Chapter 33
Acid-Base Equilibria
1
Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.
Solutions of a Weak Acid or Base
• The simplest acid-base equilibria are
those in which a single acid or base
solute reacts with water.
– we will first look at solutions of pure weak
acids or bases.
– Next, we will also consider solutions of
salts, which can have acidic or basic
properties as a result of the reactions of
their ions with water. Salts can be neutral,
acidic, or basic.
2
33.1 Acid-Ionization Equilibria
• Acid ionization (or acid dissociation) is
the reaction of an acid with water to
produce hydronium ion (hydrogen ion) and
the conjugate base anion (hydrolysis).
HC 2 H 3O 2 (aq )  H 2O(l )


H 3O (aq)  C2 H 3O 2 (aq)
<<100%
K<<1
– Because acetic acid is a weak electrolyte, it ionizes to a
small extent in water, hence double arrow; not 100% like
strong acid which wrote single arrow.
3
Acid-Ionization Equilibria
• For a weak acid, the equilibrium concentrations
of ions in solution are determined by the acidionization constant (also called the aciddissociation constant, Ka).
– Consider the generic monoprotic acid, HA.
HA(aq )  H 2O(l )


H 3O (aq )  A (aq )
4
Acid-Ionization Equilibria
HA(aq)  H 2O(l )
H 3O  (aq)  A  (aq)
– Recall that the concentrations of pure solids and pure
liquids are not included in the K expression.
– The corresponding equilibrium expression is:


[ H 3O ][ A ]
K
Ka 
[ HA]
– however, since we are referring to acid
hydrolysis (acid + H2O) the constant is known
as Ka , the acid-ionization constant.
5
Experimental Determination of Ka
• Nicotinic acid is a weak monoprotic acid with
the formula HC6H4NO2. A 0.012 M solution of
nicotinic acid has a pH of 3.39 at 25°C.
Calculate the acid-ionization constant for this
acid at 25°C.
– It is important to realize that the solution was made
0.012 M in nicotinic acid, however, some molecules
ionize making the equilibrium concentration of
nicotinic acid less than 0.012 M, just a small amount
less.
7
A Problem To Consider
– Let x be the moles per liter of product formed.
HC6 H 4 NO2 (aq)  H 2O(l )
Initial[]
Change[]
Equilibrium[]

H 3O  (aq)  C6 H 4 NO2 (aq)
0.012
- x
0.012 - x

0
0
+x
+x
x
x

[ H 3O ][C6 H 4 NO2 ]
Ka 
[ HC6 H 4 NO2 ]
8
A Problem To Consider
– We can obtain the value of x from the given
pH.

x  [ H 3O ]  anti log(  pH )  10
 pH
9
A Problem To Consider
– Substitute this value of x in our equilibrium expression.


[ H 3O ][C6 H 4 NO2 ]
Ka 
[ HC6 H 4 NO2 ]
ionization <<100%
K<<1
x2
(0.00041) 2
5
Ka 

 1.4 10
(0.012  x) (0.01159)
– Note that
HW 28
(0.012  x)  (0.012  0.00041)  0.01159  0.012
the concentration of unionized acid remains
virtually unchanged. Important point to
remember for later.
code:
ka
10
33.1.1 Percent Ionzation
• The degree of ionization of a weak
electrolyte is the fraction of molecules
that react with water to give ions.
– Electrical conductivity or some other colligative
property can be measured to determine the degree
of ionization.
– With weak acids, the pH can be used to determine
the equilibrium composition of ions in the solution
because it gives the concentration of hydronium ions
at equil. pH of solution  [H O+]  amount ionized
3
eq
11
– To obtain the degree of ionization:
% ionization 
[ H  ]eq
[acid ]o
x 100%
note: % is unit
0.00041
% ionization 
x100%  3.4%
0.012
– weak acids typically less than 5%. The lower the %ionized, less H+
formed, the lower Ka, weaker the acid. Ka is measure of strength of
acid or Kb if base. Strength based on ionization not pH directly.
Lower pH more acidic, does not mean stronger acid
12
Ka is a measure of strength of acid. It indicates the amount ionized to form
hydronium ions. The more hydronium ions formed, naturally the more
acidic and lower pH. However, strength of acid is measure of amount
ionized, not conc of acid. Conc can vary which ultimately will vary the pH
of solution. To determine the strength of an acid you must compare Ka
values not pH of solution. pH is dependent on Ka as well as conc. of
solution. Must compare apples and apples.
0.1M HC2H3O2
Ka = 1.8 x 10-5
pH = 2.87
0.1M HF
Ka = 6.9x10-4
pH = 2.08
10M HC2H3O2 Ka = 1.8 x 10-5 pH = 1.87
0.1M HF
Ka = 6.9x10-4 pH = 2.08
Which stronger acid? HF, larger Ka
13
33.1.2 pH of Weak Acids
• Once you know the value of Ka, you can
calculate the equilibrium
concentrations of species HA, A-,
and H3O+ for solutions of different
molarities.
– The general method for doing this is same way we
did Kc or Kp except we can simplify the math
because of percent ionize is so small compared to
initial conc of acid; remember 0.012-x=0.012. This
wasn't the case for Kc or Kp problems so can't
neglect in those type problems
14
Calculations With Ka
• Note that in our previous example, the degree of
ionization was very small as compared to the
concentration of nicotinic acid.
• It is the small value of the degree of ionization that
allows us to ignore the subtracted x in the
denominator of our equilibrium expressions to
simplify our calculations.
• The degree of ionization of a weak acid depends on
both the Ka and the concentration of the acid solution
15
Calculations With Ka
• How do you know when you can use
this simplifying assumption?
– It can be shown that if the acid concentration, Ca,
divided by the Ka exceeds 100, that is,
if [acid]
Ka
 100
– then this simplifying assumption of ignoring the
subtracted x gives an acceptable error of less than 5%
otherwise must do quadratic formula. Another way
look at it factor of two - three difference in power of 10
is needed between Ka and conc of acid.
16
ex. Calculate the pH in a 1.00 M solution of nitrous acid, HNO2,
for which Ka = 6.0 x 10-4 at 25oC. acid, base, or salt? acid strong or weak? weak
HNO2 + H2O
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[HNO2]
1.00
-x
1.00 - x
H3O+ + NO2-
[H3O+]
0
+x
x
[NO2-]
0
+x
x
=
Does answer make sense?
Yes, acid pH
17
A Problem To Consider
• Really two equil: Kw as well but neglect most
of the time as long as [H+] is greater than
10-6 M from the acid.
• Really [H+]tot = [H+]acid +[H+]H2O (water so
small neglect)
HNO2 + H2O
H2O + H 2O
H3O+ + NO2H3O+ + OH-
[H+]tot = [H+]HNO2 +[H+]H2O
[H+]tot = 0.024 M + <10-7M = 0.024 M
18
ex. Calculate the pH of 10.00 mL of 1.00 M solution of nitrous acid , HNO2, diluted
to 100.0 mL with water for which Ka = 6.0 x 10-4 at 25oC. acid, base, or salt? acid
H3O+ + NO2-
HNO2 + H2O
strong or weak?
weak
[HNO2]
[H3O+]
[NO2-]
Initial, [ ]o
0.100
0
0
Change, D[ ]
-x
+x
+x
x
x
Equilibrium, [ ]eq
0.100 - x
Does answer make sense?
Yes, acid pH
19
A Problem To Consider
• What is the pH at 25°C of a 3.6 x 10-3 M
solution of acetyl salicylic acid (aspirin),
HC9H7O4? The acid is monoprotic and
• Ka=3.3 x 10-4 at 25°C.
acid, base, or salt?
acid
strong or weak?
weak
20
A Problem To Consider
HC9 H 7O4 (aq )  H 2O(l )
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
3.6 x 10-3
-x
3.6 x 10-3 - x

H 3O  (aq)  C9 H 7O4 (aq)
0
0
+x
+x
x
x
21
A Problem To Consider
– Note that
which is less than 100 or 10-3 vs 10-4 not factor 3
difference; therefore, won't work, so we must
solve the equilibrium equation exactly .
22
– If we substitute the equilibrium concentrations and the
Ka into the equilibrium constant expression, we get
HC9 H 7O4 (aq )  H 2O(l )
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
3.6 x 10-3
-x
3.6 x 10-3 – x


H 3O (aq)  C9 H 7O4 (aq)
0
+x
x
0
+x
x
23
– You can solve this equation exactly by using the
quadratic formula.
– Rearranging the equation to put it in the form
ax2 +
bx
4
+
c
6
=0
x  ( 3.3  10 )x  (1.2  10 )  0
2
 b  b  4ac
x
2a
2
 (3.3 10 4 )  (3.3 10 4 ) 2  4(1)(1.2 10 6 )
x
2(1)
x  9.4  10 4 M
x  1.3  10 3 M
24
A Problem To Consider
– Taking the positive value and neglect water

4

x  [ H 3O ]  9.4 10 M  [C9 H 7O4 ]
– Now we can calculate the pH.
4
pH   log(9.4  10 )  3.03
Does answer make sense?
Yes, acid pH
HW 29
code: weak
25
Calculate the pH of a solution that has a concentration of 1.0 x 10-8 M NaOH.
NaOH  Na+
+ OH1.0 x 10-8 M 1.0 x 10-8 M
pOH = - log [OH-] = - log 1.0 x 10-8 = 8.00
pH = 14.00 – pOH = 14.00 – 8.00 = 6.00
Does the answer make sense?
No, it is a base - what’s wrong?
The principle reaction is the water not the NaOH; therefore,
you can’t neglect the water. Must rework problem with
water equilibrium and include the additional base from
NaOH (pH = 7.02).
26
33.1.3 pH of Polyprotic Acids
• Some acids have two or more protons
(hydrogen ions) to donate in aqueous solution.
These are referred to as polyprotic acids.
– Sulfuric acid, for example, can lose two protons in
aqueous solution.


H 2SO 4 (aq )  H 2O(l )  H 3O (aq )  HSO4 (aq )

HSO4 (aq )  H 2O(l )
H 2O(l )  H 2O(l )

2
H 3O (aq )  SO 4 (aq )
H 3O  (aq )  OH  (aq )
[H3O+]tot = [H3O+]H2SO4 + [H3O+]HSO4- + [H3O+]H2O
Typically worry about principle rxn only
27
Polyprotic Acids
– For a weak diprotic acid like carbonic acid, H2CO3,
several simultaneous equilibria must be
considered.
< 5%
H 2CO 3 (aq )  H 2O(l )

HCO3 (aq )  H 2O(l )
H 2O(l )  H 2O(l )


H 3O (aq )  HCO 3 (aq )
<< 5%

2
H 3O (aq )  CO 3 (aq )
H 3O  (aq )  OH  (aq )
[H3O+]tot = [H3O+]H2CO3 + [H3O+]HCO3- + [H3O+]H2O
Once again, typically worry about principle rxn only.
i.e. NaHCO3 would use the second equation.
28
Polyprotic Acids
– Each equilibrium has an associated acid-ionization
constant.– For the loss of the first proton


H 2CO 3 (aq )  H 2O(l )
H 3O (aq )  HCO 3 (aq )


[H 3O ][HCO 3 ]
K a1 
 4.3  107
[H 2CO 3 ]
– For the loss of the second proton

2

HCO3 (aq )  H 2O(l )
H 3O (aq )  CO 3 (aq )
2

[H 3O ][CO 3 ]
11
K a2 

4
.
8

10

[HCO 3 ]
29
Polyprotic Acids
– In general, the second ionization constant, Ka2, for
a polyprotic acid is smaller than the first ionization
constant, Ka1. Harder to pull proton off charged
species. Total H+ is sum of all equil H+ but usually
neglect all but the principal rxn.
– In the case of a triprotic acid, such as H3PO4, the
third ionization constant, Ka3, is smaller than the
second one, Ka2.
HPO42Ka3
H2PO4-
<<
Ka2
H3PO4
<<
Ka1
30
Polyprotic Acids
– When several equilibria occur at once, it might
appear complicated to calculate equilibrium
compositions.
– However, reasonable assumptions can be made
that simplify these calculations as we show in the
next example.
31
A Problem To Consider
• Ascorbic acid (vitamin C) is a diprotic acid,
H2C6H6O6. What is the pH of a 0.100 M
solution?
Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.
H 2C6 H 6O6 (aq)  H 2O(l )

HC6 H 6O6 (aq)  H 2O(l )


H 3O (aq)  HC6 H 6O6 (aq)
2

H 3O (aq)  C6 H 6O6 (aq)
– For diprotic acids, Ka2 is so much smaller than Ka1
that the smaller amount of hydronium ion produced in
the second reaction can be neglected.
32
A Problem To Consider
H 2C6 H 6O6 (aq)  H 2O(l )
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq

H 3O  (aq)  HC6 H 6O6 (aq)
0.100
-x
0.100 - x

0
0
+x
+x
x
x

[ H 3O ][ HC6 H 6O6 ]
K a1 
[ H 2C6 H 6O6 ]
33
A Problem To Consider
2
x
5
[ H 3O ][ HC6 H 6O6 ] 

7
.
9

10
K a1 
(0.100  x)
[ H 2C6 H 6O6 ]


– Assuming that x is much smaller than 0.10
(1265>100), you get
x 2  (7.9  10 5 )  (0.100)  7.9  10 6
3


x  2.8 10 M  0.0028 M  [ H 3O ]  [ HC6 H 6O6 ]
pH   log(0.0028)  2.55
34
A Problem To Consider
• Ascorbic acid (vitamin C) is a diprotic acid,
H2C6H6O6. What is the pH of a 0.10 M
solution? What is the concentration of the
ascorbate ion, C6H6O62- in previous example?
The acid ionization constants are Ka1 = 7.9 x
10-5 and Ka2 = 1.6 x 10-12.
– The ascorbate ion, C6H6O62-, which we will call y, is
produced only in the second ionization of H2C6H6O6.

HC6 H 6O6 (aq)  H 2O(l )
from 1st equil

2
H 3O (aq)  C6 H 6O6 (aq)
some from 1st eq
none to start
35
H 2C6 H 6O6 (aq)  H 2O(l )


H 3O (aq)  HC6 H 6O6 (aq)
0.0028 M
from 1st equil
0.0028 M
– Assume the starting concentrations for HC6H6O6- and
H3O+ to be those from the first equilibrium.

HC6 H 6O6 (aq)  H 2O(l )
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
2
H 3O  (aq)  C6 H 6O6 (aq)
0.0028
0.0028
0
-y
+y
+y
0.0028-y
0.0028+y
y
2
Ka2
[ H 3O  ][C6 H 6O6 ]


[ HC 6 H 6O6 ]
36
– Substituting into the equilibrium expression
(0.0028  y )( y )
12
 1.6  10
(0.0028  y )
– Assuming y is much smaller than 0.0028 (1.75 x 109 >
100), the equation simplifies to
(0.0028)( y )
12
 1.6  10
(0.0028)
– Hence,
2
y  [C6 H 6 O 6 ]  1.6 10
– typically, the concentration of the anion ion
equals Ka2 if starting with reactants of first equil.
12
M
HW 30
[H3O+]tot = 0.0028 MKa1 + 1.6 x 10-12 Ka2 + <1 x10-7 H2O = 0.0028 M
code: multiple
37
33.2 Base-Ionization Equilibria
• Equilibria involving weak bases are
treated similarly to those for weak acids.
– Ammonia, for example, ionizes in water as
follows.
NH 3 (aq )  H 2O(l )

NH 4 (aq )  OH  (aq )
38
Base-Ionization Equilibria
NH 3 (aq )  H 2O(l )

NH 4 (aq )  OH  (aq )
– Recall that the concentrations of pure solids and pure
liquids are not included in the K expression.
– The corresponding equilibrium expression is:


4
[ NH ][OH ]
Kb 
[ NH 3 ]
– however, since we are referring to base
hydrolysis (base + H2O) the constant is known
as Kb , the base-ionization constant.
39
Base-Ionization Equilibria
• Equilibria involving weak bases are treated similarly to those for
weak acids. Larger Kb , the stronger base, ionize more OH-,
more basic.
– In general, a weak base B with the base ionization
HB  (aq )  OH  (aq )
B(aq )  H 2O(l )
has a base ionization constant equal to

HW 31

[HB ][OH ]
K b
[B ]
code: plate
40
A Problem To Consider
• What is the pH of a 0.20 M solution of
pyridine, C5H5N, in aqueous solution?
The Kb for pyridine is 1.4 x 10-9.
acid, base, or salt?
base
strong or weak?
weak
– As before, we will follow the three steps in
solving an equilibrium.
1. Write the equation and make a table of
concentrations (ICE).
2. Set up the equilibrium constant expression.
3. Solve for x = [OH-].
41
A Problem To Consider
C5 H 5 NH  (aq )  OH  (aq )
C5 H 5 N(aq )  H 2O(l )
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[C5H5NH+]
0
[OH-]
0
- x
+x
+x
0.20 - x
x
x
[C5H5N]
0.20
42
A Problem To Consider
C5 H 5 NH  (aq )  OH  (aq )
C5 H 5 N(aq )  H 2O(l )
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[C5H5NH+]
0
[OH-]
0
- x
+x
+x
0.20 - x
x
x
[C5H5N]
0.20
43
A Problem To Consider
– Note that
Cb
Kb
 0.20
1.4  10
9
 1.4 10
8
which is much greater than 100 (also factor of 8
difference between the powers of 10 of Kb [2.0 x 10-1]
and concentration of base [1.4 x 10-9]), so we may use
the simplifying assumption that (0.20-x)  (0.20).
44
A Problem To Consider
C5 H 5 NH  (aq )  OH  (aq )
C5 H 5 N(aq )  H 2O(l )
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[C5H5NH+]
0
[OH-]
0
- x
+x
+x
0.20 - x
x
x
[C5H5N]
0.20
45
A Problem To Consider
– Solving for pOH

5
pOH   log[OH ]   log(1.7  10 )  4.77
– Since pH + pOH = 14.00
pH  14.00  pOH  14.00  4.77  9.23
Does answer make sense?
Yes, base pH
or combine equations into one:
HW 32
code: base
46
33.3 Acid-Base Properties of Ions and
Salts
• One of the successes of the BrønstedLowry concept of acids and bases was
in pointing out that some ions can act
as acids or bases.
– Consider a solution of sodium cyanide, NaCN.


NaCN(s ) 
 Na (aq )  CN (aq )
H 2O
Cations: neutral or acidic
Anion: neutral or basic
– A 0.1 M solution has a pH of 11.1 and is therefore
fairly basic.
47
Acid-Base Properties of a Salt Solution
– Sodium ion, Na+, is unreactive with water,
but the cyanide ion, CN-, reacts to produce
HCN and OH- making it a basic solution

CN (aq )  H 2O(l )

HCN(aq )  OH (aq )
– From the Brønsted-Lowry point of view, the
CN- ion acts as a base, because it accepts
a proton from H2O.
48
Acid-Base Properties of a Salt Solution
– You can also see that OH- ion is a product, so you
would expect the solution to have a basic pH. This
explains why NaCN solutions are basic.
CN  (aq )  H 2O(l )
HCN(aq )  OH  (aq )
NaCN is a basic salt.
– The reaction of the CN- ion with water is referred to
as the hydrolysis of CN-.
49
Acid-Base Properties of a Salt Solution
• The hydrolysis of an ion is the reaction of an
ion with water to produce the conjugate acid
and hydroxide ion or the conjugate base
and hydronium ion.
– The CN- ion hydrolyzes to give the
conjugate acid and hydroxide.

CN (aq )  H 2O(l )

HCN(aq )  OH (aq )
50
Acid-Base Properties of a Salt Solution
– The hydrolysis reaction for CN- has the form of a base
ionization so you write the Kb expression for it.

CN (aq )  H 2O(l )

HCN(aq )  OH (aq )

[ HCN ][OH ]
K b

[CN ]
Note: need Kb of CN- but probably given Ka HCN (ionization
constant tables contain molecular species), will need to convert.
51
Acid-Base Properties of a Salt
Solution
• An example of an cation that undergoes hydrolysis is
the ammonium ion


NH 4Cl  NH 4 (aq)  Cl (aq)
– The NH4+ ion hydrolyzes to the conjugate
base (NH3) and hydronium ion.

NH 4 (aq )  H 2O(l )
NH 3 (aq )  H 3O  (aq )

[ NH 3 ][H 3O ]
Ka 

[ NH 4 ]
– A 0.1 M solution has a pH of 5.1 and is therefore acidic.
52
33.3.1 Salts: Acidic, Basic, or Neutral
• How can you predict whether a
particular salt will be acidic, basic, or
neutral and undergoes hydrolysis?
– The Brønsted-Lowry concept illustrates the
inverse relationship in the strengths of conjugate
acid-base pairs.
– Consequently, the anions of weak acids are
good proton acceptors.
– Anions of weak acids are basic.
53
Predicting Whether a Salt is Acidic,
Basic, or Neutral
– On the other hand, the anions of strong acids (good
proton donors) have virtually no basic character, that is,
they do not hydrolyze. Therefore, conjugate base of strong
acids have do not undergo hydrolysis - neutral.
– For example, the Cl- ion, which is conjugate to the strong
acid HCl, shows no appreciable reaction with water. If did,
strong acid breaks up 100% and hydroxide produced
cancels with hydronium of strong acid and make neutral
solution


HCl (aq )  H 2O(l )  Cl  H 3O

Cl (aq )  H 2O(l )  HCl  OH
in essence,

 Cl   H 3O 
Cl  (aq )  H 2O(l )  no reaction
54
Predicting Whether a Salt is Acidic,
Basic, or Neutral
– Small highly charged (attract O more weakens O-H bond)
cations have an attraction for the O on water and cause the
production of H+ while large low charged cations do not.
Low charged large ions are group I and II metals (cations of
strong bases) and do not undergo hydrolysis - neutral.
– For example,
In summary,

Na (aq )  H 2O(l )  no reaction
If cation is that of the strong base (Na+, K+, Li+, Sr2+, Ba2+, Ca2+),
it will have neutral hydrolysis; otherwise, cation will have acidic
hydrolysis
If anion of monoprotic strong acid (NO3-, Cl-, Br-, I-, ClO4-), it
will have neutral hydrolysis; otherwise, anion will have basic
hydrolysis
55
• Salts may be acidic/basic/neutral and are composed of
cations (positive ions) and anions (negative ions)
cation anion
Acidic or neutral
Cations of strong bases
are neutral; otherwise,
cation contributes acidity
basic or neutral
Anions of monoprotic strong
acids are neutral; otherwise,
anion contributes basicity
Na+, K+, Li+, Ca2+, Sr2+, Ba2+
add neutral; rest of cations add
acidity to salt
Cl-, Br-, I-, NO3-, ClO4- add
neutral; rest of anions add
basicity to salt
56
Predicting Whether a Salt is Acidic,
Basic, or Neutral
• To predict the acidity or basicity of a
salt, you must examine the acidity or
basicity of the ions composing the salt.
Depending on the two components (cation/anion)
the overall salt will be acidic/neutral/basic :
nn  n
an  a
nb  b
ab  depends on which is larger Ka or Kb
57
Predicting Whether a Salt is Acidic,
Basic, or Neutral


NH 4Cl ( s)  NH 4 (aq)  Cl (aq)
neutral  acidic salt
pH < 7
acid


NaCN ( s ) 
 Na (aq )  CN (aq )
neutral
basic  basic salt
pH > 7
KC2H3O2(s)  K+(aq) + C2H3O2-(aq)
neutral
basic

basic salt
pH > 7
58
Predicting Whether a Salt is Acidic, Basic, or Neutral
acid neutral  acidic
• AlCl 3
• Zn(NO3)2 acid neutral  acidic
• KClO4
neutral neutral  neutral
• Na3PO4
neutral basic  basic
• LiF
neutral basic  basic
HW 33
code: important
• NH4F (Ka = 5.6x10-10 > Kb = 1.4x10-11)
acid basic  acidic
• NH4ClO (Ka = 5.6x10-10 < Kb = 3.6x10-7)
acid basic  basic
• NH4C2H3O2 (Ka =
59
5.6x10-10
= Kb =
5.6x10-10)
acid basic  neutral
33.3.2 pH of Salt Solution
• To calculate the pH of a salt solution
would require the Ka of the acidic cation
or the Kb of the basic anion.
– The ionization constants of ions are not listed
directly in tables because the values are easily
related to their conjugate species.
– Thus the Kb for CN- is related to the Ka for HCN.
60
The pH of a Salt Solution
• To see the relationship between Ka and Kb for
conjugate acid-base pairs, consider the acid
ionization of HCN and the base ionization of
CN-.
HCN(aq )  H 2O(l )
H 3O  (aq )  CN  (aq )
CN  (aq )  H 2O(l )
HCN(aq )  OH  (aq )
H 2O(l )  H 2O(l )

Ka
Kb

H 3O (aq)  OH (aq) Kw
61
HCN(aq )  H 2O(l )

CN (aq )  H 2O(l )
H 2O(l )  H 2 O(l )


H 3O (aq )  CN (aq )

Ka
HCN(aq )  OH (aq )
Kb
H 3O  (aq)  OH  (aq)
Kw
– When two reactions are added, their equilibrium
constants are multiplied.
– Therefore,
K a  K b  K w  1.00 x1014 @ 25o C
Calculate the Ka of NH4+ given the Kb of NH3 is 1.8 x 10-5 at 25oC
code: kw
HW 34
62
The pH of a Salt Solution
• For a solution of a salt in which only
one ion hydrolyzes, the calculation of
equilibrium composition follows that of
weak acids and bases.
– The only difference is first must obtain the
Ka or Kb for the ion that hydrolyzes.
63
Attack of hydrolysis problem
When attacking pH (hydrolysis problems) always ask yourself
1) reaction or single solute
2) acid, base, or salt
3) if acid or base, is it strong (straight forward) or weak (Ka or Kb
problem)
if salt, write dissociation eq and examine cation/anions involved to
determine if acid or base hydrolysis is available then Ka or Kb
problem.
100%
strong
acid
single solute
weak
Ka
strong
100%
neutral
base
weak
soluble
reaction
neutralization
cation SB
Kb
salt
cation
otherwise Ka
neutral
anion
insoluble
Ksp
Anion of monoprotic SA
otherwise Kb
64
A Problem To Consider
• What is the pH of a 0.10 M NaCN solution
at 25 °C? The Ka for HCN is 4.9 x 10-10.
acid, base, or salt?
basic salt
soluble or insoluble?
soluble
NaCN(s)  Na+(aq) + CN-(aq)
0.10 M
0.10 M
CN- (aq) + H2O(l)
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[CN-]
0.10
-x
0.10 - x
[HCN]
0
+x
x
HCN(aq) + OH-(aq)
[OH-]
0
+x
x
65
A Problem To Consider
Does answer make sense?
Yes, basic salt
66
example: What is the pH of 0.10 M NH4I solution? acid, base, or salt?
acidic salt
-5
Kb NH3 = 1.8 x 10
soluble or insoluble? soluble
NH4I(s)  NH4+(aq) + I-(aq)
0.10 M
0.10 M
NH4+ (aq) + H2O(l)
H3O+(aq) + NH3 (aq)
[NH4+]
[H3O+]
[NH3]
Initial, [ ]o
0.10
0
0
Change, D[ ]
-x
+x
+x
Equilibrium, [ ]eq
0.10 - x
x
x
HW 35
Does answer make sense?
Yes, acidic salt
code: salts
67
33.4 The Common Ion Effect
HC 2 H 3O 2 (aq )  H 2O(l )

Add NaC2H3O2

H 3O (aq)  C2 H 3O 2 (aq)
Initial: 0
pH will increase
0
• The common-ion effect is the shift in
an ionic equilibrium caused by the
addition of a solute that provides an ion
common to the equilibrium.
pH of acetic acid is lower than the pH of acetic
acid/sodium acetate because of the common ion effect.
68
The Common Ion Effect

NaC 2 H 3O2  Na (aq)  C2 H 3O2 (aq )
HC 2 H 3O 2 (aq )  H 2O(l )
yM


H 3O (aq)  C2 H 3O 2 (aq)
x
pH
x+ y M
– The equilibrium composition would shift to the left
and the degree of ionization of the acetic acid is
decreased, meaning less hydronium formed and
higher pH of solution as compared to pure acid.
– This repression of the ionization of acetic acid by
sodium acetate is an example of the common-ion
effect. Another source of ion coming from an
additional solute added to the solution.
69
33.4.1 Buffers
• When common ion effect involves weak acid/conjugate base (or
vice versa), it is referred to as a buffer. Buffer is a solution
characterized by the ability to resist changes in pH when limited
amounts of acid or base are added to it.
– Buffers contain either a weak acid and its conjugate base or a
weak base and its conjugate acid. Typically conj is in form of
salt. The components differ by one proton.
HC2H3O2 / NaC2H3O2
HCN / NaCN
H2CO3 / KHCO3
NH4Cl / NH3
H3PO4 / Na2HPO4 buffer?
No, differ by 2 protons
HCl / NaCl buffer?
No, strong acid
– Thus, a buffer contains both an acid species and a base species in
equilibrium which allows it to absorb any added acid or base and
keep pH of solution relatively constant
70
Buffers
– Consider a buffer with equal molar amounts of HA
and its conjugate base A-.
NaA  Na  (aq )  A (aq )
HA(aq)  H 2O(l )


H 3O (aq)  A (aq)
replace consumed A- and maintain pH
(slight decrease in pH due to additional hydronium)
If add small amount acid – react with acid or base component? base


A (aq)  H 3O (aq)
H 2O(l )  HA(aq)
neutralization
If add small amount of base – reacts with acid component , produce A-, and
shifts to left
replace consumed HA and maintain pH
(slight increase in pH due to decrease in hydronium)
71
An aqueous solution is 0.25 M in formic acid, HCHO2 and 0.18 M in
sodium formate, NaCHO2. What is the pH of the solution. The Ka
for formic acid is 1.7 x 10-4. neutralization or buffer? buffer
NaCHO2(s)  Na+(aq) + CHO2-(aq)
0.18 M
0.18 M
HCHO2 (aq) + H2O(l)
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[HCHO2]
0.25
-x
0.25 - x
H3O+(aq) + CHO2- (aq)
[H3O+]
0
+x
x
[CHO2-]
0.18
+x
0.18 +x
buffer, similar conc
can neglect x
Note: 0.25 M HCHO2
pH = 2.18
0.25 M HCHO2/0.18 M NaCHO2
pH = 3.63
72
A solution is prepared to be 0.15 M HC2H3O2 and 0.20 M NaC2H3O2.
What is the pH of this solution at 25oC? Ka HC2H3O2 @ 25oC = 1.8 x 10-5
neutralization or buffer? buffer
NaC2H3O2 (s)  Na+(aq) + C2H3O2 -(aq)
0.20 M
0.20 M
HC2H3O2 (aq) + H2O(l)
H3O+(aq) + C2H3O2 - (aq)
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[HC2H3O2 ]
0.15
-x
0.15 - x
[H3O+]
0
+x
x
[C2H3O2-]
0.20
+x
0.20 +x
Note: 0.15 M HC2H3O2
pH = 2.78
0.15 M HC2H3O2 /0.18 M NaC2H3O2
pH = 4.87
73
Calculate the pH of 75 mL of a buffer solution (0.15 M HC2H3O2 / 0.20 M
NaC2H3O2 pH = 4.87) to which 9.5 mL of 0.10 M HCl is added. Ka HC2H3O2
@ 25oC = 1.8 x 10-5
Step 1: Calculate the initial mmols of each component
Step 2: Neutralization reaction
NaC2H3O2(aq) + HCl(aq )  HC2H3O2(aq) + NaCl (aq)
mmolo
Dmmol
mmolneu
15.00
-0.95
14.05
0.95
-0.95
0.00
basic salt
11.25
+0.95
12.20
0
+0.95
0.95
weak acid
neutral salt
Step 3: New concentrations for species affecting pH
74
Step 3: New concentrations for species affecting pH
Step 4: Hydrolysis
NaC2H3O2 (s)  Na+(aq) + C2H3O2 -(aq)
0.166 M
0.166 M
HC2H3O2 (aq) + H2O(l)
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[HC2H3O2]
0.144
-x
0.144 - x
H3O+(aq) + C2H3O2 - (aq)
[H3O+]
0
+x
x
[C2H3O2-]
0.166
+x
0.166 +x
75
What was the change in pH to the buffer by additional HCl?
Calculate the pH if we added the same 9.5 mL of 0.10 M HCl to 75 mL of pure H2O.
HCl (aq)  H+(aq) + Cl -(aq)
0.0112 M 0.0112 M
HW 36
code: buffers
buffer holds pH relatively
constant when small amounts of
acid or base or added
76
Buffers
– Two important characteristics of a buffer are its
buffer capacity and its pH.
– Buffer capacity depends on the amount of acid
and conjugate base present in the solution;
typically,
1 pH unit.
– The next topic illustrates how to calculate the pH
of a buffer by special eq but suggest you do it just
like we did to keep with thought process and not
memorize a equation.
77
The Henderson-Hasselbalch Equation
• How do you prepare a buffer of given pH?
– A buffer must be prepared from selecting a conjugate
acid-base pair in which the Ka of the acid is
approximately equal to the desired H3O+
concentration.
– Note: you can add extra conjugate base, higher pH, or
less base, lower pH, to get to exact pH wanted.
78
The Henderson-Hasselbalch Equation
– Consider a buffer of a weak acid HA and its
conjugate base A-.

HA(aq )  H 2O(l )

H 3O (aq )  A (aq )
– The acid ionization constant is:


[H 3O ][ A ]
Ka 
[HA ]
– By rearranging, you get an equation for the H3O+
concentration.
[HA ]
[H 3O ]  K a  
[A ]

79
The Henderson-Hasselbalch Equation
– Taking the negative logarithm of both sides of the
equation we obtain:
[ HA]
- log[H 3O ]   log( K a   )
[A ]

[ HA]
- log[H 3O ]   log( K a )  ( log  )
[A ]

– The previous equation can be rewritten

[A ]
pH  pK a  log
[HA ]
80
The Henderson-Hasselbalch Equation
– More generally, you can write
[base, A ]
pH  pK a  log
[acid , HA]
Previous example: An aqueous solution is 0.25 M in formic acid,
HCHO2 and 0.18 M in sodium formate, NaCHO2 , had pH 3.63. The
Ka for formic acid is 1.7 x 10-4.
– This equation relates the pH of a buffer to the
concentrations of the conjugate acid and base. It is known
as the Henderson-Hasselbalch equation (for buffers
only). Also way to work previous buffer problem. Note
pH = pKa when conc base=acid
81
• Let’s prepare a buffer with pH of 4.90
– So to prepare a buffer of a given pH (for example, pH 4.90)
we need a conjugate acid-base pair with a pKa close to the
desired pH or 10-4.90 to find Ka close to 1.3 x 10-5 (desired
hydronium conc. equivalent to pH desired)
HC2H3O2 Ka = 1.8 x 10-5
H2CO3
Ka = 4.3 x 10-7
HNO2
Ka = 4.5 x 10-4
want Ka close to [H3O+]
– The Ka for acetic acid is 1.8 x 10-5, and its pH/pKa
is 4.74 (calculated using 1.8 x 10-5 as [H3O+]) if
equal conc of acid/salt are used.
– You could get a buffer of pH 4.90 by increasing the
ratio of [base]/[acid], meaning more basic salt.
82
What if I wanted to use 0.100 M HC2H3O2 (Ka = 1.8 x 10-5), what
concentration of NaC2H3O2 would I need to obtain a buffer with pH =4.90.
HC2H3O2 (aq) + H2O(l)
H3O+(aq) + C2H3O2 - (aq)
[base, A ]
pH  pK a  log
[acid , HA]
pKa = -log 1.8 x 10-5 = 4.74

[C H O ]
4.90  4.74  log 2 3 2
0.100

[C2 H 3O2 ]
4.90  4.74  0.16  log
0.100

[
C
H
O
0.16
2
3 2 ]
10 
0.100

100.16  0.100  [C2 H 3O2 ]

1.45  0.100  0.14 M  [C2 H 3O2 ]  [ NaC 2 H 3O2 ]
HW 37
code: mixture 83
33.5 Acid-Base Titrations or Mixtures
• An acid-base titration curve is a plot of the pH of a
solution of acid (or base) against the volume of
added base (or acid).
H 2CO3  NaOH
H 2O  NaHCO3
HCl + NaOH  NaCl + H2O
titrant
NaHCO 3  NaOH
H 2O  Na 2CO3
___________________________
H 2CO3  2 NaOH
2 H 2O  Na 2CO3
– Such curves are used to gain insight into the
titration process.
– You can use titration curves to choose an
appropriate indicator that will show when the
titration is complete.
84
Figure: Curve for the titration of a
strong acid by a strong base.
end pt
end pt
Ebbing, D. D.; Gammon, S. D. General 85
Chemistry, 8th ed., Houghton Mifflin,
New York, NY, 2005.
• Figure shows a curve for the titration of HCl with NaOH.
– Note that the pH changes slowly until the titration
approaches the equivalence point.
– The equivalence point is the point in a titration when
a stoichiometric amount of reactant has been added;
contrary to end point (observable point).
– At this equivalence point of SA and SB, the pH of the
solution is 7.0 because it contains a neutral salt, NaCl,
that does not hydrolyze. NaOH + HCl  NaCl + H2O
– To detect the end point, you need an acid-base
indicator that changes color beyond equivalence
point (titration error).
– Phenolphthalein can be used because it changes
color in the pH range 8.2-10.
86
Previous example equivalent pt was at 7 but not all neutralization reactions
eq pts are at 7. During a titration or just mixing an acid and base together to
make a solution, the resulting solution will contain a salt and water
Acid + Base  salt + water
One of the chemical properties of acids and bases is that they neutralize one
another; however, that doesn’t mean that the product will be neutral.
The misconception is that if the acid and base are in stoich proportions that
the resulting solution is neutral. This is not true. The salt formed may be a
acidic, basic, or neutral salt and will dictate the pH of the solution. Common
sense can help you predict the pH of a mixture.
SA + SB  neutral salt
only true neutralization pH =7
WA + SB  basic salt
original acid and base neutralize but
product has basic properties and basic pH
SA + WB  acidic salt
original acid and base neutralize but
product has acidic properties and acidic pH
87
When working acid-base titration problems or mixtures treat
them all the same but must be able to examine the components
remaining after neutralization to determine what affects pH or not
(strong, weak, salts). They are all done the same. The problem will
dictate where you go. We can derive titration curve by doing a
calculation at different points and draw curve or measure with pH
meter to measure and draw curve.
1.) Calculate mmols initially
2.) write the neutralization reaction and find mmols of
components left after neutralization
3.) find new conc. of species that affect pH
4.) do hydrolysis of main species that affect pH (like earlier in chapter)
88
A Problem To Consider
• Calculate the pH of a solution in which 50.00 mL of 1.000 M
NaOH is added to 50.00 mL of 1.000 M HCl.
Step 1: Calculate the initial mmols of each component
Step 2: neutralization reaction
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
mmolo
50.00
50.00
0
Dmmol
-50.00
-50.00
+50.00
mmolneu
0.00
0.00
50.00
neutral salt
Step 3: New concentrations for species affecting pH
Sodium Chloride is a neutral salt; therefore, no further calculations are needed,
pH = 7.00. Note: this only happened because we are mixing the stoichiometric
amounts of a strong acid and a strong base. If we mix a weak and a strong species,
the equivalence point will not be at 7 as we will see in later examples.
89
• Calculate the pH of a solution in which 49.99 mL of 1.000 M
NaOH is added to 50.00 mL of 1.000 M HCl.
Step 1: Calculate the initial mmols of each component
Step 2: neutralization reaction
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
mmolo
50.00
49.99
0
Dmmol
-49.99
-49.99
+49.99
mmolneu
0.01
0.00
49.99
Strong acid
neutral salt
Step 3: New concentrations for species affecting pH
Step 4: Hydrolysis
HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq)
90
• Calculate the pH of a solution in which 50.01 mL of 1.000 M
NaOH is added to 50.00 mL of 1.000 M HCl.
Step 1: Calculate the initial mmols of each component
HW 38
code: strong
Step 2: neutralization reaction
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
mmolo
Dmmol
mmolneu
50.00
-50.00
0.00
50.01
-50.00
0.01
Strong base
0
+50.00
50.00
neutral salt
Step 3: New concentrations for species affecting pH
Step 4: Hydrolysis
NaOH(aq)  Na+ (aq) + OH- (aq)
91
• Calculate the pH of a solution in which 20.00 mL of 1.000 M NaOH is
added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5
Step 1: Calculate the initial mmols of each component
Step 2: neutralization reaction
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) +
mmolo
Dmmol
mmolneu
50.00
-20.00
30.00
Weak acid
20.00
-20.00
0.00
H2O (l)
0
+20.00
20.00
Basic salt
BUFFER
Step 3: New concentrations for species affecting pH
92
Step 4: Hydrolysis
NaC2H3O2 (s)
 Na+ (aq) + C2H3O2- (aq)
0.2857 M
HC2H3O2 (aq) + H2O (l)
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[HC2H3O2]
0.4286
-x
0.4286 - x
0.2857 M
H3O+ (aq) + C2H3O2- (aq)
[H3O+]
0
+x
x
[C2H3O2-]
0.2857
+x
0.2857 +x
93
• Calculate the pH of a solution in which 50.00 mL of 1.000 M NaOH is
added to 50.00 mL of 1.000 M HC2H3O2. Ka HC2H3O2= 1.8x10-5
Step 1: Calculate the initial mmols of each component
Step 2: neutralization reaction
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) +
mmolo
Dmmol
mmolneu
50.00
-50.00
0.00
50.00
-50.00
0.00
H2O (l)
0
+50.00
50.00
Basic salt
Step 3: New concentrations for species affecting pH
94
Step 4: Hydrolysis
NaC2H3O2 (s)
 Na+ (aq) + C2H3O2- (aq)
0.5000 M
C2H3O2- (aq) + H2O (l)
Initial, [ ]o
Change, D[ ]
Equilibrium, [ ]eq
[C2H3O2-]
0.5000
-x
0.5000 - x
0.5000 M
HC2H3O2(aq) + OH- (aq)
[HC2H3O2]
0
+x
x
[OH-]
0
+x
x
Notice at eq pt and pH is not 7
because weak/strong
95
• Calculate the pH of a solution in which 51.00 mL of 1.00 M NaOH is
added to 50.00 mL of 1.00 M HC2H3O2. Ka HC2H3O2= 1.8x10-5
Step 1: Calculate the initial mmols of each component
Step 2: neutralization reaction
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) +
mmolo
Dmmol
mmolneu
50.0
-50.0
0.00
51.0
-50.0
1.0
Strong base
H2O (l)
0
+50.0
50.0
Basic salt
Step 3: New concentrations for species affecting pH
[OH-] from basic salt is extremely small as
compared to strong base ; therefore, neglect
WA/conj basic salt – buffer
conj acidic salt/WB - buffer
SA/WA
SB/WB
SA/SA – add [H3O+]
SB/SB - add [OH-]
acid/base -neutralization rxn
96
Step 4: Hydrolysis
NaOH (aq) 
Na+ (aq)
+ OH- (aq)
Notice the pH - 9.22 and [OH-] - 1.7 x 10-5 M changed tremendously compared to the
previous example. This indicates that neglecting the contribution of NaC2H3O2 would
not affect the overall answer.
HW 39
code: titration
97