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Chapter 16: Aqueous Ionic
Equilibrium
Mrs. Brayfield
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16.2: Buffers
Most solutions will rapidly change pH upon the addition
of an acid or base
Those that do not are called a buffer
A buffer is a solution that resists pH changes by neutralizing
added acid or added base
Buffers contain significant amounts of both a WEAK acid
and its conjugate base
Or a weak base and its conjugate acid
Blood is a very common buffer
It resists pH changes very well or else you would die easily
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Buffers
The reason why you
must have a weak
acid/base and its
conjugate is because
the weak acid/base
does not dissociate
much
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Buffers
Why can’t we use strong acids/bases in buffers?
It takes much more base to change the pH of a weak acid
solution because there is a large reservoir of undissociated
weak acid
Calculate the concentration of the starting acid for HCl and
HC2H3O2 (Ka = 1.8x10-5)when the pH of both solutions is 2.00
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Common Ion Effect
The common ion effect is the tendency for a common ion
to decrease the ionization of a weak acid or weak base
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Calculating the pH of a Buffer
Calculate the pH of a buffer solution that is 0.200M
HC2H3O2 and 0.100M NaC2H3O2
𝐾𝑎 =𝐶2𝐻3𝑂2
− [𝐻+]
[𝐻𝐶2𝐻3𝑂2]= 1.8 × 10−5 =
(.1 + 𝑥)(𝑥)
.2 − 𝑥
[HC2H3O2] [H+] [C2H3O2-]
I 0.200 0 0.100
C – x +x +x
E .2 – x x .1 + x
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pH of Buffers
To find the pH of buffers, we just rearrange Ka:
𝐾𝑎 =𝐻3𝑂
+ [𝐴−]
[𝐻𝐴]
𝐻3𝑂+ = 𝐾𝑎
[𝐻𝐴]
[𝐴−]
We can now take the log of both sides to find the pH:
− log 𝐻3𝑂+ = −𝑙𝑜𝑔𝐾𝑎 − 𝑙𝑜𝑔
[𝐻𝐴]
[𝐴−]
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝐴−]
[𝐻𝐴]
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
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pH of Buffers
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
This is called the Henderson-Hasselbalch equation and
we can easily calculate the pH of a buffer solution from
initial concentrations, as long as x is small (we can make
our assumption)
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pH of Buffers Example
Calculate the pH of a buffer solution that is 0.250M HCN
and 0.170M KCN. For HCN Ka = 4.9 x 10-10 (pKa = 9.31).
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Henderson-Hasselbalch Equation
So, when can we use this equation?
Basically whenever we can approximate
You should look for the initial concentrations to be at least 102
to 103 LARGER than the Ka
Also, since the log is a ratio of concentrations [HA]/[A–],
amounts of acid/base in moles can be substituted for
concentration (since volumes cancel – see page 606)
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Base Buffers
Same calculations as with our acid buffers:
𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
Just make sure that you change Kb to Ka
𝐾𝑤 = 𝐾𝑎 × 𝐾𝑏
OR
14 = 𝑝𝐾𝑎 + 𝑝𝐾𝑏
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Base Buffers Example
Calculate the pH of a solution containing 0.60M NH3 and
0.30M NH4Cl (Kb = 4.75).
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16.3: Buffer Effectiveness
For a buffer to be effective it must neutralize moderate
amounts of added acid or base
There are two factors that influence the effectiveness of a
buffer:
1. Relative amounts of the acid and conjugate base
2. Absolute concentrations of the acid and conjugate base
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Buffer Effectiveness
We do not need to calculate the change in pH of a buffer
BUT we can see on the example on page 610 that buffers
with equal amounts of acid and conjugate base is more
resistant to pH change
Therefore it is a more effective buffer
As a guideline an effective buffer MUST have a
[base]/[acid] ratio in the range of 0.10 to 10
Basically the two concentrations should not differ by more
than a factor or 10 to be effective
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Buffer Capacity
Buffer capacity is the amount of acid or base that can be
added to a buffer without destroying its effectiveness
Buffer capacity increases with increasing concentrations
of the buffer components
There are more species to deal with changes in pH
Buffer capacity also increases as the concentrations of the
buffer components get closer to each other
The overall capacity of the buffer increases
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Last Note on Buffers
Without being given any data on the initial concentrations
we can make some inferences about those values:
If pH < pKa, the acid has a higher concentration than the base
If pH > pKa, the base has a higher concentration than the acid
Homework Problems: #1, 2, 4, 8, 10, 12, 14, 16, 20, 28, 30,
32
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16.4: Titrations and pH Curves
An acid-base titration is where a basic (or acidic) solution
of unknown concentration is reacted with an acidic (or
basic) solution of known concentration
The known solution is slowly added to the unknown solution
while pH is monitored
We can monitor pH using indicators, or a substance whose color
depends on the pH
As the acid and base react, they neutralize each other
This is called the equivalence point (when the number of moles of
base EQUAL the number of moles of acid)
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Titrations
This is called a pH curve:
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Titrations
Buret
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Strong Acid and Strong Base Titration
Calculations
Find the pH after adding different amounts of 0.100M NaOH to 25mL of 0.100M HCl:
Initial pH
pH after adding:
5.00mL NaOH
10.0mL NaOH
15.0mL NaOH
20.0mL NaOH
25.0mL NaOH
30.0mL NaOH
35.0mL NaOH
40.0mL NaOH
50.0mL NaOH
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SA and SB Titration Calculations
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Titration Calculation Example 1
Calculate the pH in the titration in the previous example
after adding a total of 60.0mL of 0.200M HNO3 (50.0mL
of 0.200M NaOH titrated with 0.200M HNO3)
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Weak Acid with Strong Base Titration
Calculations
Find the pH after adding different amounts of 0.100M NaOH to 25mL of 0.100M HCHO2:
Initial pH
pH after adding:
5.00mL NaOH
10.0mL NaOH
15.0mL NaOH
20.0mL NaOH
25.0mL NaOH
30.0mL NaOH
35.0mL NaOH
40.0mL NaOH
50.0mL NaOH
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WA and SB Titration Calculations
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Titration Calculation Example 2
Determine the pH at the equivalence point for the
titration between HNO2 and KOH in the previous
example (40mL 0.10M HNO2 titrated with 0.20M KOH)
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Indicators
Titrations can be monitored with either a pH meter or
an indicator
With an indicator we rely on when that indicator turns
colors, otherwise known as the endpoint, to determine
the equivalence point
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Indicators
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Many Different Indicators…
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Homework Problems: #36, 38, 40, 43, 46, 50, 52, 54
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16.5: Solubility Equilibria and Ksp
Remember the solubility rules from a long time ago…
All sodium, potassium, ammonium, and nitrate salts are soluble
in water
AP does not require you to know the rest of the rules
Using equilibrium, we now can better understand the
solubility of an ionic compound:
𝐶𝑎𝐹2 𝑠 ↔ 𝐶𝑎2+ 𝑎𝑞 + 2𝐹− 𝑎𝑞
𝐾𝑠𝑝 = 𝐶𝑎2+ [𝐹−]2
Where Ksp is the solubility product constant
Remember that solids are omitted from the equilibrium
expression
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Ksp
Ksp is a measure
of the solubility
of a compound
See Appendix IIC
for a complete
table of values
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Molar Solubility
Molar solubility (S) is the solubility in units of moles per
liter (mol/L), which can be calculated from Ksp
For example:
𝐴𝑔𝐶𝑙 𝑠 ↔ 𝐴𝑔+ 𝑎𝑞 + 𝐶𝑙− 𝑎𝑞
Where S is the concentration of AgCl that dissolves
(molar solubility)
[Ag+] [Cl–]
I 0.00 0.00
C + S + S
E S S
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𝐾𝑠𝑝 = 𝐴𝑔+ [𝐶𝑙−]
𝐾𝑠𝑝 = 𝑆2
𝑆 = 𝐾𝑠𝑝 = 1.77 × 10−10
𝑆 = 1.33 × 10−5𝑀
So we can say that the molar solubility of AgCl is 1.33 x 10-5M
[Ag+] [Cl–]
I 0.00 0.00
C + S + S
E S S
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Molar Solubility Example 1
Calculate the molar solubility of Fe(OH)2 in pure water
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Molar Solubility Example 2
The molar solubility of AgBr in pure water is 7.3 x 10-7M.
Calculate Ksp.
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Common Ion Effect
When a solution already has one of the compounds ions
in solution, the solubility of that compound is affected
We simply use Le Châtelier's principle to explain the
equilibrium shift
The solubility of any compound is lower in solution containing
a common ion that in pure water
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Common Ion Effect Example
Calculate the molar solubility of CaF2 in a solution
containing 0.250M Ca(NO3)2.
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Effect of pH on Solubility
Treat this just like a common ion effect problem:
The solubility of an ionic compound with a strongly basic
or weakly basic anion increases with increasing acidity
(decreasing pH)
For example OH–, S2–, CO32–
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16.6: Precipitation
We know that precipitation can occur when two
solutions are mixed and produces an insoluble compound
BUT even compounds that are “insoluble” dissolve, if only
slightly, in water
Going back to Q…
We can look at Q and compare it to Ksp to determine if
any more solid will dissolve
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Q
Remember from chapter 14:
If Q < Ksp, the solution is unsaturated and more solid can
dissolve
If Q = Ksp, the solution is saturated and no more solid will
dissolve
If Q > Ksp, the solution is supersaturated and solid will
precipitate out
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Q Example
If we have solutions of 0.0600M Pb(NO3)2 and 0.0158M
NaBr, will a precipitate form in the mixed solution?
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Homework Problems: #56, 58, 60, 62, 70, 72
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16.7: Complex Ion Equilibria
A complex ion contains a central metal ion that is bound
to one or more ligands
A ligand is a neutral molecule or ion that acts as a Lewis
base with the central metal ion
For the complex ion in Ag(H2O)2+, water is the ligand
Why learn about complex ions?
If we have a barely soluble compound, we can introduce a
ligand to the solution to get the compound to dissolve more
(remove one product to shift equilibrium – Le Châtelier's)
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Review problems: #80, 82, 86, 90
http://www.youtube.com/watch?v=8Fdt5WnYn1k&list=PL
8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr&index=32