids.
• SO3 (g) + H2O(l)
H2SO4(aq)
• Ionic oxides dissolve in water to produce
bases.
• CaO(s) + H2O(l)
Ca(OH)2(aq)
A
B
E
B
APPLYING
EQUILIBRIUM
THE COMMON ION
EFFECT
• When the salt with the anion of a weak acid is added
to that acid,
• It reverses the dissociation of the acid.
• Lowers the percent dissociation of the acid.
• The same principle applies to salts with the cation of a
weak base..
• The calculations are the same as last chapter.
BUFFERED SOLUTIONS
• A solution that resists a change in pH.
• Either a weak acid and its salt or a weak base and its
salt.
• We can make a buffer of any pH by varying the
concentrations of these solutions.
• Same calculations as before.
• Calculate the pH of a solution that is .50 M HAc and .25
-5
M NaAc (Ka = 1.8 x 10 )
ADDING A STRONG ACID
OR BASE
• Do the stoichiometry first. (BAAM)
• A strong base will grab protons from the weak acid
reducing [HA]0
• A strong acid will add its proton to the anion of the salt
reducing [A-]0
• Then do the equilibrium problem.
• What is the pH of 1.0 L of the previous solution when
0.010 mol of solid NaOH is added?
+
-
GENERAL EQUATION
• Ka = [H ] [A ]
[HA]
+
• so [H ] = Ka [HA]
[A ]
+
-
• The [H ] depends on the ratio [HA]/[A ]
• taking the negative log of both sides
-
• pH = -log(Ka [HA]/[A ])
-
• pH = -log(Ka)-log([HA]/[A ])
-
• pH = pKa + log([A ]/[HA])
THIS IS CALLED THE HENDERSONHASSELBACH EQUATION
-
• pH = pKa + log([A ]/[HA])
• pH = pKa + log(base/acid)
• Calculate the pH of the following mixtures
• 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium
-4
lactate (Ka = 1.4 x 10 )
• 0.25 M NH3 and 0.40 M NH4Cl
• (Kb = 1.8 x 10
-5)
BUFFER CAPACITY
• The pH of a buffered solution is determined by the ratio
[A ]/[HA].
• As long as this doesn’t change much the pH won’t
change much.
• The more concentrated these two are the more H
and OH the solution will be able to absorb.
• Larger concentrations bigger buffer capacity.
+
BUFFER CAPACITY
• Calculate the change in pH that occurs when 0.010
mol of HCl(g) is added to 1.0L of each of the following:
• 5.00 M HAc and 5.00 M NaAc
• 0.050 M HAc and 0.050 M NaAc
• Ka= 1.8x10
-5
BUFFER CAPACITY
• The best buffers have a ratio
• This is most resistant to change
-
• True when [A ] = [HA]
• Make pH = pKa (since log1=0)
-
[A ]/[HA] = 1
TITRATIONS
• Millimole (mmol) = 1/1000 mol
• Molarity = mmol/mL = mol/L
• Makes calculations easier because we will rarely add
Liters of solution.
• Adding a solution of known concentration until the
substance being tested is consumed.
• This is called the equivalence point.
• Graph of pH vs. mL is a titration curve.
STRONG ACID WITH
• Do the stoichiometry.
STRONG
BASE
• There is no equilibrium .
• They both dissociate completely.
• The titration of 50.0 mL of 0.200 M HNO3 with 0.100 M
NaOH
• Analyze the pH
WEAK ACID WITH
STRONG BASE
• There is an equilibrium.
• Do stoichiometry.
• Then do equilibrium.
• Titrate 50.0 mL of 0.10 M HF
with 0.10 M NaOH
-4)
(Ka = 7.2 x 10
TITRATION CURVES
•
Strong acid with strong Base
•
Equivalence at pH 7
pH
7
mL of Base added
Weak acid with strong Base
 Equivalence at pH >7

pH
>7
mL of Base added
Strong base with strong acid
 Equivalence at pH 7

pH
7
mL of Base added
Weak base with strong acid
 Equivalence at pH <7

pH
<7
mL of Base added
SUMMARY
• Strong acid and base just stoichiometry.
• Determine Ka, use for 0 mL base
• Weak acid before equivalence point
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak acid at equivalence point Kb
• Weak base after equivalence - leftover strong base.
SUMMARY
• Determine Ka, use for 0 mL acid.
• Weak base before equivalence point.
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak base at equivalence point Ka.
• Weak base after equivalence - leftover strong acid.
SOLUBILITY EQUILIBRIA
Will it all dissolve, and if not, how much?
• All dissolving is an equilibrium.
• If there is not much solid it will all dissolve.
• As more solid is added the solution will
become saturated.
• Solid
dissolved
• The solid will precipitate as fast as it
dissolves .
• Equilibrium
GENERAL EQUATION
+ stands for the cation (usually metal).
• Nm stands for the anion (a nonmetal).
• M
• But the concentration of a solid doesn’t change.
+a
-b
• Ksp = [M ] [Nm ]
• Called the solubility product for each compound.
WATCH OUT
• Solubility is not the same as solubility product.
• Solubility product is an equilibrium constant.
• it doesn’t change except with temperature.
• Solubility is an equilibrium position for how
much can dissolve.
• A common ion can change this.
CALCULATING KSP
• The solubility of iron(II) oxalate
FeC2O4 is 65.9 mg/L
• The solubility of Li2CO3 is 5.48 g/L
CALCULATING
SOLUBILITY
• The solubility is determined by equilibrium.
• Its an equilibrium problem.
• Calculate the solubility of SrSO4, with a Ksp
-7
of 3.2 x 10 in M and g/L.
• Calculate the solubility of Ag2CrO4, with a
-12
Ksp of 9.0 x 10
in M and g/L.
RELATIVE SOLUBILITIES
• Ksp will only allow us to compare the solubility of solids
the at fall apart into the same number of ions.
• The bigger the Ksp of those the more soluble.
• If they fall apart into different number of pieces you
have to do the math.
COMMON ION EFFECT
• If we try to dissolve the solid in a solution with either the
cation or anion already present less will dissolve.
-7
• Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10
in M and g/L in a solution of 0.010 M Na2SO4.
-7
• Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10
in M and g/L in a solution of 0.010 M SrNO3.
PRECIPITATION
+a
-b
• Ion Product, Q =[M ] [Nm ]
• If Q>Ksp a precipitate forms.
• If Q<Ksp No precipitate.
• If Q = Ksp equilibrium.
-3
• A solution of 750.0 mL of 4.00 x 10 M Ce(NO3)3 is
-2
added to 300.0 mL of
2.00 x 10 M KIO3. Will
-10M)precipitate and if so,
Ce(IO3)3 (Ksp= 1.9 x 10
what is the concentration of the ions?
SELECTIVE
PRECIPITATIONS
• Used to separate mixtures of metal ions in solutions.
• Add anions that will only precipitate certain metals at a
time.
• Used to purify mixtures.
• Often use H2S because in acidic solution Hg
Bi
+3, Cu+2, Sn+4 will precipitate.
+2, Cd+2,
SELECTIVE PRECIPITATION
-
-2
• In Basic adding OH solution S will increase so more
soluble sulfides will precipitate.
• Co
+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH) , Al(OH)
3
3
B
D
A
D
E
Lab
Primary Learning Objective (LO)
Big Idea 1: Atoms and Elements
1. Spectroscopy: What Is
the Relationship Between
the Concentration of a
Solution and the Amount of
Transmitted Light Through
the Solution?
2. Spectrophotometry: How
Can Color Be Used to
Determine the Mass
Percent of Copper in
Brass?
3. Gravimetric Analysis:
What Makes Hard Water
Hard?
LO 1.15 The student can justify the selection of a
particular type of spectroscopy to measure properties
associated with vibrational or electronic motions of
molecules.
4. Titration: How Much Acid
Is in Fruit Juice and Soft
Drinks?
LO 1.20 The student can design, and/or interpret data
from, an experiment that uses titration to determine the
concentration of an analyte in a solution.
LO 1.16 The student can design and/or interpret the
results of an experiment regarding the absorption of light
to determine the concentration of an absorbing species
in a solution.
LO 1.19 The student can design, and/or interpret data
from, an experiment that uses gravimetric analysis to
determine the concentration of an analyte in a solution.
Big Idea 2: Structure and
Properties of Matter
5. Chromatography: Sticky
Question: How Do You
Separate Molecules That
Like to Stay Together?
LO 2.10 The student can design and/or
interpret the results of a separation
experiment (filtration, paper
chromatography, column
chromatography, or distillation) in
terms of the relative strength of
interactions among and between the
components.
6. Bonding in Solids: What’s
in That Bottle?
LO 2.22 The student is able to design or
evaluate a plan to collect and/or
interpret data needed to deduce the
type of bonding in a sample of a solid.
Big Idea 3: Chemical Reactions
7. Stoichiometry: Using the Principle
That Each Substance Has Unique
Properties to Purify a Mixture: An
Experiment in Applying Green
Chemistry to Purification
8. Redox Titration: How Can We
Determine the Actual Percentage of
H2O2 in a Commercial (Drugstore)
Bottle of Hydrogen Peroxide?
9. Physical and Chemical Changes:
Can the Individual Components of
Pain Away Be Used to Resolve
Consumer Complaints?
LO 3.5 The student is able to design a plan in
order to collect data on the synthesis or
decomposition of a compound to confirm the
conservation of matter and the law of definite
proportions.
LO 3.3 The student is able to use stoichiometric
calculations to predict the results of performing a
reaction that is assumed to go to completion in
the laboratory, and/or to analyze deviations from
the expected results.
LO 3.9 The student is able to design and/or
interpret the results of an experiment involving a
redox titration.
LO 3.10 The student is able to evaluate the
classification of a process as a physical change,
chemical change, or ambiguous change based
on both macroscopic observations and the
distinction between rearrangement of covalent
interactions and non-covalent interactions.
Big Idea 4: Kinetics
10. Kinetics: Rate of Reaction:
How Long Will That Marble
Statue Last?
LO 4.1 The student is able to design and/or
interpret the results of an experiment
regarding the factors (i.e., temperature,
concentration, surface area) that may
influence the rate of a reaction.
11. Kinetics: Rate Laws: What Is
the Rate Law of the Fading of
Crystal Violet Using Beer’s
Law?
LO 4.2 The student is able to analyze
concentration vs. time data to determine
the rate law for a zeroth-, first-, or secondorder reaction. In cases in which the
concentration of any other reactants
remains essentially constant during the
course of the reaction, the order of a
reaction with respect to a reactant
concentration can be inferred from plots of
the concentration of reactant versus time.
Big Idea 5:
Thermodynamics
12. Calorimetry: The
Hand Warmer Design
Challenge: Where
Does Heat Come
From?
LO 5.7 The student is able to
design and/or interpret the
results of an experiment in
which calorimetry is used to
determine the change in
enthalpy of a chemical
process (heating/cooling,
phase transition, or chemical
reaction) at constant pressure.
Big Idea 6: Equilibrium
13. Equilibrium: Can We Make the
Colors of the Rainbow? An Application
of Le Châtelier’s Principle
LO 6.9 The student is able to use LeChâtelier’s principle
to design a set of conditions that will optimize a desired
outcome, such as product yield.
14. Acid-Base Titration: How Do the
Structure and the Initial Concentration
of an Acid and a Base Influence the pH
of the Resultant Solution During a
Titration?
LO 6.13 The student can interpret titration data for
monoprotic or polyprotic acids involving titration of a
weak or strong acid by a strong base (or a weak or
strong base by a strong acid) to determine the
concentration of the titrant and the pKa for a weak
acid, or the pKb for a weak base.
15. Buffering Activity: To What Extent Do
Common Household Products Have
Buffering Activity?
LO 6.20 The student can identify a solution as being a
buffer solution, and explain the buffer mechanism in
terms of the reactions that would occur on addition of
acid or base.
16. Buffer Design: The Preparation and
Testing of an Effective Buffer: How Do
Components Influence a Buffer’s pH
and Capacity?
LO 6.18 The student can design a buffer solution with a
target pH and buffer capacity by selecting an
appropriate conjugate acid-base pair and estimating
the concentrations needed to achieve the desired
capacity.