Chem 106 3-22-2011
Chapter 18:
Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations
Sect 4-5 Ionic solubility
Sect 6-7 Complex Formation
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1
The net ionic equation for the reaction of
KOH(aq) and HNO3(aq) is…
1.
2.
3.
4.
K+ + NO3- → KNO3(aq)
K+ + OH- + H3O+ → H2O(l) + K+
H3O+ + OH- → 2 H2O(l)
HNO3 + KOH → 2 H2O(l) + KNO3(aq)
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2
The net ionic equation for the reaction of
KOH(aq) and HNO3(aq) is…
1.
2.
3.
4.
K+ + NO3- → KNO3(aq)
K+ + OH- + H3O+ → H2O(l) + K+
H3O+ + OH- → 2 H2O(l)
HNO3 + KOH → 2 H2O(l) + KNO3(aq)
ANSWER=(3) You need to be able to recognize KOH as a
strong base, and HNO3 as a strong acid. The neutralization
occurs with K+ and NO3- as spectator ions.
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Section 18.1
Common Ion Effect
In acid-base, and other equilibria involving ionic components,
you can manipulate the position of the equilibrium by adding or
removing one ionic reactant or product.
We can use an ICE table to quantify this effect.
Or, qualitatively, the effect is predicted by L’Chatelier.
ClCH2CO2H +
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H 2O 
ClCH2CO2-
+
H3O+
4
Review. We have seen what happens when, say, 0.10 M
chloroacetic acid is dissolved in water (Ka = 1.35 x 10-3):
ClCH2CO2H +
H2O 
ClCH2CO2-
+
H3O+
I
0.10 M
0
0
C
-x
x
x
E
0.10 -x
x
x


[ HCO2 ][ H 3O ]

Ka
[ HCO2 H ]
0.10  x  0.10
st
x  0.0116 (1 approx)
x2
1.35 x10 
0.10  x
x  0.0109 (2 approx )
rd
x  0.0110 (3 approx)
pH   log(0.0110)  1.96
3
nd
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(Use
quadratic or method of successive approximations if Co < 100 x Ka . Here (100)Ka = 0.135 M.
Now let’s see what happens when we add 0.040 M sodium
chloroacetate (NaClCH2CO2) to 0.10 M chloroacetic acid:
ClCH2CO2H +
H2O 
ClCH2CO2-
+
H3O+
I
0.10 M
0.040
0
C
-x
x
x
E
0.10 - x
0.040 + x
x


[ HCO2 ][ H 3O ]

Ka
[ HCO2 H ]
(0.040  x) x
3

1.35 x10
0.10  x
(0.040) x
3

1.35 x10
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0.10
x  0.00338 (1 approx)
nd
x  0.00301 (2 approx)
st
x  0.00305 (3 approx)
pH   log(0.00305)  2.52
rd
6
Now, think about this in terms of L’Chatelier’s Principle:
If you add more chloroacetate ClCH2CO2-, the reaction will
go to the LEFT to use up the excess ClCH2CO2-.
ClCH2CO2H +
H2O 
ClCH2CO2-
+
H3O+
Therefore [H3O+] decreases, and pH goes UP.
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9
Section 18.2
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Buffers
10
Definition
A buffer is …an aqueous solution containing a
mixture of a weak acid and its conjugate base,
(or a weak base and its conjugate acid).
The function of a buffer is to absorb H+ or OH- ions,
minimizing the change in pH that might otherwise occur.
HA + H2O
B + H2O
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 A- + H3O+
 BH+ + OH-
(Notice that either
equilibrium can
absorb H+ or OH- ions)
11
The Henderson-Hasselbalch Equation
is used to describe buffer solutions
(these have appreciable amounts of both HA and A- , like
at least 5% of the total concentration as one or the other).
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Henderson-Hasselbalch Equation
Example buffer solutions, which contain a weak acid and
conjugate base of the weak acid (or weak base and conjugate
acid of the weak base).
CH3COOH + H2O

CH3COO-
Acetic acid
acetate
HNO2 + H2O

Nitrous acid

Ammonium
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NO2-
+ H3 O +
nitrite
NH4+ + H2O
F- +
+ H3O+
H2O
Fluoride
NH3
+ H3O+
ammonia

HF + OH-
hydrofluoric acid
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CH3COOH + H2O
Acetic acid
(HOAc)

CH3COO-
+ H3O+
acetate
(OAc-)

equilibrium constant definition
Take log10 of both sides
log of product = sum of logs
Mult both sides by -1
pX = -log10(x)
Solve for pH
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
products [OAc ][ H 3O ]

Ka 
reactants
[ HOAc]
 [OAc  ][ H 3 O  ] 

log K a  log 
[ HOAc]


 [OAc  ] 


log K a  log[ H 3 O ]  log 
 [ HOAc] 
 [OAc  ] 

 log K a   log[ H 3 O ]  log 
 [ HOAc] 
 [OAc  ] 

pK a  pH  log 
 [ HOAc] 

 [OAc  ] 

pH  pK a  log 
 [ HOAc] 
14
Logic of the H-H equation
 [ A ] 
pH  pK a  log

 [ HA] 
Please remember
this equation and
know how to use it.
Say 1/5 [A-], 4/5 [HA]
Say [A-] = [HA]
Say 2/3 [A-] and 1/3 [HA]
Then log(1/4) = -0.60
Then log(1) = 0
Then log(2/1) = 0.30
pH = pKa - 0.60
pH = pKa + 0 = pKa
pH = pKa+ 0.30
or [H3O+] > Ka
or [H3O+] = Ka
or [H3O+] < Ka
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Most buffers are made up with equal or near-equal concentrations of
acid and conjugate base. This has maximum ability to absorb H+ or
OH- ions.
Buffers can be prepared 3 ways.
1) Start with the acid form, and add ½ mole NaOH per mole acid.
2) Start with the conjugate base, and add ½ mole HCl per mole
base.
3) Add equal amounts of acid and conjugate base.
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Say you prepared a buffer using equal moles of sodium nitrite NaNO2
and nitrous acid HNO2 (Ka = 3.2 x 10-4). What is the pH of the resulting
solution? (Try using pencil and paper, and not a calculator…)
3.0
3.5
4.2
4.5
14
6
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4.
5
4.
2
3.
5
1
3.
1.
2.
3.
4.
24
17
Say you prepared a buffer using equal moles of sodium nitrite NaNO2
and nitrous acid HNO2 (Ka = 3.2 x 10-4). What is the pH of the resulting
solution? (Try using pencil and paper, and not a calculator…)
1.
2.
3.
4.
3.0
3.5
4.2
4.5
pH = pKa + log(A-/HA) = pKa
pH = -log(3.2 x 10-4)
pH = -[log3.2 + log(10-4)]
pH = -(~0.5 - 4)
pH = -(-3.5) = 3.5
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1. Calculate the stoichiometry (moles) first.
0.250 L HCN 
0.482 mol HCN
 0.1205 mol HCN
L HCN
HCN + OH-  CNI 0.1205 0.0593
C -0.0593 -0.0593
E 0.0612
~0
+ H2O
0
0.0593
0.0593
2. Strong bases react completely with
weak acids.
HCN + OH-  CN
+ H2O
1 mol CN
 0.0593 mol CN 
0.0593 mol NaOH 
mol NaOH
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[CN ] 
0.0593 mol
 0.2372 M
0.250 L
[ HCN ] 
0.0612 mol
 0.2448 M
0.250 L
19
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20
[CN  ]
pH  pK a  log
[ HCN ]
pH   log( 4.0 x10 10 )  log
0.2372
0.2448
pH  9.398  log(0.9690)
pH  9.398  ( 0.01370 )
pH  9.384
Notice that if [HA] > [A-] , the pH
must be LESS THAN the pKa.
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Demo





Vernier pH electrode/LoggerPro software
30. mL water
3.0 mmoles sodium acetate (0.25 g)
3.0 M HCl (aq) ( 3.0 mmol/mL)
Pipetman 1000 (delivers 1.000 mL max)
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Observations:
 1.00 mL 3.0 M HCl + 30. mL water: pH ~ 1.5 (This is about 0.10 M strong acid,
which should give pH = 1.0.)
 0.24 g NaCH3CO2 (sodium acetate) + 30 mL water: pH ~ 7.8. (This should be about
8.7 if you do the calculation.)
 Adding 0.50 mL of 3.0 M HCl to the sodium acetate solution. pH ~ 4.8 (This should
be 4.74 since it is a buffer with about 0.05 M acetate and 0.05 M acetic acid where
pH = pKa).
 Adding another 0.50 mL of 3.0 M HCl to the above solution gives pH ~ 3.0. (This is
now essentially a solution of 0.10 M in acetic acid, and the calculated pH = 2.87).
Why is the pH not the same as the first dilution?? Because the acetate reacts with,
and “holds”, the H+ in its weak acid structure.
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HClO +
0.352
-x
0.352 – x
OH-  ClO- + H2O
x
-x
~0
0
x
x

[A ]
pH  pK a  log
[ HA]
We know these


[ HA] [ A ]  0.352 M
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25
(Substituting real numbers gives less writing.)
[ A ]
pH  pKa  log
[ HA]
[ A ]
pH  pKa  log
[ HA]
10
pH  pKa
[ A ]
 1.3614 
[ HA]
[ A ]
[ HA] 
1.3614

[ HA]  [ A ]  0.352 M
[ A ]
 [ A  ]  0.352 M
1.3614
 1

[ A  ]
 1  0.352 M
 1.3614 
0.352 M
0.352 M
[ A ] 

1
1.7345
1
1.3614
[ A  ]  0.2029 M

0.2029 mol OH
 0.250 L  0.0507 mol OH 
L
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27
18.3 Titration Calculations
Adding OH- of a known concentration to
acid solution until moles OH- = moles H+
(“neutralization reaction”)
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2 HI(aq) + Ca(OH)2(aq)

CaI2(aq) +
2 H2O(l)
0.0166 L Ca (OH ) 2

0.0166 L Ca (OH ) 2 
0.229 mol Ca (OH ) 2
2 mol HI
1 L HI


 0.022747 L HI
L Ca (OH ) 2
mol Ca (OH ) 2 0.274 mol HI
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? L HI
29

0.0166 L Ca (OH ) 2
0.0166 L Ca (OH ) 2 
? L HI
L Ca (OH ) 2


1 L HI
 ? L HI
0.0166 L Ca (OH ) 2 
mol Ca (OH ) 2

L Ca (OH ) 2
0.0166 L Ca (OH ) 2 
mol Ca (OH ) 2
1 L HI


 ? L HI
L Ca (OH ) 2
mol Ca (OH ) 2 mol HI
0.0166 L Ca (OH ) 2 
mol Ca (OH ) 2
mol HI
1 L HI


 ? L HI
L Ca (OH ) 2
mol Ca (OH ) 2 mol HI
.0166 L Ca (OH ) 2 
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
1 L HI
 ? L HI
mol HI
0.229mol Ca (OH ) 2
2 mol HI
1 L HI


 0.0227 47 L HI
L Ca (OH ) 2
mol Ca (OH ) 2 0.274mol HI
30
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http://www.chem.uoa.gr/applets/Applet_Index2.htm
This shows the equation for the entire titration curve (pH vs. volume of OH- titrant) for 25
different acids. This equation will NOT be on the next exam.
However, we have studied several sections of the typical titration curve:
1. Acid by itself: For a strong acid, [H3O+] = molarity of acid.
For a weak acid use the ICE table.
2. Acid + conjugate base: Use the Henderson-Hasselbalch Equation.
3. At the equivalence point: pH = 7.00 for strong acid vs. OH- titration.
Use ICE table for hydrolysis of the conjugate base of a weak acid.
4. Past the equivalence point: This is now just dilute OH- , and so pH = -log(Kw/[OH-]).
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Titrating a weak acid (1.0 M, Ka = 1 x 10-6) with strong base NaOH
HA + OH-  A- + H2O
Acid 2
“pH of NaOH solutions”
14
12
“buffer of weak acid & its
conjugate base”
10
pH
“pH of
conjugate base”
8
6
4
“pH of
weak acid”
2
titration
equivalence pt
0
0.00
“[weak acid]=[conjugate base]
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pH = pKa”
0.50
1.00
1.50
Mol OH-/mol HA
33