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Unit E: Equilibrium, Acids and Bases and Intro to Thermodynamics
Imagine a sink with a tap on. If the sink is plugged, it will fill up with water
and eventually overflow. If it is unplugged the water will drain out.
What if you turn the tap on full?
The water level will rise, but then stay constant over time as the amount of
water going into the sink = the amount draining out. The actual water
molecules in the sink are always changing, but the water level remains the
same. This is an example of a DYNAMIC EQUILIBRIUM.
For a Dynamic Equilibrium: rate in = rate out
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Lab E.1: Equilibrium Concepts
Results:
Experiment
final "biliverdin concentration"
final "bilirubin concentration"
(# cheerios on the "left")
(# cheerios on the "right")
Equilibrium ratio =
bilirubin/biliverdin
1
2
3
4
5
6
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Lab E.1: Equilibrium Concepts
B
A
1/6 models k1
B
A
1/2 models k­1
k1 is the rate constant of the "forward" reaction from left to right
k­1 is the rate constant of the "reverse" reaction from right to left
Results:
Experiment
final "biliverdin concentration"
final "bilirubin concentration"
(# cheerios on the "left")
(# cheerios on the "right")
Equilibrium ratio =
bilirubin/biliverdin
1
75
25
25/75 = 1/3
2
75
25
1/3
3
75
25
1/3
4
150
50
1/3
5
225
75
1/3
6
169
56
1/3
67(1/6) = 11
50(1/6) = 8
50
67
50
33(1/2) = 17
50(1/2) = 25
75(1/6) = 12.5
75
33
25
25(1/2) = 12.5
73(1/6) = 12
73
27
27(1/2) = 14
75(1/6) = 12.5
75
25
25(1/2) = 12.5
The numbers of molecules turning
from A into B is the same and will
remain the same as long as the
conditions (the 1/6 and 1/2) remain
the same. The overall numbers of
molecules of "A" and "B" are also
stable, although the individual
molecules are constantly turning
from one into the other and back
again.
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Lab E.1: Equilibrium Concepts
Analysis:
What do you notice about the number of "molecules" going from one side
of the equilibrium to the other after several iterations in each experiment?
How is the ratio of biliverdin : bilirubin related to the ratio of forward and
reverse rate constants, k1 and k­1?
B
A
1/6 models k1
rate1 = k1[A]1
rate­1 = k­1[B]1
at Equilibrium:
B
A
1/2 models k­1
rate1 = rate­1
so:
k1[A]e = k­1[B]e
rearranging:
[B]e
=
[A]e
k1
k­1
Either of these expressions is how we define the Equilibrium Constant, Keq
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Phase Equilibria
Equilibria exist all around us.
When a solute dissolves in a solvent it will be in equilibrium between
dissolved and solid states at or above its saturation point. Imagine
dissolving table salt in a glass of water. If you add more salt than the water
can dissolve, you create a dynamic equilibrium:
NaCl(s)
Na+(aq) + Cl­(aq)
Overall, the saltiness of the solution will not change, but the actual ions that
are dissolved, or in solid state will constantly change.
Why do you need to keep the cap on a carbonated beverage in order to
keep it from going "flat"?
The same type of thing happens when you have a substance changing
physical states ­ depending on the temperature the amount of a given
substance (in a closed system) will be in equilibrium between two (actually
three) states:
H2O(g)
H2O(l)
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Practice:
pg. 422 # 1 ­ 5
pg. 428 # 1 ­ 5
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We know from thermochemistry that there are exothermic and
endothermic reactions.
ENDOTHERMIC
(a.k.a. transition state)
A
A
B
B
Don't these two reactions look like they could be mirror images of each
other?
What if they were? What is stopping both reactions from happening at the
same time?
In short: NOTHING
In reality, reactions don't just go "forward" from reactants to products.
Under the correct conditions, those products can turn back into
reactants.
This property of reactions is called REVERSIBILITY. A reversible reaction
eventually reaches a DYNAMIC EQUILIBRIUM, a state in which a certain
number of reactant molecules are turning into products, but at the same
time, that same number of product molecules are turning back into
reactants.
The direction of the FORWARD and REVERSE reactions are just a
matter of point of view.
A CHEMICAL EQUILIBIRUM is said to be DYNAMIC because both the
forward and reverse reactions keep happening, but the overall numbers
of reactants and products don't change over time...
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The Equilibrium Constant
r1 = k1[A]m[B]n
e
r­1 = k­1[C]p[D]q
e
Imagine for the reaction: m A + n B
pC+qD
Keq = 3.01 x 10­3. What can you say about:
1. The forward and reverse rates if the system is at equilibrium?
these two rates are what are actually equal at equilibrium
2. The forward and reverse rate constants?
kforward < kreverse (because Keq < 1)
3. The concentration of products vs. reactants at equilibrium?
[products]e < [reactants]e (because Keq < 1)
4. What will happen to the rates of forward and reverse reaction initially, and once
equilibrium is re­established if the concentration of reactants is increased?
(A & B)
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Keq
We rarely use rate constants directly to investigate Keq. It is far more common to
use the concentrations of reactants and products (easier to measure).
In the reaction on the previous page, we will assume (and for all reactions in this
unit) that the reaction happens in one step (hence, the rate­determining step).
The expression for Keq is therefore:
mA+nB
pC+qD
rateforward =
ratereverse =
where e denotes that the concentrations are measured at equilibrium.
Note the exponents of each chemical species are all = to their stoichiometric
coefficients in the balanced equation. Since we are going to assume that every
reaction occurs in one step, this makes sense in terms of the rate laws of the forward
and reverse reactions.
Ex:
For the (one step) reaction:
A+2B
3C
The expression for the equilibrium constant would be:
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Keq
1. nA + B
mC + D in a single step. What is the Keq expression?
2. If ethanol and ethanoic acid react to form an ester in a single step, what is
an expression for the Keq using concentrations?
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LeChatelier's Principle
We saw in Lab E.1 that if you perturb (or "stress") a chemical system at
equilibrium, the system will automatically shift the concentrations of the
reactants and products so that:
1. the ratio of reactants to products returns to the Keq.
2. the rates of the forward and reverse reactions are again equal (although
not necessarily the same as they were before the "stress").
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Example:
2 A (g)
B (g)
purple
colourless
∆H = +250 kJ
What is the effect of (i.e. will there be more purple colour or less than
before):
Decreasing the [A]?
Decreasing the [B]?
Decreasing the pH?
Decreasing the
temperature?
Increasing the
pressure?
Increasing the
temperature?
Decreasing the
pressure?
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Example:
2 A (g)
B (g)
purple
colourless
∆H = +250 kJ
What is the effect of (i.e. will there be more purple colour or less than
before):
Decreasing the [A]?
Assuming we start at equilibrium, if we decrease the amount of A, we will
initially lose some purple colour.
Then the system will want to turn some existing B into A to replace that which
was lost. This will then regenerate some purple colour.
At the new equilibrium point, we will have the same ratio of A to B as before
the stress, but fewer overall molecules of each, so therefore less A than at the
first equilibrium and a bit less purple colour.
Decreasing the [B]?
B is taken away, so the system will want to replace it until the ratio of A to B is the
same as it was. The only source of B is A, so we will use up A to replace B.
Therefore we will have less A and less purple colour at the re­established
equilibrium.
Decreasing the pH?
Increasing the
pressure?
Decreasing the
temperature?
Increasing the
temperature?
Decreasing the
pressure?
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2 A (g)
June 05, 2023
B (g)
∆H = +250 kJ
Decreasing the pH?
This means increasing the acidity of the a solution, or specifically increasing [H+].
If neither A nor B is H+, then decreasing the pH will have no effect on the
concentrations of A or B, nor on the amount of purple colour.
Increasing the pressure?
We are stressing any side of the equilibrium that has gas molecules on it.
We are stressing the side of the equilibrium with more molecules of gas the
most. So, the system will react by decreasing the concentration (amounts) of
gaseous species on the side of the equilibrium with more molecules of gas.
In this reaction there are more A molecules in the gas state, therefore we will
alleviate the pressure increase the most by turning some A into B.
Increasing the pressure favours the side of the equilibrium with fewer
moles of gas.
The equilibrium constant does not change.
So here we lost some A, therefore we lose some purple colour.
Decreasing the pressure?
Here we are stressing the side of the equilibrium with the fewest
moles of gas the most. Another way of looking at it is that we are
decreasing the concentration of the side with more gas more, so
we make more of the species with more moles of gas.
Decreasing the pressure favours the side of the equilibrium
with more moles of gas.
Decreasing the temperature?
Increasing the temperature?
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B (g)
2 A (g)
∆H = +250 kJ
Increasing the temperature?
Re­write the reaction with the enthalpy in as a reactant or product. Here:
2 A + 250 kJ
B
Here since the reaction is endothermic the
heat is like one of the reactants...
What would happen if we increased the amount of any other reactant?
We would make more B to compensate.
The only way to make more B is to use up A.
We'll use up A until we have offset the increase in the "reactant" heat.
Thermodynamically this means that we are trying to re­equilibrate the heat as well
as the concentrations of A and B.
We end up with less A and more B than before ­ in a NEW equilibrium. The K­eq
actually changes when we change the temperature. Here, we'd end up with less
purple colour, and we'd also increase the value of Keq since Keq = [B]/[A]2 and now
[B] is higher than before and [A] is lower
Decreasing the temperature?
We're decreasing one of the reactants, but not one that appears in the Keq
expression. We will make more A to compensate for the loss of the other
reactant (heat), therefore increase the purple colour, and decrease the value
of Keq
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LeChatelier's Principle: Changing Concentrations (demo)
If we use Cobalt chloride we can see what happens to an actual chemical
system. Cobalt chloride exists in equilibrium in aqueous solution:
[Co(H2O)6]2+(aq) + 4 Cl­(aq)
[CoCl4]2­(aq) + 6 H2O(l)
∆H > or < 0
We'll do a few things to perturb this equilibrium. What do you think will
happen when we:
1. Add HCl
2. Add CaCl2
3. Add H2O
4. Add AgNO3
5. Add H2SO4 (a dehydrating agent)
6. Heat/Cool the solution.
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LeChatelier's Principle: Changing Temperature or Pressure
(demo)
If we use Nitrogen dioxide in equilibrium with dinitrogen teroxide we can see
what happens to an actual chemical system in the gas phase if we change
temperature or pressure.
2 NO2 (g)
N2O4 (g)
∆H = ­57 kJ
What is the effect of:
Increasing the volume?
Increasing the temperature?
Decreasing the volume?
Decreasing the temperature?
http://www.youtube.com/watch?v=PxJbp1SzGjY
http://www.dnatube.com/video/6698/Effect­of­Temperature­on­an­Equilibrium­Reaction
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Practice with LeChatelier's Principle
For the reaction:
2 CrO42­ (aq) + 2 H+(aq)
Cr2O72­(aq) + H2O(l)
What would be the effect of:
Lowering the pH?
Raising the pH?
http://www.youtube.com/watch?v=zP9qEiaL4kQ
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Practice with LeChatelier's Principle
Recall that raising the pressure in a reaction in which all or some of the compounds
are gases will cause the equilibrium to shift to make more of the compounds on the
side of the equilibrium with fewer moles of gas.
Lowering the pressure will favour the side with more moles of gas.
For the theoretical reaction:
A(g) + 2 B(g)
C(g)
where A is a red molecule, what will be the effect of:
Raising the pressure?
Lowering the pressure
Adding more C?
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Practice with LeChatelier's Principle
For the theoretical reaction:
A+B
C+D
[C][D]
Keq =
[A][B]
A is a Red coloured compound, and D is a Blue coloured compound.
Let's say that the equilibrium constant, Keq for this reaction is 4.56. What colour is this
reaction mixture at equilibrium?
What if
[C] = 20 M at equilibrium
[D] = 1 M at equilibrium
[A] = 4.39 M at equilibrium and
[B] = 1 M at equilibrium
but What if
[C] = 2 M at equilibrium
[D] = 10 M at equilibrium
[A] = 4.39 M at equilibrium and
[B] = 1 M at equilibrium
We actually can't know what colour we begin with with only this info....
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A+B
June 05, 2023
C+D
What would happen if we added more A?
[C][D]
Keq =
[A][B]
Keq = 4.56
What would happen if we added more D?
What would happen if we added more B?
What would happen if we added more C?
What would happen if we took away some B?
What would happen if we took away some A?
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C+D
A+B
Keq =
[C][D]
[A][B]
Keq = 4.56
What would happen if we added more A?
It will just turn a darker shade of the same colour. Initially the mixture would be more red due to the added
A. The system would then shift to use up A to relieve the stress. You cannot use up A without also:
­using up B
­making both C and D
So, in the end, the ratio of all the reactants and products will not change from the original equilibrium. Same
colour, different shade (darker due to more molecules).
What would happen if we added more D?
It will just turn a darker shade of the same colour. Initially the mixture would be more blue due to the added
D. The system would then shift to use up D to relieve the stress. You cannot use up D without also:
­using up C
­making both A and B
So, in the end, the ratio of all the reactants and products will not change from the original equilibrium. Same
colour, different shade (darker due to more molecules).
What would happen if we added more B?
It will turn more blue. Initially the mixture would not change colour due to the added B (colourless). The
system would then shift to use up B to relieve the stress. You cannot use up B without also:
­using up A
­making both C and D
So, in the end, you will have used up A in the attempt to relieve the stress of the added B, and made more C
and D in the process, therefore less red A, more blue D.
What would happen if we added more C?
Adding more C makes it like this:
C D
A B
If we start like this at equilibrium:
C
CC D
A B
So to re­equilibrate we need to get rid of C,
but the only way to do that is to use up D and
also make more A and B
So overall:
A B
A B
C
C
D
What would happen if we took away some B?
equilibrium
stress
A B
C D
A
C D
response to regain
equilibrium (original
ratios of reactants and
products)
B
make more B
C
AB
result of reaction
to stress
D
more red
What would happen if we took away some C?
more blue
What would happen if we took away some D?
A B
C D
A B
C
A
B
C
D
mixture is now less blue, more red
D
D and A are in the same relative amounts as in
the original equilibrium, therefore the colour
doesn't change, but the overall shade is
lighter.
What would happen if we took away some A?
same colour, lighter shade
A
A
A
B
B
C D
C D
more blue
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A+B
June 05, 2023
C+D
[C][D]
Keq =
[A][B]
Keq = 4.56
What would happen if we added more C?
If we start like this at equilibrium:
Adding more C makes it like this:
A B
So to re­equilibrate we need to get rid of C,
but the only way to do that is to use up D and
also make more A and B
So overall: More red
C D
A B
C D
CC
A B
A B
C
C
D
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A+B
June 05, 2023
[C][D]
Keq =
[A][B]
C+D
Keq = 4.56
What would happen if we took away some B?
If we start like this at equilibrium:
Taking away B makes it like this:
Re­equilibrating makes it like this:
So overall: More red
C D
A B
C D
A
A
B
C
D
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A+B
June 05, 2023
[C][D]
Keq =
[A][B]
C+D
Keq = 4.56
What would happen if we took away some A?
If we start like this at equilibrium:
A B
C D
B
C D
Taking away A makes it like this:
Re­equilibrating makes it like this:
A
B
C
D
What does this look like? Hmmmm.
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LeChatelier's Principle
To recap: after a system at equilibrium is stressed...
1. the ratio of reactants to products returns to the Keq as long as the
temperature is constant.
2. the rates of the forward and reverse reactions are again equal (although
not
necessarily the same as they were before the "stress").
There are a number of ways to "stress" a chemical system at
equilbrium. The
most common are to:
1. change the concentration (increase or decrease) of a reactant or
product
(equilibrium system). This can be done directly (just physically add
more of a
reactant or product), or by introducing a secondary reaction that adds
or
removes one of the chemicals.
If one of the reactants or products is H+ then increasing or decreasing
the pH is effectively changing the concentration of something (pH up, [H
+] down, pH down, [H+] up
2. change the pressure of the equilibrium system. This only has an
effect if
one or more of the reactants or products is a gas. Recall PV = nRT, i.e.
you
can change the pressure of a system by changing the volume.
3. change the temperature of the environment.
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Chemical Equilibrium: Quantitative Analysis
We can use the Keq expression to:
1. Determine what the concentrations of reactants and products will be once
equilibrium has been established in a system that isn't at equilibrium yet.
2. Determine what the new concentrations of reactants and products will be if a
system is already at equilibrium and experiences a stress.
There are a few steps to figuring these things out.
1. Determine the initial concentrations of each chemical.
2. Write out your equilibrium expression (Keq). Sub in the initial concentrations. This is
called the "Q ratio". If the resulting Q ratio is greater than Keq, you currently have too
high a concentration of products and will therefore need to make reactants in order to
reach equilibrium. If the number is less than Keq you will make products.
3. Set up an I.C.E. table to keep track of your initial conditions, the changes each
chemical species will undergo to get to equilibrium, and the concentrations of each
chemical at equilibrium (you won't have exact numbers there, but algebraic
expressions).
4. Write out your Keq equation subbing in all known values and the expressions for
each concentration at equilibrium.
5. Use whatever means necessary to solve the equation (always try to make
approximations if appropriate).
6. Use the solution of the equation to determine the equilibrium concentrations of each
chemical species (you will often have a choice of values to use ­ you will need to use
your reasoning skills to choose the correct solution).
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Chemical Equilibrium: Quantitative Analysis
For the following reaction, determine the equilibrium concentrations of PCl5, PCl3,
and Cl2 at equilibrium if you start with 1.0 mol/L of PCl5. The Keq for this reaction is
12.5.
PCl5 (g)
PCl3 (g) + Cl2 (g)
[PCl3]e[Cl2]e
Keq =
[PCl5]e
= 12.5
where the subscript e represents the concentration at equilibrium.
It might be obvious in this example, but you should always examine the
initial state of reactants to see how concentrations will change to acheive
equilibrium. We define Q here as an expression that will say what side of
the equilibrium we are currently on:
[PCl5]i = 1.0 M
Q=
[PCl3]i[Cl2]i
[PCl3]i = 0 M
[Cl2]i = 0 M
[PCl5]i
=
(0)(0)
(1)
= 0 < Keq
Since Q < Keq we'll need to create more products in
order to achieve equilibrium.
ICE table:
[PCl5]
[PCl3]
[Cl2]
Initial
1
0
0
Change
­x
+x
+x
1­x
x
x
Equilibrium
Keq =
(x)(x)
1­x
depends on stoichiometry in rxn
and Keq expression
= 12.5
x2 = 12.5(1 ­ x)
0 = x2 + 12.5x ­ 12.5
x=
­12.5 +/­ √(12.5)2 ­ 4(1)(­12.5)
2(1)
x = 0.9307... or something negative.
[PCl3]e = 0.93 M
[Cl2]e = 0.93 M
[PCl5]e = 1 ­ 0.93 = 0.069 M
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Chemical Equilibrium: Quantitative Analysis
For the following reaction, determine the equilibrium concentrations of PCl5,
PCl3, and Cl2 if you start with 1.0 mol/L each of PCl3 and Cl2. The Keq for this
reaction is 12.5.
PCl3 (g) + Cl2 (g)
PCl5 (g)
Keq =
[PCl5]i = 0 M
Q=
[PCl3]e[Cl2]e
[PCl3]i[Cl2]i
[PCl3]i = 1.0 M
[Cl2]i = 1.0 M
= 12.5
[PCl5]e
=
[PCl5]i
(1)(1)
(0)
= undef. > Keq
Since Q > Keq we'll need to create more reactants in
order to achieve equilibrium.
[PCl5]
[PCl3]
[Cl2]
Initial
0
1
1
Change
+x
­x
­x
x
1­x
1­x
Equilibrium
Keq =
(1 ­ x)(1 ­ x)
= 12.5
x
x2 ­ 2x + 1 = 12.5x
x2 ­ 14.5x + 1 = 0
x=
14.5 +/­ √(­14.5)2 ­ 4(1)(1)
2(1)
x = 0.069... or x = 14.43... (makes no sense)
[PCl3]e = 1 ­ 0.069...
[Cl2]e = 1 ­ 0.069...
= 0.93 M
= 0.93 M
[PCl5]e = 0.069 M
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LeChatelier's Principle: Quantitative Analysis
If you now add 1.0 mol of PCl3 to the reaction that was at equilibrium, what
will the new concentrations of each molecule be once equilibrium is re­
established?
PCl5 (g)
[PCl5]i = 0.069 M
Q=
PCl3 (g) + Cl2 (g)
[PCl3]i[Cl2]i
[PCl3]i = 0.93 + 1.0 M
[PCl5]i
=
(1.93)(0.93)
(0.069)
= 25.9... > Keq
Since Q > Keq we'll need to create more reactants in
order to achieve equilibrium. Does this make sense in
terms of "stressing" the equilibrium?
[Cl2]i = 0.93 M
[PCl5]
[PCl3]
Initial
0.069
1.93
0.93
Change
+x
­x
­x
Equilibrium
0.069 + x 1.93 ­ x
Keq =
[Cl2]
0.93 ­ x
(1.93 ­ x)(0.93 ­ x)
= 12.5
0.069 + x
1.7949 ­ 2.86x + x2 = 0.8625 + 12.5x
x2 ­ 15.36x + 0.9324 = 0
x=
15.36 +/­ √(­15.36)2 ­ 4(1)(0.9324)
2(1)
x = 0.0609... or x = 15.... (makes no sense)
[PCl3]e = 1.93 ­ 0.069... [Cl2]e = 0.93 ­ 0.0609...
= 1.9 M
= 0.87 M
[PCl5]e = 0.069 + 0.0609...
= 0.13 M
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Chemical Equilibrium: Using Approximations
550 mL of a 0.25 M solution of A is added to 450 mL of a 0.50 M solution of B and is
allowed to come to equilibrium with molecule C. What are the equilibrium
concentrations of each molecule? The Keq is 5.0 x 10­6.
2C
A+B
Keq =
[C]e2
[A]e[B]e
= 5.0 x 10­6
The only wrinkle in this problem is that you have to calculate the initial
concentrations of each molecule before you begin the actual problem...
V 1C1 = V 2 C 2
[A]i = (0.25 mol/L x 0.55 L)/(0.55 L + 0.45 L) = 0.1375 mol/L
[B]i = (0.50 mol/L x 0.45 L)/(0.55 L + 0.45 L) = 0.225 mol/L
Q is obviously > Keq
Approximations: If [A]i and [B]i (and any other reactants/products) are large
compared to Keq we can safely approximate that they don't change in
concentration significantly as they approach equilibrium.
Rule of thumb: [X]i/Keq > 100 we can approximate the the change in [X] is
negligible when calculating using the Keq.
[A]
[B]
[C]
Initial
0.1375
0.225
0
Change
­x
­x
2x
0.1375 ­ x
Equilibrium ~0.1375
Keq =
(2x)2
0.225 ­ x
~0.225
2x
= 5.0 x 10­6
(0.1375)(0.225)
4x2 = 1.54... x 10­7
x = 1.9665... x 10­4
[C]e = 2x = 2(1.9665...x 10­4) = 3.9 x 10­4 M
[A]e = 0.1375 ­ 1.9665 x 10­4 = 0.14 M
[B]e = 0.225 ­ 1.9665 x 10­4 = 0.22 M
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Problem:
When 0.150 mol of propanoic acid and 0.0550 mol of ethanol are dissolved in
1.00 L of water at 95 ºC, one makes 2.01x10­6 mol of the corresponding ester.
What is the Keq for the condensation reaction?
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Problem:
When 0.150 mol of propanoic acid and 0.0550 mol of ethanol are dissolved in
1.00 L of water at 95 ºC, one makes 2.01x10­6 mol of the corresponding ester.
What is the Keq for the condensation reaction?
H3CCH2COOH(aq) + H3CCH2OH(aq)
Keq =
H3CCH2COOCH2CH3 (aq) + H2O(l)
[ester]
[acid][EtOH]
[acid]
[EtOH]
Initial
0.150
0.055
0
Change
­x
­x
x
0.150 ­ x
0.055 ­ x
Equilibrium
[ester]
2.01 x 10­6
x must = 2.01 x 10­6 mol/L
Keq =
2.01 x 10­6
(0.15 ­ 2.01 x 10­6)(0.055 ­ 2.01 x 10­6)
Keq = 2.44 x 10­4
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Practice:
Review:
pg. 437 # 6, 7
pg. 462 # 3 ­ 5
pg. 437 ­ 438 # 1 ­ 11
pg. 462 # 1 ­ 3
pg. 442 # 1
pg. 465 # 1, 2
pg. 444 ­ pg. 445 # 2 ­ 6
pg. 466 # 3, 4
pg. 447 # 7
pg. 472 # 5, 6
pg. 448 ­ 449 # 1 ­ 9
pg. 476 # 7, 8
pg. 457 ­ 458 # 1 ­ 8
pg. 480 # 9, 10
pg. 481 # 1 ­ 8
pg. 459 ­ 460 # 1 ­ 12
pg. 461 # 1, 2
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Heterogeneous Equilibrium
When two or more chemical species involved in an equilibrium are in different
phases, we define the equilibrium constant slightly differently, depending on
the case. The phases you will have to deal with in this course are: solid, liquid,
gas and aqueous. In this course we will only be dealing mathematically with
two kinds of heterogeneous equilibria:
1. Solid/Aqueous equilibria
2. Equilibria involving dissolved species and the production/use of H2O(l).
Some general examples of heterogeneous equilibria
are:
1. a solid dissolving in water
2. an acid or base reacting with water
3. a phase change (solid ­ liquid, liquid ­ gas)
4. An aqueous reaction that uses or produces water
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The Solubility Product Constant (Ksp)
If a compound isn't very soluble in water, it will create an equilibrium between
dissolved and undissolved forms when placed in water. This equilibrium will occur
after as much solid has dissolved in the water as possible. This state is called
SATURATION. After saturation, adding more solid to the water won't cause more
ions to dissolve, and you will simply see the solid collect at the bottom of the flask.
This excess solid is called a PRECIPITATE.
Often, we talk of precipitates as being the products of a reaction. This happens
when two aqueous solutions combine and form a product that isn't soluble in water.
(Note, this can also happen with two non­aqueous solutions, but you'll learn about
that much later).
In order to determine the concentration of an ion in solution at saturation, we need to
solve a problem similar to the Keq problems we have been solving, except due to the
simplified nature of the Ksp expression (i.e. there will never be a reactant included,
therefore there will never be a denominator in the expression), the problems are
easier to solve mathematically.
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The Solubility Product Constant (Ksp)
Ksp is a type of Keq that we use in a heterogeneous equilibrium in which a solid is
dissolving in water.
You may recall using a solubility chart in grade 11 chemistry that told you which
kinds of compounds were soluble in water, and which weren't (check your text's
appendices). This appendix is a bit crude for us. Everything is at least a tiny bit
soluble in water...
Recall the reaction:
NaOH(s)
Na+(aq) + OH­(aq)
This reaction is really an equilibrium, where the reverse reaction is practically
negligible compared to the forward reaction (i.e. NaOH is soluble in water).
How would we write the "Keq" for this reaction? It might be:
But what is the molar concentration of a solid in water? Does it change with the
amount of water present (i.e. can you dilute a solid?)?
Doesn't make much sense. The "concentration" of a solid (moles/volume) could
be thought of as its density (mass/volume). But this property doesn't change if
put in water, so a Keq including it doesn't make much sense.
To fix this problem, we replace the concentration of the solid with its density, and
then combine this constant with the equilibrium constant, Keq. We then simply re­
define this product of constants as "the solubility product constant", Ksp.
In this particular case, Ksp will be a very large number because practically all of the
NaOH(s) added to the water will dissolve and become Na+ and OH­. In the back of
your textbook there is a table of Ksp values, mostly for compounds that aren't very
soluble in water (i.e. with smaller Ksp values).
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Calculating Equilibrium Concentrations (a.k.a. Solubilities of
the ions) Using Ksp
What are the equilibrium (or saturation) concentrations of the ions in the reaction
(NOTE: this questions could also be stated "what are the SOLUBILITIES of Mg2+
and F­ when in solution together?"):
­
Mg2+(aq) + 2 F (aq)
MgF2 (s)­
The Ksp = 7.4 x 10­11
where the subscript "e" denotes that the concentrations are at equilibrium, just like
in previous Keq questions, but, in the case of this heterogeneous equilibrium,
means when as much Mg2+ and F­ have dissolved in the water as possible (a.k.a. at
saturation)
Using ICE tables to solve Ksp problems:
Initial: We can think of a solid being put into water as the initial state (the state
where we'd be measuring Q in Keq type questions). The Ksp type questions are
always a bit easier though as you can think of the situation as having started with
nothing dissolved, i.e. [Mg2+]i = [F­]i = 0 M. This is how you can start any Ksp type
question.
Change: The question then becomes, how many moles of each ion are made per
mole of solid dissolving? This will be the change the chemical species in which we
are interested will undergo to reach equilibrium (saturation). In this case, we will
­
make one mole of Mg2+ per mole MgF2 dissolved, and we will make 2 moles of F
per mole MgF2. Since we don't actually know the number of moles that will
­
2+
dissolve, we can use +x and +2x for the change in concentration that Mg and F
will undergo to get to equilibrium, respectively.
Equilibrium (or Saturation): This will be the initial + the change, as it was in Keq
­
questions. Here it will be x and 2x for Mg2+ and F respectively.
[Mg2+]
[F­]
I
0
0
C
+x
+2x
E
x
2x
As in Keq questions, we sub in the algebraic expressions of our equilibrium
concentrations into the Ksp equation:
So, the saturation concentrations, or SOLUBILITIES of these ions when produced
by dissolving MgF2 are:
[Mg2+]e = x = 2.6 x 10­4 M
[F­]e = 2x = 5.3 x 10­4 M
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What are the saturation concentrations of the ions made by dissolving
Iron(III)hydroxide if the Ksp for this compound is 4.5 x 10­9?
Fe3+(aq) + 3 OH­(aq)
Fe(OH)3­ (s)
Ksp = 4.5 x 10­9
Ksp = [Fe3+][OH­]3
[Fe3+]
[OH­]
Initial
0
0
Change
+x
+3x
Equilibrium
x
3x
Ksp = (x)(3x)3
4.5 x 10­9 = 27x4
x = 0.0036 M
So, [Fe3+]e = x = 0.0036 M
and [OH­]e = 3x = 3 x 0.0036 = 0.011 M
39
0 Thermodynamics Master.notebook
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What are the saturation concentrations of the ions made by dissolving
Nickel(II)chloride if the Ksp for this compound is 5.5 x 10­9?
Ni2+(aq) + 2 Cl­(aq)
NiCl2 (s)
Ksp = 5.5 x 10­9
Ksp = [Ni2+][Cl­]2
[Ni2+]
[Cl­]
Initial
0
0
Change
+x
+2x
Equilibrium
x
2x
Ksp = (x)(2x)2
5.5 x 10­9 = 4x3
x = 0.0011 M
So, [Ni2+]e = x = 0.0011 M
and [Cl­]e = 2x = 2 x 0.0011 = 0.0022 M
40
0 Thermodynamics Master.notebook
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If the concentration of phosphate is 1.53 x 10­5 M at saturation when you
dissolve Iron(II)phosphate, what is the Ksp?
41
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If the concentration of phosphate is 1.53 x 10­5 M at saturation when you
dissolve Iron(II)phosphate, what is the Ksp?
Fe3(PO4)2 (s)
3 Fe2+(aq) + 2 PO43­(aq)
Ksp = [Fe2+]s3[PO43­]s2
[Fe2+]
[PO43­]
I
0
0
C
+3x
+2x
E
3x
2x
Ksp = (3x)3(2x)2
and 2x = 1.53 x 10­5 M
so x = 7.65 x 10­6 M
Ksp = [3(7.65 x 10­6)]3(1.53 x 10­5)2
= 2.829... x 10­24
= 2.83 x 10­24
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Practice:
pg. 486 # 1 ­ 4
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June 05, 2023
Predicting Precipitation
If 100.0 mL of 0.100 mol/L CaCl2 (aq) is mixed with 100.0 mL of 0.0400 mol/L
Na2SO4 (aq) at 20oC, will a Double Displacement reaction occur? Ksp for
CaSO4 (s) at 20oC is 3.6 x 10­5.
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Predicting Precipitation
If 100.0 mL of 0.100 mol/L CaCl2 (aq) is mixed with 100.0 mL of 0.0400 mol/L
Na2SO4 (aq) at 20oC, will a "successful" Double Displacement reaction
occur? Ksp for CaSO4 (s) at 20oC is 3.6 x 10­5.
CaCl2 (aq) + Na2SO4 (aq) ­­­­> CaSO4 (s) + 2 NaCl(aq)
In a solubility chart we'd blindly predict that a precipitate would occur, but
because of this equilibrium:
CaSO4 (s)
Ca2+(aq) + SO42­ (aq)
it's possible we wouldn't have enough CaSO4 to reach saturation (and
equilibrium). If that is the case we won't actually see any precipitate.
We need to determine the concentrations of Ca2+ and SO42­ ions in the
solution after the two solutions are mixed together, and then use those to
compare with the Ksp value.
[Ca2+]before any precipitate would be made: 0.1 LCaCl2 x 0.1 mol/L x 1 molCa2+/ 1 molCaCl2
= 0.01 molCa2+
0.01 molCa2+ / total volume = 0.01/0.2
= 0.05 mol/L
2­
[SO4 ]before any precipitate would be made: 0.1 LNa2SO4 x 0.04 mol/L x 1 molSO4/ 1 molNa2SO4
= 0.004 molSO4
0.004 molSO4 / total volume = 0.004/0.2
= 0.02 mol/L
Our "Q" value for this situation is the same as it was for non­heterogeneous
equilibria, i.e. Ksp but using these initial concentrations
Q = 0.05(0.02) = 0.001
If Q > Ksp then we have too much dissolved ions for them to all stay in solution,
otherwise if Q < Ksp we are below our saturation point.
Ksp = 3.6 x 10­5, which is < Q, so we are above saturation and would make
a precipitate in this reaction.
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Would a precipitate of PbSO4 (s) form if 255 mL of 0.00016 M Pb(NO3)2 (aq)
is mixed with 456 mL of 0.00023 M Na2SO4 (aq)? Ksp of PbSO4 = 1.8 x 10­8.
46
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Would a precipitate of PbSO4 (s) form if 255 mL of 0.00016 M Pb(NO3)2 (aq)
is mixed with 456 mL of 0.00023 M Na2SO4 (aq)? Ksp of PbSO4 = 1.8 x 10­8.
PbSO4(s) + 2 NaNO3 (aq)
maybe
depends on where we are in the equilibrium
i.e. are we at or
Pb2+(aq) + SO42­(aq)
PbSO4(s)
above saturation?
Pb(NO3)2 (aq) + Na2SO4 (aq)
47
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Would a precipitate of PbSO4 (s) form if 255 mL of 0.00016 M Pb(NO3)2 (aq)
is mixed with 456 mL of 0.00023 M Na2SO4 (aq)? Ksp of PbSO4 = 1.8 x 10­8.
PbSO4(s) + 2 NaNO3 (aq)
maybe
depends on where we are in the equilibrium
i.e. are we at or
Pb2+(aq) + SO42­(aq)
PbSO4(s)
above saturation?
Pb(NO3)2 (aq) + Na2SO4 (aq)
[Pb2+]i = 1.6E­4MPb(NO3)2 x 0.255 L /0.711 LTotal x 1 molPb / 1 molPb(NO3)2
= 5.738...E­5 M
[SO42­]i = 2.3E­4MNa2(SO4)2 x 0.456 L /0.711 LTotal x 1 molSO4 / 1 molPbSO4
= 1.475...E­4 M
Q = [Pb2+]i[SO42­]i = (5.738...E­5)(1.475...E­4)
= 8.4647...E­9
Ksp = [Pb2+]s[SO42­]s = 1.8 x 10­8
Here Q < Ksp so we aren't at the saturation point yet and there will be no
precipitate.
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Common Ion Effect and Molar Solubility
What is the molar solubility of PbCl2 (s) if the Ksp is 1.2 x 10­5 at SATP?
Pb2+(aq) + 2 Cl­(aq)
PbCl2 (s)
Ksp = [Pb2+]s[Cl­]s2
[Pb2+]
[Cl­]
I
0
0
C
+x
+2x
E
x
2x
Ksp = (x)(2x)2
1.2 x 10­5 = 4x3
x = 0.01443...
What is the molar solubility of PbCl2 (s) in a solution that already contains
0.2 mol/L NaCl(aq) (at SATP)?
This problem is identical to the molar solubility problem above, but with
one twist ­ the initial concentration of Cl­(aq) (that would normally have
come from PbCl2 dissolving) is not 0 M, because there'd be 0.2 M Cl­(aq)
already in the solution from the NaCl.
PbCl2 (s)
Pb2+(aq)
[Pb ]i = 0 M
2+
+
2 Cl­(aq)
[Cl­]i = 0.2 M
The ICE table looks almost the same as before...
[Pb2+]
[Cl­]
I
0
0.2
C
+x
+2x
E
x
0.2 + 2x
Ksp = [Pb2+]s[Cl­]s2
Ksp = 1.2 x 10­5 (from App. C8 pg. 802)
so
1.2 x 10­5 = (x)(0.2 + 2x)2
but [Cl­]i / Ksp = 0.2/1.2E­5 >> 100, so
1.2 x 10­5 = (x)(0.2)2
x = 3 x 10­4
So the molar solubility of PbCl2 (s) in 0.2 M NaCl(aq) is about 3.0 x 10­4
Compare this to above. Does it make sense that it's much lower?
Also
[Pb2+]s = 3.0 x 10­4 M
[Cl­]s = 2x = 6.0 x 10­4 M, but we already had 0.2 M Cl­ from the NaCl, so the
[Cl­]total = 0.2 + 0.0006 = 0.2006 M. You can see how insignificant the PbCl2's
contribution is...
In case you are interested, DESMOS gives the solution for x in the
unsimplified cubic equation as:
x
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pg. 802
1. What is the molar solubility of Silver (I)
bromide?
2. Will a precipitate occur if you add 25.0
mL of 1.2 x 10­5 M AgNO3 to 35.0 mL of
1.8 x 10­6 M MgBr2?
3. What is the molar solubility of Silver (I)
bromide if it is dissolved in a 0.00350 mol/L
solution of AlCl3?
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0 Thermodynamics Master.notebook
1. AgBr(s)
Ag+(aq) + Br­(aq)
June 05, 2023
Ksp = [Ag+]e[Br­]e
= 5.4 x 10­13
so
[Ag+]
I
0
C +x
x
E
[Br­]
0
+x
x
and
Q = [Ag+]i[Br­]i
=0
Ksp = 5.4 x 10­13 = (x)(x)
x = 7.348...x 10­7 M
This is the molar solubility
This is also the [Ag+] at sat.
This is also the [Br­] at sat.
2. Q = [Ag+]i[Br­]i
If Q > Ksp we have too many "products", i.e. too
many ions in solution so some will precipitate.
­5
[Ag+]i = 0.025 L x 1.2 x 10 mol/L
0.025 L + 0.035 L
[Br­]i =
0.035 L x 1.8 x 10­6 mol/L x 2
0.025 L + 0.035 L
= 2.1 x 10­6 M
= 5 x 10­6 M
Q = 1.05 x 10­11
Q > Ksp so too many ions are in solution initially and we'll have some ppt out.
3. Same as before. Adding Al3+ or Cl­ doesn't change the AgBr
equilibrium.
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Putting it all Together: Determining the Mass of a Precipitate
50.0 mL of a 0.0500 M solution of Pb(NO3)2 is added to 50.0 mL a 0.0250 M solution
of KI. What is the mass of the precipitate (if any):
Pb(NO3)2 (aq) + 2 KI(aq)
PbI2 (s)
PbI2 (s) + 2 KNO3 (aq)
Pb2+(aq) + 2 I­(aq)
The Ksp of PbI2 is = 8.5 x 10­9
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Grade 11 style: assume PbI2 is 100% insoluble. This means the amount
made is based only on the limiting reactant.
Limiting Reactant:
Pb(NO3)2 (aq) + 2 KI(aq)
PbI2 (s) + 2 KNO3 (aq)
nPb(NO3)2 needed = 0.05 LKI x 0.025 molKI / LKI x 1 molPb(NO3)2 / 2 molKI =
0.000625 molPb(NO3)2
nPb(NO3)2 present = 0.05 LPb(NO3)2 x 0.05 molPb(NO3)2 / LPb(NO3)2 = 0.0025 molPb(NO3)2
We would make PbI2 based on nKI
nPbI2 = 0.05 LKI x 0.025 molKI x 1 molPbI2 / 2 molKI = 0.000625 molPbI2
m = 0.000625 mol x 461 g/mol = 0.288125 g of PbI2
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0 Thermodynamics Master.notebook
PbI2 (s)
June 05, 2023
Pb2+(aq) + 2 I­(aq)
Since PbI2 is NOT 100% insoluble, i.e. Ksp ≠ 0 (Ksp = 8.5 x 10­9), we need to
determine the amount of PbI2 that will dissolve and subtract that from the
amount we calculated based on 100% insolubility.
Ksp = [Pb2+]s[I­]s2
Q = [Pb2+]i[I­]i2
[Pb(NO3)2]i = 0.05 M x 0.05/(0.05 + 0.05) = 0.025 M
[KI]i = 0.025 M x 0.05/(0.05 + 0.05) = 0.0125 M
Q = (0.025)(0.0125)2 = 3.9...x10­6 > Ksp so we actually do
make a ppt here.
For the equilibrium [I­]i = 0 M (all was used to make the PbI2 bc/ I­ was limiting).
For the equilibrium [Pb2+]i = (0.0025 ­ 0.000625)/0.1 = 0.01875 M
I
[Pb2+]
[I­]
0.01875
0
Ksp = (0.01875 + x)(2x)2
0.01875/Ksp >>
8.5 x 10 = (0.01875)(2x)
100 we can
approximate
x = 3.3665...x10­4 M
that 0.01875 + x
= 0.01875
nPbI2 that have dissolved = 3.3665...x10­4 M x 0.1 L
­9
C
E
+x
0.01875 + x
+2x
2x
2
= 3.3665...x 10­5 mol
The # moles of PbI2 left = 0.000625 ­ 3.3665...x 10­5 = 5.9133...x10­4
mass of PbI2 remaining in solid form = 5.9133...x 10­4 x 461 = 0.2726... g
= 0.273 g
Compare that to the 0.288 g or so we'd have predicted last year!
% error = (0.273 ­ 0.288)/0.288 x 100% = ­5.3%
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Note: If you have graphing software or advanced algebra skills you can
solve this entire problem a different way, if all you are interested in is the
mass of ppt.
1. Check your Q value to make sure you need to do anything.
2. Calculate initial concentrations of ions that will make a ppt. and think of
them as the reactants and the solid product as the product. Using the entire
initial concentrations will account for any excess/common ion effects.
3. ICE table it, use graphing software to solve for "x". This is the # moles/L
of your precipitate.
Pb2+(aq) + 2 I­(aq)
Ksp = [Pb2+]s[I­]s2
PbI2 (s)
Q = [Pb2+]i[I­]i2
[Pb(NO3)2]i = 0.05 M x 0.05/(0.05 + 0.05) = 0.025 M
[KI]i = 0.025 M x 0.05/(0.05 + 0.05) = 0.0125 M
Q = (0.025)(0.0125)2 = 3.9...x10­6 > Ksp so we actually do
make a ppt here.
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Pb2+(aq) + 2 I­(aq)
0.025
C
­x
E
0.025 ­ x
PbI2 (s)
[I­]
[Pb2+]
I
June 05, 2023
0.0125
­2x
Ksp = [Pb2+]s[I­]s2
8.5 x 10­9 = (0.025 ­ x)(0.0125 ­ 2x)2
Graph & get roots.
0.0125 ­ 2x
0.025 M is rejected ­ not
enough I­ to make that much
ppt.
0.06557 M is rejected ­ not
enough I­ to make that much
ppt.
x = 0.00591632 M
Only root that can work.
# moles ppt. = 0.00591623 mol/L x 0.1 L
mass ppt = 0.000591623 mol x 461 g/mol
mPbI2 = 0.272738... g
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What is the mass of the precipitate formed (if any) when a 50.00 mL
solution of 0.0500 M magnesium chloride is added to 75.00 mL of 0.0500 M
lead (II) nitrate?
Ksp's are in appendix C8, pg. 802
1. Calc. Initial concentrations of ions that would combine to make a ppt. and
check the Q expression. If it's > Ksp you will make a ppt.
2. Determine limiting reactant and maximum mass of ppt you'd make if
the ppt was 100% insoluble.
3. What ion is in excess? What would it's "initial" concentration be after
you make all the ppt. possible? Set up ICE table to determine the [ions] or
# moles that will dissolve at saturation, and subtract from the max
amount. Calculate new # moles and mass of ppt.
Practice:
pg. 489 # 5, 6
pg. 492 # 7 ­ 12
Review:
pg. 493 # 1 ­ 13
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What is the mass of the precipitate formed (if any) when a 50.00 mL
solution of 0.0500 M magnesium chloride is added to 75.00 mL of 0.0500 M
lead (II) nitrate?
K 's are in appendix C8, pg. 802
sp
PbCl2 (s) + Mg(NO3)2 (aq)
MgCl2 (aq) + Pb(NO3)2 (aq)
maybe, if enough Pb2+
and Cl­ are in solution
If we are at saturation: PbCl2 (s)
From Appendix C8, Ksp = 1.2 x 10­5
Ksp = [Pb2+]s[Cl­]s2
Pb2+(aq) + 2 Cl­(aq)
What's our Q?
Q = [Pb2+]i[Cl­]i2
[Pb2+]i = (0.075 L x 0.05 mol/L)/(0.075 L + 0.050 L) = 0.03 M
[Cl­]i = [(0.050 L x 0.05 mol/L) x 2 molCl­/molMgCl2]/(0.075 L + 0.050 L) = 0.04 M
Q = (0.03)(0.04)2 = 4.8 x 10­5 which is > Ksp so a ppt will form.
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Limiting reactant:
nPb2+ needed = 0.05 L x 0.05 molMgCl2/LMgCl2 x 1 molPb(NO3)2/1 molMgCl2 = 0.0025 mol
nPb2+ present = 0.075 x 0.05 = 0.00375 mol
So Pb(NO3)2 is in excess and MgCl2 is LIMITING
If PbCl2 was 100% insoluble:
nPbCl2 made = 0.05 L x 0.05 molMgCl2/LMgCl2 x 1 molPbCl2 / 1 molMgCl2 = 0.0025 mol
But Pb(NO3)2 is NOT 100% insoluble. Nothing is. So we have to take the Ksp
into account.
We are starting with 0.00375 mol ­ 0.0025 mol = 0.00125 mol Pb2+ ions in
excess, so these will be in solution "before" we dissolve the PbCl2
[Pb2+]i = 0.00125 mol / 0.125 L = 0.01 M
I
C
E
[Pb2+]
0.01
+x
0.01 + x
[Cl­]
0
+2x
2x
Ksp = 1.2 x 10­5 = (0.01 + x)(2x)2
Approximate? 0.01 / 1.2 x 10­5 >> 100,
so we can say that 0.01 + x = 0.01
1.2 x 10­5 = (0.01)(2x)2
x = 0.01732... mol/L
Mass of PbCl2 left as a ppt.
multipling this by 0.125 L = 0.00216... mol of
PbCl2 that will dissolve
0.0025 mol ­ 0.00216... mol x 278.1 g/mol
= 0.0931... g
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Practice:
Review:
pg. 489 # 5, 6
pg. 493 # 1 ­ 13
pg. 492 # 7 ­ 12
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Equilibria Involving Acids and Bases
Acid: A molecule that gives an H+ ion to water. If it is a strong acid, this reaction
will not be reversible. If it is a weak acid, there will be a heterogeneous equilibrium.
HA(aq) + H2O(l)
Cl­ + H3O+
Strong: HCl + H2O
Weak: CH3COOH + H2O
acid
A­(aq) + H3O+(aq)
base
CH3COO­ + H3O+
conjugate base conjugate acid
Ka = large #
Ka = 1.8 x 10­5
Just like Keq, Ka, the acid dissociation constant has an expression at equilibrium
where the concentrations of the products are divided by the concentrations of the
reactants.
Like the Ksp situation, we have a heterogeneous equilibrium here (aqueous acids
and conjugate bases, liquid water). We turned the Keq of a solid dissolving into a Ksp
by multiplying the Keq constant by another constant, the density of the solid, leaving
everything else in the same phase in the expression.
Here, we have the concentration of water in the denominator. This water is in an
aqueous system, so what we are really saying by the [H2Ol] is "the concentration of
water in water". Whatever this is (try and figure it out ­ remember that the units for
concentration are moles per litre) it is constant at constant T and P, so we can
simply combine the two constants and create a new constant for the heterogeneous
equilibrium that only has species in the same phase on the right of the equation.
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What is the pH of a 0.100 M solution of HCl?
HCl(l)
H+(aq) + Cl­(aq)
=
HCl(l) + H2O(l)
H3O+(aq) + Cl­(aq)
So, since pretty much all the HCl dissociates in water the "equilibrium"
concentrations of H3O+ (or H+ for short) and Cl­ are 0.100 M and HCl is 0 M
pH = ­log[H+]
pH = ­log(0.1)
pH = 1
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ex. What is the pH of a 0.100 M solution of acetic acid, Ka = 1.80 x 10­5?
acetic acid = HOAc
H3O+(aq) + OAc­(aq)
conj. acid conj. base
(acetate)
[OAc­]e [H3O+]e
HOAc(aq) + H2O(l)
acid
base
Ka =
[HOAc]e
Before the acetic acid is able to react with the water, its concentration is
0.100 M, and the concentration of all products is 0 M prior to equilibrium.
[HOAc]
[H3O+]
I
0.100
0
0
C
­x
+x
+x
E
0.1 ­ x
x
x
[OAc­]
Approximation? Check if [HOAc]i/Ka > 100? (Yes, 0.1/1.8 x 10­5 >> 100)
So, we can assume that 0.1 ­ x
1.8 x 10­5 =
0.1
(x)(x)
0.1
x = 0.00134... = [H3O+]e
pH = ­log[H3O+]e
pH = 2.87
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Base: In water, a base can be defined as any molecule that increases the
concentration of OH­(aq) ions. A weak base will be in equilibrium, and a strong base
won't.
There are two kinds of aqueous bases:
1. an ionic compound that dissolves in water to release OH­ ions
2. an molecule that reacts with water, taking an H+ from water to leave OH­ as a
product.
Type 1:
X(OH)
ex. (strong)
(weak)
X+ + OH­
KOH
K+ + OH­
Pb(OH)2
Ksp = large #
Pb2+ + 2 OH­
Ksp = 1.42 x 10­20
no conjugate acids or bases
Type 2:
B­ + H2O
ex. (strong)
(weak)
BH + OH­
NH2­ + H2O
NH3 + OH­
Kb = large #
NH3 + H2O
NH4+ + OH­
Kb = 1.78 x 10­5
base
acid
conjugate
conjugate base
acid
*note the changing
role of H2O reacting
with acids or bases...
For type 1 bases, we use the Ksp expression to solve problems.
ex.
For type 2 bases, we use a Kb expression, similar to the Ka expression:
and since [H2Ol] is a constant also...
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ex. What are the pH's of 0.500 M solutions of:
a. NaOH Ksp > 1000
b. Mg(OH)2 Ksp = 5.6 x 10­12
c. NH2­
Kb > 1000
d. H3CNH2 Kb = 1.78 x 10­5
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ex. What are the pH's of 0.500 M solutions of:
a. NaOH Ksp > 1000
b. Mg(OH)2 Ksp = 5.6 x 10­12
c. NH2­
Kb > 1000
d. H3CNH2 Kb = 1.78 x 10­5
a.
NaOH(s)
0.500 M
Na+(aq) + OH­(aq)
0.500 M
0.500 M
pOH = ­log[OH­]
pOH = 0.30...
pH + pOH = 14 (we'll see why soon)
pH = 14 ­ 0.30...
pH = 13.7
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b.
Mg(OH)2 (s)
June 05, 2023
Mg2+(aq) + 2 OH­(aq)
Ksp = 5.6 x 10­12
Ksp = [Mg2+]e [OH­]2e
[Mg2+]
[OH­]
I
0
0
C
+x
+2x
E
x
2x
5.6 x 10­12 = (x)(2x)2
5.6 x 10­12 = 4x3
x = 1.118...x 10­4
[OH­]e = 2x = 2.237... x 10­4
pOH = ­log[OH­]
pOH = 3.65...
pH + pOH = 14
pH = 14 ­ 3.65...
pH = 10.3497... = 10
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c.
NH2­ + H2O
0.5 M
NH3 + OH­
0.5 M
0.5 M
So, at "equilibrium" [OH­] = 0.5 M
pOH = ­log(0.5)
pH = 14 ­ 0.3
pH = 13.7
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d.
June 05, 2023
H3CNH2 (aq) + H2O(l)
H3CNH3+(aq) + OH­(aq)
[H3CNH3+]e [OH­]e
Kb =
[H3CNH3+]e
[MeNH2] [MeNH3+] [OH­]
I
0.500
0
0
C
­x
+x
+x
E
0.5 ­ x
x
Kb = 1.78 x 10­5
x
0.5/1.78 x 10­5 >> 100, so we can
approximate that 0.5 ­ x 0.5
1.78 x 10­5 =
(x)(x)
0.5
x = [OH­]e = 0.00298328...
pOH = ­log(0.00298...)
pH = 14 ­ 2.525...
pH = 11.5
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Practice:
pg. 526 # 1 ­ 4
pg. 532 # 1 ­ 3
pg. 537 # 4 ­ 7
pg. 540 # 8 ­ 11
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Relationship Between Ka and Kb
Consider the weak base reaction:
NH4+ + OH­
NH3 + H2O
and the weak acid reaction of the conjugate acid of NH3, NH4+
NH4+ + H2O
NH3 + H3O+
Ka =
[H3O+]e[NH3]e
[OH­]e[NH4+]e
Kb =
[NH3]e
[NH4+]e
Ka x Kb =
[H3O+]e[NH3]e
x
[OH­]e[NH4+]e
[NH4+]e
[NH3]e
Ka x Kb = [H3O+]e[OH­]e
= Kw
this is called the "water dissociation constant"
@ SATP, Kw = 1 x 10­14, and [OH­]e = [H3O+]e
so [H3O+]e[OH­]e = 1 x 10­14
[H3O+]e2 = 1 x 10­14
[H3O+]e = 1 x 10­7
so @ SATP, pH = ­log[H3O+]e
= ­log(1 x 10­7)
=7
and similarly pOH = 7
This is where the formula pH + pOH = 14 comes from
The net reaction between an acid, it's conjugate base and water at
equilibrium is always
H3O+ + OH­
2 H2 O
So at equilibrium [H3O+] = 1 x 10­7 M. This is why "neutral" pH = 7
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1. What is the pH of a 0.0225 M solution of acetic acid (HOAc) if it's
Ka = 1.80 x 10­5?
2. Use the Ka above to determine the pH of a 0.0225 M solution of
acetate (OAc­).
3. Given the Ka above, could you determine the Kb of acetic acid?
What reaction would that describe?
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1. What is the pH of a 0.0225 M solution of acetic acid (HOAc) if it's
Ka = 1.80 x 10­5?
HOAc + H2O
H3O+ + OAc­
[H3O+]e[OAc­]e
Ka =
[HOAc]e
Q = 0/0.0225 = 0 < Keq so we will proceed towards products to get to
equilibrium.
[HOAc]
[H3O+]
[OAc­]
Approximation?
i
0.0225
0
0
c
­x
+x
+x
x
x
e
0.0225/1.8x10­5 >> 100
so
0.0225 ­ x = 0.0225
0.0225 ­ x
Ka = 1.8x10­5 =
(x)(x)
0.0225
x = 6.3639...x10­4 = [H3O+]e
pH = ­log(6.3639...x10­4)
= 3.20
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2. Use the Ka above to determine the pH of a 0.0225 M solution of
acetate (OAc­).
KaKb = Kw where the acid and base are a conjugate pair, like HOAc and OAc­.
Kb = Kw/Ka = 1x10­14/1.8x10­5 = 5.55...x10­10
The base reaction in water is:
OAc­ + H2O
Kb =
HOAc + OH­
[OH­]e[HOAc]e
[OAc­]e
Q = 0/0.0225 = 0 < Keq so we will proceed towards products to get to
equilibrium.
[OAc­]
[OH­]
[HOAc]
Approximation?
i
0.0225
0
0
c
­x
+x
+x
e
0.0225/5.5x10­10 >> 100
so
0.0225 ­ x = 0.0225
0.0225 ­ x
x
x
Kb = 5.55...x10­10 =
(x)(x)
0.0225
x = 3.5355...x10­6 = [OH­]e
pOH = ­log(3.5355...x10­6)
= 5.45..
pH = 14 ­ pOH = 8.55
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3. Given the Ka above (of acetic acid), could you determine the Kb of
acetic acid? NO
What reaction would that describe?
HOAc + H2O
H2OAc + OH­
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Neutralization and Titration
Recall: Strong acid + Strong base neutralization reaction:
HCl (aq) + NaOH (aq)
H2O (l) + NaCl (aq)
Ionic equation:
H+(aq) + Cl­(aq) + Na+(aq) + OH­(aq)
H2O(l) + Na+(aq) + Cl­(aq)
Net ionic:
H+(aq) + OH­(aq)
H2O (l)
pH = 7.0
Titration is the experimental method by which a controlled
volume of one solution with known concentration is added
to a controlled volume of another of unknown
concentration.
In so doing, we can determine the unknown concentration
(if we know the reaction the chemicals in the two solutions
will undergo).
During a titration, one usually measures a volume of
solution of unknown concentration into a flask, and then
fills a burette with the solution of known concentration.
The burette solution is added ("titrated") into the unknown
in a controlled way (using the burette).
Note: you could just as easily titrate the unknown solution
into the known ­ it will depend on the type of END POINT
you are trying to reach.
One titrates one solution into another until an END POINT
is reached. The end point is a visual cue that you have
added enough of one solution to the other. This end point
can involve the formation of a precipitate, colour change
or any other readily visible chemical phenomenon.
Neutralization titrations involve titrating an acid with a base (or vice versa). The acid is
added to the base until all the base has been neutralized (or vice versa). Once all the
original acid (or base) has been neutralized, we reach what is called the
"EQUIVALENCE POINT", i.e. the point at which an equivalent number of moles of acid
have been added to the base.
Ideally, the END POINT will occur at the EQUIVALENCE POINT...
In order to visualize the equivalence point in a neutralization titration, we need to add
an INDICATOR. The indicator is itself a weak acid or base that changes colour when
the pH of a solution reaches a certain value.
There are many indicators that cover various ranges of pH. One of the most commonly
used in neutralization titration is Bromothymol Blue, which is yellow below pH 6.0,
green between 6.0 and 7.3, and blue above 7.3. Because it changes colour around pH
7.0, it is a good indicator of the Equivalence Point for a neutralization between a strong
acid and strong base (which produces a pH of 7.0 once all the base has been
neutralized by the acid, or vice versa).
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Titration Curves
Graphing the pH of a titration of a strong acid like HCl with a strong base like
NaOH looks like:
Titration of 100 mL of 1 M HCl
end point of
phenolphthalein
9.3
pH 7.0
x
end point of BTB
x
neutralization point
100 mL
Volume of 1 M NaOH added (mL)
When equal moles of base have been added to the acid the pH is 7.0.
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Titration Curves
Graphing the pH of a titration of a strong base like NaOH with a strong acid
like HCl looks like:
Titration of 100 mL of 1 M NaOH
pH 7.0
x
neutralization point
100 mL
Volume of 1 M HCl added (mL)
When equal moles of acid have been added to the base the pH is 7.0.
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What is the pH of 50 mL of a 1.0 M NaOH solution into which you have
titrated 25 mL of 1.0 M HCl ?
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What is the pH of 50 mL of a 1.0 M NaOH solution into which you have
titrated 25 mL of 1.0 M HCl ?
NaOH + HCl
H2O + NaCl
13.5
x
pH
All the added HCl will neutralize the NaOH.
We start with 0.05 L x 1.0 mol/L NaOH, or 0.05 mol
We add 0.025 L x 1.0 mol/L HCl, or 0.025 mol.
25
VHCl (mL)
We neutralized 0.025 mol of the NaOH.
We are left with 0.05 ­ 0.025 = 0.025 mol NaOH.
The new volume of solution is 0.05 L + 0.025 L = 0.075 L
New concentration of OH­ is 0.025mol/0.075 L = 0.333 mol/L
pOH = ­ log[OH­]
pOH = 0.48
pH = 14 ­ pOH
pH = 14 ­ 0.48
pH = 13.5
The original pH of the NaOH solution was:
pH = 14 ­ pOH
pH = 14 ­ (­log[OH­])
pH = 14 + log (1)
pH = 14
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
49 mL of 1.0 M HCl?
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
49 mL of 1.0 M HCl?
nNaOH = 0.05 L x 1.0 mol/L = 0.05 mol
nHCladded = 0.049 x 1.0 mol/L = 0.049 mol
nNaOH left = 0.05 ­ 0.049 = 0.001 mol
new Volume = 0.05 + 0.049 L = 0.099 L
CNaOH = 0.001 mol / 0.099 L = 0.0101...mol/L
pOH = ­log(0.0101...) = 1.9956...
13.5
x
12
pH
x
25
49
VHCl (mL)
pH = 14 ­ pOH = 12.00...
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
49.9 mL of 1.0 M HCl?
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
49.9 mL of 1.0 M HCl?
nNaOH = 0.05 L x 1.0 mol/L = 0.05 mol
nHCladded = 0.0499 x 1.0 mol/L = 0.0499 mol
nNaOH left = 0.05 ­ 0.0499 = 0.0001 mol
new Volume = 0.05 + 0.0499 L = 0.0999 L
CNaOH = 0.0001 mol / 0.0999 L = 0.001001...mol/L
pOH = ­log(0.001001...) = 2.999...
13.5
12
pH
11
x
x
x
25 49
49.9
VHCl (mL)
pH = 14 ­ pOH = 11.00...
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
50 mL of 1.0 M HCl?
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
50 mL of 1.0 M HCl?
nNaOH = 0.05 L x 1.0 mol/L = 0.05 mol
nHCladded = 0.0499 x 1.0 mol/L = 0.05 mol
nNaOH left = 0.05 ­ 0.0499 = 0 mol
new Volume = 0.05 + 0.05 L = 0.0999 L
So there is only water left, and the pH comes from
the equilibrium of the dissociation of water itself.
13.5
x
x
x
x
12
11
pH 7
25
50
VHCl (mL)
2 H2O
H3O+ + OH­
[H3O+]e = 1.0 x 10­7
pH = 7
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
51 mL of 1.0 M HCl?
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What's the pH of 50 mL of 1.0 M NaOH to which you have added
51 mL of 1.0 M HCl?
nNaOH = 0.05 L x 1.0 mol/L = 0.05 mol
nHCladded = 0.051 x 1.0 mol/L = 0.051 mol
There is now no NaOH left, but there will
be some excess HCl in the solution
nHClexcess = 0.001 L x 1.0 mol/L = 0.001 mol
13.5
x
x
x
x
12
11
pH
7
2
x
25
V = 0.05 + 0.051 = 0.101 L
CHCl = 0.001 / 0.101 = 0.0099... mol/L
51
VHCl (mL)
pH = ­log[H+] = ­log(0.0099...)
pH = 2.004...
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Neutralization and Titration:
Weak Acids and Strong Bases
If you titrate a weak acid with a strong base (or vice versa), the conjugate base of the
weak acid is itself a weak base...
ex.
H3CCOOH + NaOH
and then H3CCOO­ + H2O
H2O + H3CCOO­ + Na+
H3CCOOH + OH­
This means that at the equivalence point, you actually have the initial conditions for a
weak base equilibrium (second reaction above), which will produce free OH­ (albeit not
much). This means that at the equivalence point of a titration of a weak acid with a
strong base, you will create a solution that is slightly basic.
1. Say we start with 50 mL of a 1.0 M solution of NaOH, and titrate it with a 1.0 M
solution of acetic acid. What will be the pH at the equivalence point (after everything
has equilibrated)?
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nNaOH titrated = 0.05 L x 1.0 mol/L = 0.05 mol
nHOAC used to get to equivalence point = 0.05 mol
H3CCOOH + NaOH
H2O + H3CCOO­ + Na+
All of the HOAc used will have turned into OAc­
This OAc­ will in turn equilibrate in solution by acting like a base with water:
H3CCOOH + OH­
H3CCOO­ + H2O
From Appendix C, the Ka for HOAc = 1.75 x 10­5.
Since Kw = KaKb, the Kb will be Kw/Ka = 1 x 10­14/1.75 x 10­5
Kb = 5.71428...x 10­10
Kb =
[HOAc]e[OH­]e
[OAc­]e
The initial [OAc­] will be 0.05 mol / (0.05 + 0.05) L = 0.5 mol/L
[HOAc]
I
0
+x
C
x
E
[OAc­]
0.5
­x
0.5 ­ x
[OH­]
0
+x
x
[OAc­] / Kb >> 100, so we can
approximate that 0.5 ­ x 0.5 for
equilibrium calculation purposes.
Kb =
(x)(x)
0.5
= 5.71428...x 10­10
x = 1.6903... x 10­5 = [OH­]e
pOH = ­log[OH­]e = 4.772...
pH = 14 ­ pOH = 9.227...
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Titration of 50 mL of 1 M NaOH with acetic acid
pH
9.2
7
x
equivalence point
50 mL
Volume of 1 M HOAc added (mL)
When equal moles of acid have been added to the base the pH is
slightly > 7.0.
Recall the End points of Bromothymol Blue (7.5) and Phenolphthalein (9.3)...
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2. What would the pH be after only 25 mL of the weak acid was added?
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2. After 25 mL of 1.0 M HOAc added we'd have added 0.025 mol.
This means that there would be 0.05 mol ­ 0.025 mol = 0.025 mol NaOH left.
H3CCOOH + NaOH
H2O + H3CCOO­ + Na+
0.025 mol
0.025 mol
The volume of solution would now be 0.075 L
CNaOH = 0.025 / 0.075 = 0.333...mol/L
pH = 14 ­ (­log(0.333...))
pH = 13.5228...
...but wait ­ do we need to take into account the weak base we've made
too? The equilibrium between OAc­ and water is still going on...
At this point in the titration we'd have an initial concentration of OAc­ of
0.025/0.075 = 0.333...mol/L as well.
Using the Kb and equilibrium calculation as before and since we now have
[OH­]i = 0.33333... M
(x)(0.333...+ x)
[HOAc] [OAc­] [OH­]
Kb =
= 5.71428...x 10­10
0.333...­
x
I
0.333...
0.333...
0
0.333.../5.71...x10­10 >>> 100, so
­x
+x
C
+x
(x)(0.333)
x
E
0.333 ­ x 0.333...x
Kb =
= 5.71428...x 10­10
0.333...
x = [OH­] because of this equilibrium = 5.71... x 10­10 mol/L
compare: pOH = ­log(0.3333333333...) vs.
pH = 13.522878745...
pOH = ­log(0.3333333339...)
pH = 13.522878746...
~5x10­9% difference, so unless we are very close to the equivalence
point, the equilibrium's contribution of OH­ is insignificant compared to the
contribution of the unreacted OH­
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Practice:
pg. 599 # 1 ­ 3
pg. 607 # 4, 5
pg. 608 # 6
pg. 611 # 7 ­ 9
pg. 620 # 1, 2
pg. 620 # 1 ­ 8
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Neutralization and Titration:
Weak Bases and Strong Acids
If you titrate a weak base with a strong acid (or vice versa), the conjugate acid of the
weak base is itself a weak acid...
ex.
NH3 + HCl
NH4+ + Cl­
NH4+ + H2O
H3O+ + NH3
This means that at the equivalence point, you actually have the initial conditions for a
weak acid equilibrium, which will produce free H3O+ (albeit not much). This means that
at the equivalence point of a titration of a weak base with a strong acid, you will create
a solution that is slightly acidic.
Say we start with 50 mL of a 0.1 M solution of NH3, and titrate it with a 0.1 M solution of
HCl. What will be the pH at the equivalence point (after everything has
equilibrated)?
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NH3 + HCl
NH4+ + Cl­
NH4+ + H2O
H3O+ + NH3
At the equivalence point we'll have added 0.05 L of HCl (same concentration,
so same # moles)
[NH4+]i = (0.05 L x 0.1 mol/L)/(0.05 L + 0.05 L)
= 0.05 M
This will now equilibrate. The Ka = 5.8 x 10­10 (appendix C)
[NH4+]
[NH3]
[H3O+]
I
0.05
0
0
C
­x
+x
+x
x
x
E
0.05 ­ x
Ka =
(x)(x)
0.05 ­ x
we can obviously approximate
Ka =
(x)(x)
0.05
x = [H3O+]e = 5.38...x 10­6
pH = 5.2688...
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Titration of 50 mL of 0.1 M NH3 with 0.1 MHCl
pH
7
5.3
x
equivalence point
50 mL
Volume of 0.1 M HCl added (mL)
When equal moles of acid have been added to the base the pH is
slightly < 7.0.
Do some research and find out which indicator might be a good one
to visualize this equivalence point.
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Practice:
pg. 588 # 1 ­ 6
pg. 589 # 7, 8
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Investigation E.2 Analysis
Mass of KHP: 2.05 g
[KHP] = 2.050 g / 204.22 g/mol / 0.1000 L = 0.10038... M (don't round!)
Trial 1: 9.7 mL
Trial 2: 9.8 mL
KHP + NaOH
Trial 3: 9.8 mL
NaKP + H2O
nKHP = nNaOH = 0.0100 L x 0.10038.... mol/L = 0.0010038... mol
[NaOH]1 = 0.0010038... mol / 0.0097 L = 0.1034... M
[NaOH]2 = 0.0010038... mol / 0.0098 L = 0.10243... M
[NaOH]2 = 0.0010038... mol / 0.0098 L = 0.10243... M
[NaOH]ave = 0.10275.... M
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Investigation E.2 Analysis
[NaOH]ave = 0.10275.... M
Literature value for Ka of Potassium bitartrate = 3.981 x 10­5
(change this # in your lab guides)
initial pH of solutions = 3.83
C4H4O62­ + H3O+
HC4H4O6­ + H2O
NaKC4H4O6 + H2O
NaOH + KHC4H4O6
Trial 1: 2.7 mL
nKHC4H4O6 = nNaOH = (0.0027 L)(0.10275 mol/L) = 0.000277...
[KHtartrate] = 0.000277 mol / 0.0100 L = 0.0277... M
Do this again for your two other best trials, take average...
NaKC4H4O6 + H2O
KHC4H4O6 + OH­
if we had measured the pH of the titrated solution it
would have been basic...
I
C
E
[HC4H4O6­]
[C4H4O62­]
[H3O+]
0.0277425
0
0
+x
+x
x
x
­x
0.0277425 ­ x
Ka =
[C4H4O62­]e[H3O+]e
[HC4H4O6­]e
(x)(x)
=
0.0277425 ­ x
pH = 3.83 = ­log[H3O+]e
[H3O+]e = x = 10­3.83 = 1.4791... x 10­4 M
Ka =
(1.4791... x 10­4)2
0.0277425 ­ 1.4791... x 10
­4
= 7.928 x 10­7
So why???
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Quantitative Equilibrium Practice Problems
1. In the reaction A(aq) + 3 B(aq) ­­­­­> C(aq) + D(aq), 50.0 mL of a 0.100 M solution
of A is added to 100.0 mL of a 0.200 M solution of B.
What are the equilibrium concentrations of all species if Keq = 1.56 x 10­4?
2. 100.0 mL of 5.00 x 10­6 M Al(NO3)3 (aq) is added to 200.0 mL of 5.00 x 10­6 M
NaCl(aq). What is the mass of the precipitate that forms if the Ksp for AlCl3 is
1.00 x 10­12?
3. What is the pH of a 0.250 M solution of NaHSO4? The Ka of HSO4­ is
1.2 x 10­2.
4. What is the pH of a 0.250 M solution of NaCN? The Ka of HCN is
6.2x 10­10.
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1. In the reaction A(aq) + 3 B(aq) ­­­­­> C(aq) + D(aq), 50.0 mL of a 0.100 M solution
of A is added to 100.0 mL of a 0.200 M solution of B.
What are the equilibrium concentrations of all species if Keq = 1.56 x 10­4?
V1C1 = V2C2
Initial concentrations:
Keq =
[A]i = (50)(0.1)/(150) = 0.0333...
[B]i = (100)(0.2)/(150) = 0.1333...
[A]
C
E
­x
[A]e[B]e3
Q=
[B]
0.0333... 0.1333...
I
[C]e[D]e
­3x
0.0333... ­ x 0.1333... ­ 3x
[C]
[D]
0
0
+x
+x
x
x
[C]i[D]i
[A]i[B]i3
=0
(x)(x)
1.56 x 10­4 =
(0.0333... ­ x)(0.1333... ­ 3x)3
0.1333.../1.56 x 10­4 = 854... > 300, so 0.1333... ­ 3x = 0.1333...
0.0333.../1.56 x 10­4 = 213... > 100, so 0.0333... ­ x = 0.0333...
1.56 x 10 =
­4
x2
(0.0333...)(0.1333...)3
x = 1.1102...x 10­5 (reject negative root)
[A]e = 0.0333... ­ 1.1102...x 10­5 = 0.0332 M
[B]e = 0.1333... ­ 3(1.1102...x 10­5) = 0.133 M
[C]e = [D]e = 1.11...x 10­5 = 1.11...x 10­5 M
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2. 100.0 mL of 5.00 x 10­6 M Al(NO3)3 (aq) is added to 200.0 mL of 5.00 x 10­6 M
NaCl(aq). What is the mass of the precipitate that forms if the Ksp for AlCl3 is
1.00 x 10­12?
1st part: Assume (pretend) all the precipitate that can be made will be made (this
is a grade 11 Limiting Reactant problem).
2nd part: Determine the amount that would dissolve using the Ksp and subtract
that from the 1st part.
Al(NO3)3 (aq) + 3 NaCl(aq) ­­­­­­­­> AlCl3 (s) + 3 NaNO3 (aq)
nAl(NO3)3 present = 0.1 L x 5E­6 mol/L = 5x10­7 mol
nAl(NO3)3 needed = 0.2 L x 5x10­6 mol/L x 1 molAl(NO3)3/3molNaCl
= 3.333...E­7 mol
So NaCl is limiting.
nAlCl3 formed "initially" = 0.2 x 5E­6 molNaCl x 1molAlCl3/3molNaCl = 3.333...E­7 mol
Al3+(aq) + 3 Cl­(aq)
Now for the second part. If we have AlCl3(s)
Excess [Al3+] = 0.1 L x 5E­6 M ­ 3.3...E­7 mol / 0.3 L
= 5.55...E­7 M
­
3+
[Cl ]
[Al ]
Ksp = [Al3+]S[Cl­]S3
5.55E­7
0
I
Ksp = 1.00 x 10­12 = (5.55E­7+ x)(3x)3
C
E
+x
+3x
5.55E­7 + x
3x
x = [Al3+]
Assumption: 5.55E­7/1E­12 >> 100, so
5.55E­7 + x = 5.55E­7
Ksp = 1.00 x 10­12 = (5.55E­7)(3x)3
x = 4.0548...x10­3
nAl3+ = 4.05...x10­3 mol/L x 0.3 L = 0.001216... mol
nAl3+ = nAlCl3, so 0.001216... mol of AlCl3 can dissolve.
Since we could have only made 3.333 x 10­7 mol based on the
amounts of reactants we have, all of this should dissolve (i.e. we
are not at saturation yet).
***NOTE: It's definitely worth your while to calculate Q before
doing all this equilibrium work. Remember that if Q < Keq you
won't make any ppt at all...
Q = [Al3+]i[Cl­]i3
[Al3+]i = 5 x 10­6 M x 0.1 L / (0.1 L + 0.2 L)
= 1.666... x 10­6 M
[Cl­]i = 5 x 10­6 M x 0.2 L / (0.1 L + 0.2 L)
= 3.333... x 10­6 M
Q = 1.666 x 10­6(3.333 x 10­6)3 = 6.17... x 10­23, so <<< Keq
i.e. we won't make any ppt yet
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3. What is the pH of a 0.250 M solution of NaHSO4? The Ka of HSO4­ is
1.2 x 10­2.
If NaHSO4­ is acting like an acid in water, the reaction would be:
HSO4­ (aq) + H2O(l) ­­­> SO42­(aq) + H3O+(aq)
Ka =
Q = 0, so:
[H3O+]e[SO42­]e
[HSO4­]e
[HSO4­] [SO42­]
0
[H3O+]
I
0.25
C
­x
+x
+x
E
0.25 ­ x
x
x
Ka = 1.2 x 10 =
­2
0
approximation?
0.25/1.2x10­2 is NOT > 100,
so we can't make an
approximation here.
x2
0.25 ­ x
0.012(0.25) ­ 0.012x = x2
0 = x2 + 0.012x ­ 0.003
2
x = ­0.012 +/­ √[(0.012) ­ 4(1)(­0.003)]
2(1)
x = 0.0490999... = [H3O+]e
pH = ­log(0.049...) = 1.3089... = 1.3
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4. What is the pH of a 0.250 M solution of NaCN? The Ka of HCN is
6.2x 10­10.
We are given the Ka of the conjugate acid to the molecule we
are starting with. What we really need is the Kb describing the
reaction:
CN­ + H2O ­­­> HCN + OH­
KaKb = Kw, so Kb for this reaction = 1x10­14/6.2x10­10 = 1.6129...x10­5
[CN­]
[HCN]
[OH­]
I
0.25
C
­x
+x
+x
E
0.25 ­ x
x
x
0
Kb =
[HCN][OH­]
[CN­]
0
0.25/1.6...x10­5 >> 100, so
0.25 ­ x = 0.25
(x)(x)
0.25
x = 0.002... = [OH­]
Kb = 1.61...x10­5 =
pOH = ­log(0.002...) = 2.697...
pH = 14 ­ pOH = 11.3... = 11
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Practice:
pg. 546 # 12 ­ 16
pg. 549 # 17 ­ 19
pg. 549 ­ 550 # 1 ­ 8
pg. 554 # 1, 2
pg. 556 # 3 ­ 5
pg. 562 # 6
pg. 568 # 7, 8
pg. 570 # 9, 10
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Neutralization and Titration Review Problems
Review problems (good exam review ­ use your textbooks for required constants):
1. What is the pH of a 0.0450 M solution of hydrogen peroxide, H2O2?
2. What is the pH of a 0.0450 M solution of pyridine, C5H5N?
3. What does it mean in terms of a chemical reaction to say that the Ka of perchloric
acid, HClO4 is "very large"? What would be the pH of a 0.0450 M solution of perchloric
acid?
4. The Kb of ammonia is 1.8 x 10­5. Write out the reaction to which this Kb is referring.
Which are the base, acid, conjugate acid and conjugate base? What is the Ka of
ammonia's conjugate acid, and to what reaction is that Ka referring? What is the
expression of this Ka?
5. If 0.000100 mol of Fe(OH)3 were placed in 50.0 mL of water, would there be a
precipitate? What would be the concentrations of Fe3+ and OH­ at saturation? How
much would the precipitate weigh in grams (if any)? What is the pH of a saturated
solution of Fe(OH)3?
6. What would be the pH of 150 mL of a 0.5 M solution of acetic acid to which
a) 0.65 mL of a 0.75 M solution of NaOH had been titrated?
b) How much NaOH solution would need to be added to get to the equivalence point?
c) What is the pH of the equivalence point?
d) Do some research ­ what would be a good indicator to use to visualize this
equivalence point (i.e. which indicator has an end point at a pH near to the equilvaence
point's pH)?
e) What is the pH of this solution after 125 mL of the NaOH solution has been added?
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Review:
pg. 579 # 1 ­ 23
pg. 631 # 1 ­ 19
pg. 633 # 1 ­ 25
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Introduction to Thermodynamics
First Law: Energy cannot be created nor destroyed, only transferred
Spontaneity: A reaction that, once started, will sustain itself without outside
input of energy
ex.
2 H2O(l) + CO2 (g)
CH4 (g) + 2 O2 (g)
exothermic and spontaneous
2 H2(g) + O2 (g)
2 H2O (l)
endothermic and non­spontaneous
NH4NO3 (s)
H2O
NH4+ (aq) + NO3­ (aq)
endothermic and spontaneous
H2O(s)
H2O(l)
endothermic and spontaneous only at
temperatures above 0oC.
So what determines whether a reaction is spontaneous, and what
implications does that have?
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Entropy
From the examples shown, the release of heat (being exothermic) helps a
reaction or physical process be spontaneous because the energy released
helps the reaction overcome the Ea further molecules need to react. So,
∆H is a factor in reactions being spontaneous, but it's not the only factor...
It looks like temperature is also a factor (melting ice above 0oC).
It also looks like there is something else at play ­ why would NH4NO3
dissolving both take up energy from the surroundings and be spontaneous?
ENTROPY change, ∆S: Entropy, in a simplistic sense, can be thought of as
the inherent DISORDER of a chemical system, or of its surroundings.
The universe overall spontaneously progresses towards a state of more and
more disorder. Think of what happens if you drop a test tube on the floor...
Chemical systems that create atoms/molecules that are physically further
apart, moving faster, or have more, smaller species show an increase in
Entropy.
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Try pg. 498 # 1
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Entropy and the 2nd Law of Thermodynamics
∆Suniverse > 0
All processes increase the entropy of the universe overall. Even if a process,
like a chemical reaction, decreases its internal entropy, the surroundings of
that process will experience an even greater increase in entropy.
∆Suniverse = ∆Ssystem + ∆Ssurroundings
How does "life" fit in with the above equations?
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Enthalpy and Spontaneity:
If the bonds in the reactants have more potential energy than the
products, the transformation from reactants to products will release
energy and tend to be more spontaneous, very much like a book falling
off your desk.
∆Ho = Hoproducts ­ Horeactants, and if ∆Ho < 0, more spontaneous
Entropy and Spontaneity:
If a chemical system becomes more disordered in the course of a
reaction or transformation, it tends to be more spontaneous, very much
like a deck of cards you throw into the air falling down all over the
ground.
∆So = Soproducts ­ Soreactants, and if ∆So > 0, more spontaneous
Temperature and Spontaneity:
Raising the temperature of a chemical or physical process tends to
increase the disorder, and give more energy to overcome Ea, so
raising the temperature tends to promote spontaneity.
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Spontaneity and Gibbs Free Energy
Free Energy is energy that can do Work.
Work, in thermodynamics, is defined as an event that changes the potential
energy of something. If you lift up a box and put it on a shelf, you have done
Positive Work by increasing the gravitational potential energy of the box
(you've given the box the potential to fall off the shelf and go boom).
Heat energy tends not to be "free" energy because it is an average of kinetic
energy of many particles all moving in random directions. It's kind of like
everyone in the class trying to move the box in different directions all at once
­ the box won't gain any energy of its own overall.
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Gibbs Free Energy, Enthalpy, Entropy and Temperature
~150 years ago a physicist named Josiah Gibbs experimented with energy
transformations in physical and chemical systems and determined a
relationship between "useful" energy, enthalpy, entropy and temperature.
This is the "Gibbs Free Energy", ∆G, equation, and is fundamental to
thermodynamics.
Gfinal ­ Ginitial = ∆G = ∆H ­ T∆S
If ∆G < 0, a process is spontaneous. If ∆G > 0, it is not spontaneous.
Take a moment and think if this makes sense based on what we know
about enthalpy, entropy, temperature and spontaneity.
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Measuring ∆G Using ∆H and ∆S
Recall the enthalpy change for the formation of a substance is relative, and we
define that change as 0 kJ if the substance is in its most common form on
Earth (like O2 (g) or C(graphite)).
The THIRD LAW of THERMODYNAMICS is involved with the measurement of
Entropy:
The Entropy of a perfectly ordered crystalline substance is 0 at
absolute 0
S = 0 when T = 0 K
This implies that the entropy of a substance depends on temperature. The
units for entropy are actually J/K.
Recall the Standard Enthalpy of a reaction can be determined by the
Standard Enthalpies of Formation:
∆Ho = Σ(nproducts∆Hof products) ­ Σ(nreactants∆Hof reactants)
(from Appendix C)
Standard Entropies of a reaction are calculated in a similar way, but
since we are comparing all entropies to an identical theoretical "zero"
level there is no individual "formation" entropy we need to reference to
for each:
∆So = Σ(nproductsSoproducts) ­ Σ(nreactantsSoreactants) (from Appendix C)
and for any chemical or physical process:
∆Go = ∆H o ­ T∆ So
Note: at standard conditions, p = 100 kPa and T = 25oC, so 298.15 K
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ex. What is the standard free energy change of the hydrogenation of ethene?
Is it spontaneous under standard conditions?
C2H4 (g) + H2 (g)
C2H6 (g)
From Appendix C, or other source:
∆Ho = 1 mol x ­84.5 kJ/mol ­ (1 mol x 51.9 kJ/mol + 1 mol x 0 kJ/mol)
∆Ho = ­136.4 kJ
∆So = 1 mol x 0.2295 kJ/molK ­ (1 mol x 0.2198 kJ/mol + 1 mol x 0.130 kJ/mol)
∆So = ­0.1203 kJ
T = 298.15 K
∆ G o = ∆H o ­ T ∆ S o
∆Go = ­136.4 kJ ­ 298.15(­0.1203)
∆Go = ­100.532555 kJ
Since ∆Go is negative this reaction is spontaneous at 25oC. If you recall in
organic chemistry the hydrogenation of an alkene to get an alkane required a
catalyst like Pd/C. This is because despite being spontaneous, there is a very
high activation energy barrier to overcome to get things started.
What that implies is that this reaction is spontaneous, but could take millions
or billions of years to get going.
Kinetics (reaction rates and specific mechanisms) have NOTHING to do with
Free Energy, or Thermodynamics in general. Free Energy, Enthalpy and
Entropy are what are called State Functions ­ they are only concerned with
the intial and final states of a reaction or process in general.
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If
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∆G and Equilibrium
∆G = ∆H ­ T∆S for any reaction A
B
how does this apply to reversible reactions?
If, at a given temperature ∆G < 0, the reaction is spontaneous in the
direction written. If ∆G > 0, the reaction must be spontaneous in the opposite
direction.
For every reaction there is a temperature at which ∆G = 0, i.e. the reaction is
equally spontaneous in both the forward and reverse directions.
If a reaction is "equally spontaneous" in both directions, the overall levels of
A and B above aren't changing, and we have a DYNAMIC EQUILIBRIUM.
∆G = 0 at Equilibrium
and Tequilbrium = ∆H/∆S
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∆G and Keq
We can show that the Gibbs free energy associated with a system getting
to equilibrium is:
∆G = ∆Go + RTlnQ
This implies that a system can do work if it is not yet at equilibrium.
If a system is already at equilibrium, Q = Keq and ∆G = 0, so:
0 = ∆Go + RTlnKeq
and
∆Go = ­RTlnKeq where R is the ideal gas constant
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Practice:
pg. 511 # 3, 4
pg. 512 # 1 ­ 19
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Review:
AP Review:
pg. 522 # 1 ­ 18
On AP Classroom try Progress Checks
for Unit 8 (FR and MCQ)
AP ONLY: # 19, 20
pg. 523 ­ 524 # 1 ­ 24
AP ONLY: # 25
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The Henderson­Hasselbach Equation and the pH of a Buffer
3.
To answer question #3 we need to think about how to approach a
system in which we start with both the acid and conjugate base pair of
a weak acid.
After 75 mL of 1.0 M HOAc added we'd have added 0.075 mol.
This means that there would be 0.05 mol of OAc­ made from the initial
neutralization of NaOH, and now an excess of 0.025 mol HOAc. The total
volume is now 0.125 L.
We now have an initial concentration of:
HOAc = 0.025/0.125 = 0.2 M
OAc­ = 0.05/0.125 = 0.4 M
We can look at this system from an acid dissociation point of view:
H3CCOOH + H2O
H3CCOO­ + H3O+
This acid/conjugate base pair will establish a new equilibrium according to the
Ka:
[OAc­]e[H3O+]e
Ka =
if we take the log of both sides:
[HOAc]e (and let OAc­ be represented by A­ and HOAc by HA)
logKa = log([A­][H3O+]/[HA])
using rules of logs:
logKa = log[A­] + log[H3O+] ­ log[HA]
­log[H3O+] = ­logKa + log[A­]/[HA]
and ­log[H3O+] is of course the pH
­logKa is something we call the pKa of the weak acid (more later)
pH = ­logKa + log[A­]/[HA]
The Ka of acetic acid is 1.75 x 10­5, and we can think of the [A­] and [HA] as
the initial concentrations of the acid and base at the instant we've titrated in
75 mL of HOAc.
So, the pH = ­log(1.75 x 10­5) + log[(0.4)/(0.2)]
pH = 5.05799...
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The Henderson­Hasselbach Equation and the pH of a Buffer
The Henderson­Hasselbach equation is:
pH = ­logKa + log[A­]/[HA]
For any reaction in which a weak acid (or base) dissociates in water
HA + H2O
A­ + H3O+
A pH BUFFER is a substance that limits pH changes in an aqueous
solution. Any weak acid or base is a BUFFER.
There are very many buffers in our bodies. These buffers help keep the
environments in our bodies from changing in acidity level too much. This is
very important because ENZYMES, which catalyze just about every
reaction in our bodies, require specific environments (temperature, pH...)
in which to function properly. Without buffers, enzymes would denature
(break down) and we'd die.
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Buffers and pH Change
What would the pH of a solution be if you added 0.05 mol HCl to 100 mL?
What would the pH of a solution be if you added 0.05 mol HCl to 100 mL of
a 1 M acetic acid solution, Ka = 1.75 x 10­5?
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Buffers and pH Change
What would the pH of a solution be if you added 0.05 mol HCl to 100 mL?
HCl is a strong acid all of which will react with H2O to make H3O+:
H3O+ + Cl­
HCl + H2O
[HCl] = [H3O+] = 0.05 mol / 0.1 L = 0.5 M
pH = ­log(0.5) = 0.301...
What would the pH of a solution be if you added 0.05 mol HCl to 100 mL of
a 1 M acetic acid solution, Ka = 1.75 x 10­5?
HOAc + H2O
H3O+ + OAc­
Here we'd have a initial equilibrium between HOAc and OAc­. The HCl will
react with the strongest base that is around first, so in this case it will
react with the OAc­ from the equilibrium before it reacts with H2O. Any
OAc­ that turns into HOAc will pull the equilibrium over to the OAc­ side
(LeChatelier's principle). Effectively, all the HCl we add will create HOAc
before H3O+.
If we add 0.05 mol HCl to the 100 mL of acetic acid buffer, we make 0.05
mol of HOAc in 0.1 L = 0.5 M HOAc.
We'll have 1 x 0.1 ­ 0.05 = 0.05 mol OAc­ left over, or 0.5 M OAc­
From the H­H equation:
pH = ­logKa + log[A­]/[HA]
pH = ­log(1.75 x 10­5) + log(0.5/0.5)
pH = 4.7569...
So the effect of adding the same amount of HCl to pure water is far
greater than adding it to a buffer solution.
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ex. a. A solution was prepared by dissolving 0.0200 moles of acetic acid,
HOAc, pKa = 4.75 in water to give 1.00 L of solution. What is the pH?
b. 0.00800 mol of NaOH was then added to this solution. What is the new
pH?
c. An additional 0.0120 mol NaOH is then added. What is the pH now?
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ex. a. A solution was prepared by dissolving 0.0200 moles of acetic acid,
HOAc, pKa = 4.75 in water to give 1.00 L of solution. What is the pH?
HOAc + H2O
OAc­ + H3O+
pKa = ­logKa
Ka = 10­pKa = 10­4.75 = 1.7782...x 10­5
[HOAc] [OAc­] [H3O+]
I
0.02
0
0
C
­x
+x
+x
x
x
E
0.02 ­ x
Ka =
(x)(x)
0.02 ­ x
approximation?
[HOAc]i / Ka = 0.02/1.778... x 10­5
= 1124 >> 100
so we can say 0.02 ­ x 0.02
1.7782...x 10­5 =
(x)(x)
0.02
x = [H3O+]e = 5.963...x 10­4
pH = 3.22
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b. 0.00800 mol of NaOH was then added to this solution. What is the new
pH?
Any NaOH added will react with the strongest acid around first, so it will turn
HOAc into OAc­, so [NaOH] = [OAc­].
[OAc­] = 0.008 M
[HOAc]new = 0.02 ­ 0.008 = 0.012 M (we can do this directly because we're in
exactly 1 L, otherwise we'd have to calculated the new # moles of HOAc and
divide by volume).
pH = pKa + log([OAc­]/[HOAc])
pH = 4.75 + log(0.008/0.012)
pH = 4.57
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c. An additional 0.012 mol NaOH is then added. What is the pH now?
We've now added 0.02 mol NaOH, so all the original HOAc will have been
neutralized and turned into the weak base OAc­. This will now re­equilibrate
in the basic reaction:
OAc­ + H2O
HOAc + OH­
Kb = Kw/Ka = 1 x 10­14 / 1.778...x 10­5 = 5.623...x 10­10
[OAc­] [HOAc] [OH­]
I
C
E
0.02
0
0
­x
+x
+x
x
x
0.02 ­ x
Kb =
(x)(x)
0.02 ­ x
we can obviously approximate
Kb =
(x)(x)
0.02
x = [OH­]eq = 3.35...x 10­6
pH = 14 ­ 5.47...
pH = 8.53
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More on the pKa of a buffer
pKa = ­ log (Ka)
The pKa of an acid indicates the pH at which it exists in exactly half acidic form and
half basic form. For example:
H3CCOOH + H2O
H3O+ + H3CCOO­
Ka = 1.8 x 10­5,
so, when in solution [H3CCOOH] = [H3CCOO­], the pH = pKa, so, since the
pKa = ­ log (Ka)
pH = ­ log 1.8 x 10­5
pH = 4.74
In a titration, the pKa indicates the pH at which exactly half of the weak acid has
been titrated (i.e. [HA] = [A­] where the source of the A­ was the strong base you
would have added.
Titration of 150 mL of 0.5 M acetic acid with 0.5 M NaOH
9.1
x
pH 7
4.74
2.52
x
BUFFER ZONE
x
125 mL
75 mL
150 mL
Volume NaOH added (mL)
The pKa is very biologically significant. It is the best pH at which a weak acid
will act as a BUFFER (a substance that enables a solution to resist changes
in pH due to added base or acid).
The best buffering pH of a weak acid is generally pKa +/­ 1 pH unit. This is
the pH range in which the pH will change the least when a base (or acid in
the case of a weak base buffer) is added.
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Note: The Henderson­Hasselbach Equation is Actually an
Approximation
It works very well for concentrations of Conjugate base and Acid that are
close to one another, but gets worse and worse at predicting the pH of a
buffer solution the larger (or smaller) the ratio of [A­]/[HA] gets.
Check this out: https://chem.libretexts.org/Bookshelves/
Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Henderson­
Hasselbach_Equation
Because we are usually using this equation to determine the pH of a
buffer, and we usually don't use buffers out of their buffering range (pKa
+/­ 1) this equation does a good enough job for most calculations we
need it for.
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pKa and the titration of Polyprotic acids
Polyprotic acids ("polyprotic" is greek for "many protons") are acids that can act as
acids with H2O more than once. They are acids that are strong enough to lose more
than one proton (H+) to water.
ex.
H2SO4 + H2O
HSO4­ + H3O+
Ka1 > 1000
HSO4­ + H2O
SO42­ + H3O+
Ka2 > 1 x 10­2
Titration of 150 mL of 0.5 M H2SO4 with 0.5 M NaOH
x equivalence point 2 (all HSO ­ is now SO ­)
4
4
7
pH
pKa2
x
x
pKa1
x
equivalence point 1 (all H2SO4 is now HSO4­)
150 mL
300 mL
Volume 0.5 M NaOH added (mL)
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Practice:
pg. 613 # 10
pg. 613 ­ 614 # 1 ­ 14
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