Calculating pH
Chapter 16 part IV
Summary of General Strategies
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Think Chemistry: focus on solution and
the components. It is usually easy to
identify one reaction that is not important.
Be systematic: Acid-base problems
require step by step approach.
Be Flexible: Although all acid-base
problems are similar, important differences
do occur. Do not force a given problem
into matching a problem you have solved
before.
Summary of General
Strategies
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Be patient: the complete solution to a
complicated problem cannot be seen
immediately in all its detail. Pick the
problem apart into workable steps.
Understand & Think: don’t just
memorize.
Calculating the pH of a
strong acid
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We need to look at the solutions
components and their chemistry.
For example: 1.0 M HCL has virtually
no HCl molecules as HCl ionizes
“completely” into H+ and Cl-.
We need to determine which
components are significant, and which
can be ignored.
Calculating the pH of a
strong acid
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Find the major species: the solution
components present in large amounts.
Back to our example: 1 M HCl
If it is 1 M than it must be a solute in
a solvent to make a solution. Solvent
is water.
Major species are H+, Cl- & H2O. HCl
is virtually dissociated and therefore
not a major species.
Calculating the pH of a
strong acid
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H+ is a major species from HCl, but
what about H+ from H2O?
H2O  H+ + OHIs this an important source of H+?
In pure water at 25°C [H+]= 1X10-7M
In water with 1.0 M HCl, water will
produce even less H+. Why?
Le Châtelier’s Principle
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Le Châtelier’s states that equilibrium
will shift to reduce stress. If  [H+] is
produced by HCl, then H2O produce
less H+ is response to that stress.
Therefore: if 1 M [H+] from HCl, then
is 1X10-7M[H+] from water significant?
No, pH=-log[1.0]
pH =0.0
Calculating the pH of a
strong acid
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Calculate the pH of 0.1 M HNO3.
Major species:
H+
NO3H2O
HNO3.
0.1M 0.1M
1X10-7M
0
pH = -log[H+]
pH = -log[0.1]
pH=1.00
Calculating the pH of a
strong acid
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Calculate the pH of 1.0x10-10M HCl
Major species:
H+
ClH2O
HCl
1.0x10-10M 1.0x10-10M 1.0x10-7M 0
pH=-log[H+]
What’s the sum of 1.0x10-10+ 1.0x10-7
pH=-log[1.0x10-7]
pH=7.00
Calculating the pH of a
Weak acid
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Find the pH of a 1.00 M solution of HF,
Ka= 7.2 X 10-4.
Remember: Weak acids do not
completely dissociate in water.
Calculations with weak acids must be
done carefully and systematically and are
the prototype to almost all equilibrium
problems. Following all the steps here
make it easier to remember all the steps
for more difficult problems.
Calculating the pH of a
Weak acid
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1. Write the major species, Remember
very little of the HF dissociates.
HF
H2O
But which ones furnish the H+?
Answer: Both
HFH+ + FKa = 7.2 X 10-4
H2O  H+ + OH- Kw = 1.0 X 10-14
Calculating the pH of a
Weak acid
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Which is the dominant source of H+?
Ka = 7.2 X 10-4 >> 1.0 X10-14
Thus the dominate source is HF.
The equilibrium expression is:
Ka = 7.2 X 10-4 = [H+][F-]
[HF]
Calculating the pH of a
Weak acid
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Write the equilibrium expression.
Make your ice table
I
C
E
What are the [equilibrium]’s?
Does it meet the 5% rule?
Are we done?
Calculating the pH of a
Weak acid
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The question asked for the pH!!
[H+] was 2.7 X10-2
pH=-log(2.7 X10-2)
pH= 1.57
Summary
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List all major species
Chose species that produces H+ and
write balanced equation.
Using Ka +Kw for the reactions find
the dominant source of H+
Write the equilibrium expression for
the dominant source of H+
ICE in terms of X
Summary continued.
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Substitute the equilibrium [ ]’s into the
equilibrium expression.
Solve for X using the “easy” way when
the Ka has a small value. HA-x=HA
Verify with the 5% rule
Calculate [H+], pH
Example
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OCl- is a strong oxidizer in bleach and
forms in chlorinated pools.
This relatively strong base forms a
weak acid with a Ka = 3.5 X 10-8
Find the pH of 0.100 M HOCl
Species, equations, ICE,
etc
Answer
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pH=4.23
What about a mixture of weak acids?
Find the pH of a mixture of
1.0 M HCN Ka= 6.2 X 10-10
5.0 M HNO2 Ka= 4.0 X 10-4
Also find the [CN-] at equilibrium
Major Species, Equations,
Ka expression, ICE
Answer: HCN, HNO2, H2O
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HCN  H+ + CN- Ka = 6.2 X 10-10
H2O  H+ + OH- Ka = 1.0 X 10-14
HNO2 H++ NO2- Ka = 1.0 X 10-4
Major [H+] is HNO2
I 5.0 0 0
C -X +X +X
E 5.0-X X X
Ka=4.0X10-4 =[H+][NO2-] ≈ X2/5
[HNO2]
Answer continued
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X= 4.5 X 10-2
5% rule (4.5 x10-2/5)X100 = 0.9%
Valid
[H+] = 0.045 pH = 1.35
Find [CN-]
HCN  H+ + CN
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This is the only source of CNKa = 6.2 X 10-10 =([H+][CN-])/[HCN]
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HCN  H+ + CNI
1.0
.045 0
C
-X
+X +X
E 1.0-X 0.045 + X
X
1.0 –X ≈ 1.0
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0.045 + X ≈0.045
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Ka = 6.2 X 10-10
And Finally
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6.2 x 10-10= 0.045 [CN-]
1.0
[CN-]=6.2 X 10-10
0.045
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[CN-]=1.4 X10-8 M
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