Chapter 16 - Acids and Bases: General Properties
 Brønsted Acid  proton donor
 Brønsted Base  proton acceptor
 Conjugate acid - base pair - an acid and its conjugate base or a
base and its conjugate acid
e.g. look at acetic acid dissociating in solution
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)


Brønsted acid
Conjugate base
look at NH3(aq) in water
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)


Brønsted base
conjugate acid
 KOH, RbOH, NaOH, are not strictly Brønsted Bases since none
of these substances accept a proton
 NaOH(aq)  Na+(aq) + OH-(aq)
OH-(aq) + H3O+(aq) ⇌ 2 H2O(l)
 Identify the acid base pairs in the following
CH3NH2(aq) + HCOOH(aq) ⇌ CH3NH3+ (aq) + HCOO-(aq)
CH3NH2  base
CH3NH3+  conjugate acid
HCOOH(aq) acid
HCO O-(aq) cong. Base
Representing protons in aqueous solution
CH3COOH(aq)
⇌ CH3COO-(aq) + H+(aq)
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
2
both representations of H+(aq) are equivalent


H7O3+ (aq), H9O4+ (aq) that have been observed
Although H3O+(aq) is the most useful when discussing Brønsted 
m, will use either H+(aq) or H3O+(aq)
Autoionization and the pH scale
 Water autoionizes to a small extent
2H2O(l) ⇌ H3O+(aq) + OH-(aq)
or H2O(l)  H+(aq) + OH-(aq)
 these are both equivalent definitions of the autoionization reaction.
Water is acting as a base and an acid in the above reaction.
Water is amphoteric.
from the preceding chapter
Kc
[ H 3 O+ ][ OH - ]
=
[ H 2 O]

[ H+ ][ OH - ] 
 or

[ H 2 O]  1

but we know [H2O] is constant
 Kc [H2O] = Kw = [H+][OH-]
 Ion product constant for water, i.e. the product of the molar
concentrations of H+ and OH- ions at a particular temperature at
298.15 K, Kw = [H+][OH-] = 1.0x10-14
 In a solution containing [H+] = [OH-] = 1.0 x 10-7 M, this solution is
neutral.
 what if we adjust [H+] in solution by adding a small amount of a
proton donor (an acid)
e.g. [H+] = 1.0*10-4M
 [OH-] = 1.0*10-14 M/1.0*10-4 M = 1.0*10-10 M
IMPORTANT
 Kw is temperature dependent
e.g. look at T = 313.15 K
Kw = 3.8*10-14
 but a neutral solution has [H+] = [OH-] = (Kw)½
 At 313.15 K, a neutral solution is one where [H+] = [OH-] = 1.9*10-7 M
(UNLESS OTHERWISE INDICATED, ALL OUR
CALCULATIONS WILL BE AT 298.15 K)
Example
 A certain dish detergent has a [OH-] = 0.000500 M, calculate [H+]
in the solution.
Kw = 1.0 x 10-14 = [H+][OH-]
 [H+] = 1.0x10-14/0.00500 M = 2*10-12
The pH scale
 Attributed to Sorenson in 1909
pH  -log [H+]
3
4
Solution Type
[H+] / M
pH range
acid solutions
[H+] > 1.0*107
pH < 7.00
basic solutions
[H+] < 1.0*107
pH > 7.00
neutral solutions
[H+] = [OH] = 1.0*10-7 pH = 7.00
-log [H+] = pH
since Kw = [H+][OH-]
log [OH-] = pOH -log Kw = {-log [H+] - log [OH-]}
or pKw = pH + pOH = 14.00
Example
 As solution of NaOH has a pOH = 1.73. Calculate the pH of the
solution, and calculate the [H+]!
pH + pOH = 14.00 = pKw
pH = 14.00 - 1.73 = 12.27
[H+] = 10-12.27 = 5.37*10-13
[H+] = 5.4*10-13 mol/L
note the significant figures.
Acid-Base Strength
 note acid strength is  number of H+ ions donated to water
 define the % ionization = (acid conc. at  m)/(initial acid conc.) *
100%
CH3COOH(aq)  CH3COO-(aq) + H+(aq) weak acid
HCOOH(aq)  HCOO-(aq) + H+(aq) weak acid
both < 5% ionized
5
 Other examples of weak acids  HF, HNO2, HCN
 Acids like HCl(aq), HBr(aq), H2SO4(aq)  all strong acids; they
are100% ionized in water.
e.g. HNO3(aq)  H+(aq) + NO3- (aq)
HClO4(aq)  H+(aq) + ClO4-(aq)
 Strong Bases  also 100% ionized in water
NaOH(aq)  Na+(aq) + OH-(aq)
Ba(OH)2(aq)  Ba2+(aq) + 2OH-(aq)
 Some bases are weak bases; they don’t ionize completely.
e.g. NH3(aq) + H2O(l)  NH(aq) + OH-(aq)
 There is a correlation between the strength of the acid and its
conjugate base and the base and its conjugate acid.
Example
HCl(aq)  H+(aq) + Cl-(aq)

conjugate base (very weak)
CH3COOH(aq)  CH3COO-(aq) + H+(aq)

conjugate base is relatively strong
NH3(aq) + H2O(l) NH4+ (aq) + OH-(aq)

relatively strong conjugate acid
NOTE THE FOLLOWING
1.
6
H+ (aq) (or H3O+(aq)) is the strongest acid that can exist in
aqueous solution. Any acid stronger than H+(aq) reacts with
water completely to produce H+(aq) and the weak conjugate
base.
e.g. HCl (aq) stronger acid than H3O+ (H+(aq)
 reacts with water completely to form H+(aq)
2.
Acids weaker than H+(aq) have the  m lying primarily to the left.
e.g. HNO2(aq) + H2O(l)  H3O+(aq) + NO2- (aq)
or HNO2(aq)  H+(aq) + NO2- (aq)
3.
The OH- ion is the strongest base that can exist in aqueous
solution. Bases stronger than OH-(aq) react with water to
produce the hydroxide ion (OH-).
e.g. NaNH2 (aq) + H2O (l)  NH3 (aq) + OH-(aq)
 NaNH2 does not exist in aqueous solution.
The Levelling Effect
 Any acid that is stronger than H+(aq) means that we have 100%
ionization of the acid. For acids like HCl(aq), HClO4(aq),
HNO3(aq), the appearance is one of equal acid strength.
 Water is said to have a levelling effect on the acid strength, in that
the solvent (in this case water) cannot differentiate among the
relative strength of all acids stronger than H+(aq), which is water’s
conjugate acid.
 How do we differentiate acid strength?
Use a solvent that is a fairly weak base.
CH3COOH (l) + H+ CH3COO+H2
 CH3COOH is a very weak base, if the acid can protonate this
weak base, it must be a strong acid.
e.g.
HCl (solv) + CH3COOH (l)  CH3COO+H2 (solv) + Cl-(solv)
HNO3 + CH3COOH (l)  CH3COO+H2 (solv) + NO3-(solv)
HClO4 + CH3COOH (l)  CH3COO+H2 (solv) + ClO4-(solv)
 Differences in these  m are easily seen.
We see HNO3 < HCl < HBr < HI < HClO4
Examples
 Calculate the pH of the following solutions.
1)
1.0 * 10-2 M HClO4
2)
0.020 M NaOH
3)
a mixture of the above solutions
Solutions
1) pH = -log [H+]
HClO4 ionizes completely
HClO4 (aq)  H+ (aq) + ClO4-(aq)
 1.0 x 10-2 M HClO4(aq)  1.0 x 10-2 M H3O+
pH = -log (1.0 x 10-2) = 2.00 (note sig. figs.)
2) 0.020 M NaOH  strong base, ionized completely
[OH-] = 0.0200
7
pOH = -log (2.00 * 10-2) = 1.70
8
 pH = 14.00 - 1.70 = 12.30
3)
When we mix the two (1.00 L of solution each, total volume = 2.00
L).
note strong acid + strong base; have to find the limiting reagent for
the system.
H3O+(aq) + OH-(aq)  2 H2O(l)
moles H+ = 0.010; moles OH- = 0.020 moles
 H3O+ is the limiting reagent; OH- is left over
moles OH- remaining = (0.0200 - 0.0100) moles
= 0.0100 moles
MOH- = 0.010 moles/2.00 L = 0.0050 moles/L

pOH = 2.30 = -log (0.0050)
pH = 14.00 - pOH = 14.00 - 2.30 = 11.70
Equilibria in Aqueous Solutions of Weak Acids/ Weak Bases
 by definition, a weak acid or a weak base does not ionize
completely in water ( <<100%). How would we calculate the pH
of a solution of a weak acid or a weak base in water?
 e.g., look at the equilibrium of HF (aq) in water.
HF (aq) ⇌ H+ (aq) + F- (aq) Ka = 7.1 x 10-4
or
HF (aq) +H2O (l) ⇌ H3O+ (aq) + F- (aq) Ka = 7.1 x 10-4
 the small value of Ka indicates that this acid is only ionized to a
small extent at equilibrium.
Ka 
9
[H ][F ]
[HF ]


Example
Calculate the pH of an aqueous solution of HF at the following
molarities
a) 0.10 M
b) 0.60 M
c) 1.00 M.
a) As with Any chemical equilibrium problem, we need an
equilibrium Data Table.
Data Table
Substance
Start
Change
m
[HF]
0.100
-x
0.100-x
[H3O+]
0
+x
x
[F-]
0
+x
x
let x = the amount of HF that dissociates.
Ka
[H  ][F  ]

[HF ]
[ x ][ x ]
[ x ]2
Ka 

[0.100  x ] [0.100  x ]

since Ka is small, we can try the assumption method.
 assume 0.100 - x  0.100
[ x ][ x ]
[ x ]2
Ka 

[0.100  x ] [0.100]
2
7.1x10  4 
x 
10
[x ]
 7.1x10  5  [ x ]2
[0.100]
7.1x10  5  8.43 x10  3 M
 Before we proceed any further, we have to check to see if
our assumption was valid!
[H  ]eq
[HF ]o
x 100%  5%
[8.43x 10 3 ]
x 100%  8.43%  5%
[0.100]
 Note: this is not a valid assumption. Therefore, we must use the
complete quadratic formula.
Ka 
[ x ][ x ]
 7.1x10  4
[0.100  x ]
7.1x10 4 (0.100  x )  x 2  7.1x10 5  7.1x10 4 x  [ x ]2
x 2  7.1x10  4 x  7.1x10  5  0
 from the quadratic formula, we find that
b2 - 4ac
x =
2ac
For this particular equation
- b+
a = 1; b = 7.1x10  4 ; c = 7.1x10  5
 from the quadratic equation, we get two roots.
root(1) = +8.06 x 10-3 M
root(2) = -8.77 x 10-3 M
 Of course, only one root has any physical significance
11
(root (1)).
 Use root (1) to calculate the pH of the solution.
pH   log(8.06x10 3 )  2.09
b) We again need an equilibrium Data Table.
Data Table
Substance
Start
Change
m
[HF]
0.600
-x
0.600-x
[H+]
0
+x
X
[F-]
0
+x
x
let x = the amount of HF that dissociates.
[H  ][F  ]
Ka 
[HF ]
[ x ][ x ]
[ x ]2
Ka 

[0.600  x ] [0.600  x ]

since Ka is small, we can try the assumption method.
 assume 0.600 - x  0.600
[ x ][ x ]
[ x ]2
Ka 

[0.600  x ] [0.600]
[ x ]2
4
7.1x10 
 4.26 x10  4  [ x ]2
[0.600]
x 
4.26 x10  4  2.06 x10  2 M
Before we proceed any further, we have to check to see if our
assumption is valid!
12
[H  ]eq
[HF ]o
x 100%  5%
[2.06x 10 2 ]
x 100%  3.44%  5%
[0.600]
 Note: this is a valid assumption. Therefore, the quadratic
formula is unnecessary.
 Calculate the pH
pH   log(2.06 x10 2 )  169
.
 We have stated that the quadratic formula is unnecessary in this
application; we will show here that the approximation and the
quadratic formula give the same answer for the pH of the
solution.
Ka 
[ x ][ x ]
 7.1x10  4
[0.600  x ]
7.1x10 4 (0.600  x )  x 2  4.26 x10 4  7.1x10 4 x  [ x ]2
x 2  7.1x10  4 x  4.26 x10  4  0
 from the quadratic formula, we find that
b2 - 4ac
x =
2ac
For this particular equation
- b+
a = 1; b = 7.1x10  4 ; c = 4.26 x10  4
 from the quadratic equation, we get two roots.
root(1) = +2.03 x 10-2 M
root(2) = -2.10 x 10-2 M
13
 Use root (1) to calculate the pH of the solution.
pH   log(2.03 x10 2 )  169
.
the same answer as when we used the approximation!
c) We again need an equilibrium Data Table.
Data Table
Substance
Start
Change
m
[HF]
1.00
-x
1.00 - x
[H+]
0
+x
x
let x = the amount of HF that dissociates.
[H3O  ][F  ]
Ka 
[HF ]
[ x ][ x ]
[ x ]2
Ka 

[100
.
 x ] [100
.
 x]

We can again try the assumption method.
 assume 1.00 - x  1.00
[ x ][ x ]
[ x ]2
Ka 

[100
.
 x ] [100
. ]
[ x ]2
4
7.1x10 
 7.1x10  4  [ x ]2
[100
. ]
x
7.1x10  4  2.66 x10  2 M
[F-]
0
+x
x
Before we proceed any further, we again have to check to see
if our assumption was valid!
[H 3O  ]eq
[HF ]o
14
x100%  5%
[2.66 x10 2 ]
x100%  2.66%  5%
[100
. ]
 Note: this is a valid assumption. Therefore, the quadratic
formula is unnecessary.
 Calculate the pH
pH   log(2.66 x10 2 )  157
.
[HF]o
0.100
0.600
1.00
pH
2.09
1.69
1.57

8.1%
3.4%
2.7%
[H  ]eq
note  = % dissociation of acid  n[acid ] x100%
o
where n = # of ionizable protons.
 As expected, when we increase the concentration of weak acid in
solution, the [H+]eq increases, and the pH decreases. However,
the % dissociation of the acid decreases! Reason. Le Chatelier’s
principle. When we increase the initial concentration of acid, the
amount of acid in the system at equilibrium increases, and this
pushes the dissociation of the acid to the left (i.e., towards
undissociated acid).
Equilibria of Weak Bases in Water
 To calculate the percentage dissociation of a weak base in
15
water, we approach the problem as in the case of the weak acid
above.
Example
Calculate the pH and the  value for the dissociation of a 0.20 M
solution of ammonia in water (Kb = 1.8 x 10-5)
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
[NH4 ][OH  ]
Kb 
[NH3 ]
note that the concentration of water is not included in the
equilibrium constant expression since it is a pure liquid.
We again need an equilibrium Data Table.
Data Table
Substance
Start
Change
m
[NH3]
0.20
-x
0.20-x
[NH4+]
0
+x
x
[OH-]
0
+x
x
let x= the amount of NH3 that dissociates.
[NH4 ][OH  ]
Kb 
[NH3 ]
[ x ][ x ]
[ x ]2
Kb 

[0.20  x ] [0.20  x ]

since Kb is small, we can try the assumption method.
 assume 0.20 - x  0.20
[ x ][ x ]
[ x ]2
Kb 

[0.20  x ] [0.20]
[ x ]2
5
18
. x10 
 3.6 x10  6  [ x ]2
[0.20]
x 
3.6 x10  6  19
. x10  3 M
Before we proceed any further, we have to check to see if our
assumption was valid!
[OH  ]eq
[NH 3 ]o
x100%  5%
[19
. x10 3 ]
x100%  0.95%  5%
[0.20]
 A valid assumption. Therefore, the quadratic formula is
unnecessary.
 Calculate the pH
pOH   log(19
. x10 3 )  2.72
pH  14.00  pOH  14.00  2.72
= 11.28
the degree of dissociation, , is calculated as above

[OH  ]eq
[NH 3 ]o
x100%
[19
. x10 3 ]
x100%  0.95%
[0.20]
Obtaining the Relationship between Ka and Kb
16
17
 e.g., look at the dissociation of CH3COOH in water.
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)
[H3O  ][CH3COO  ]
Ka 
[CH3COOH ]
 e.g., look at the reverse reaction, the hydrolysis (reaction of the
substance with water) of CH3COO-.
CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)
[OH  ][CH3COOH ]
Kb 
[CH3COO  ]
 For the overall reaction.
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq) Ka
CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH-(aq)
Kb
2 H2O (l) ⇌ H3O+ (aq) + OH- (aq)
 The equilibrium constant for this reaction is the ion-product
constant of water Kw = [H3O+][ OH-].
 From our rules for the equilibria of multiple reactions.
Kw = Kb Ka
 Hence, if we know the strength (i.e., the magnitude of the
dissociation constant) of either the acid or the base, we
automatically know the strength of the conjugate base of the acid
or the conjugate acid of the base.
18
Example
The equilibrium constant for the dissociation of acetic acid in water
is Ka = 1.8 x 10-5. Calculate the dissociation constant for the
conjugate base of acetic acid.
CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)

conjugate base
Kw = Kb Ka
1.0 x 10-14 = Kb (1.8 x 10-5)
Kb = 1.0 x 10-14 / 1.8 x 10-5
= 5.6 x 10-10
 Compare with the Kb value of the formate ion (HCOO-).
Ka (HCOOH) = 1.7 x 10-4
Kb (HCOO-) = 1.0 x 10-14 / 1.7 x 10-4
= 5.9 x 10-11
Kb(HCOO-) < Kb(CH3COOH)
 Since acid (or base strength) increases as the value of Ka (or Kb)
increases, we clearly see that the stronger acid (HCOOH) has the
weakest conjugate base (HCOO-), whereas, the weaker acid
(CH3COOH) has a stronger conjugate base (CH3COO-).
Diprotic/Polyprotic Acids
 Look at the following system.
H2C2O4 (aq) + H2O (l) ⇌ HC2O4- (aq) + H3O+ (aq)
Ka1
HC2O4- (aq) + H2O (l) ⇌ C2O42- (aq) + H3O+ (aq)
Ka2
 In general, we find that for the dissociation of diprotic and
polyprotic acids, the magnitudes of the dissociation constants
decrease in the direction
Ka1 > Ka2 > Ka3 etc.
19
For oxalic acid in water,
Ka1= 6.5 x 10-2
Ka2= 6.1 x 10-5
 Since Ka1>> Ka2, the [H3O+] (and the pH) in the solution is due
primarily to the first dissociation ONLY.
Salts of Conjugate Acids/Conjugate Bases
 Look at the dissolution of CH3COONa in water.
CH3COONa (aq)  Na+ (aq) + CH3COO- (aq)
But we know that the acetate ion, CH3COO- (aq), will hydrolyze in
aqueous solution according to the following reaction.
CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH-(aq)
Kb (CH3COO-)
 Therefore, the dissolution of the salt of a conjugate base will give
a basic solution.
 Similarly for the dissolution of HCOONa in water.
HCOONa (aq)  Na+ (aq) + HCOO- (aq)
Hydrolysis of the formate ion, HCOO- (aq), will occur.
HCOO- (aq) + H2O (l) ⇌ HCOOH (aq) + OH-(aq)
20
Kb (HCOO-)
 Also gives a basic solution.
Calculate the pH of a solution of 0.100 M CH3COONa.
CH3COONa (aq)  CH3COO- (aq) + Na+ (aq)
CH3COO- (aq) + H2O  CH3COOH (aq) + OH- (aq)
This is the basic hydrolysis reaction for sodium acetate.
[ CH 3 COOH ][ OH  ]
Kb 
 5 .6 x10 10

[ CH 3 COO ]
Data Table
[CH3COO-]
[CH3COOH]
[OH-]
Start
0.100
0
0
Change
-x
+x
+x
m
0.10 - x
x
x
let x = the amount of OH- produced by the hydrolysis reaction.
[ x ][ x ]
Kb 
 5 .6 x10 10
[ 010
.  x]
assume 0.10 - x  0.10
21
2
Kb 
x
 5 .6 x10 10
[ 010
. ]
x 2 = 5 .6 x10 11
x = 5 .6 x10 11  7 .48 x10 6
pOH   log( 7 .48 x10 6 )
pOH  5 .12
pH  14 .00  pOH
= 14.00 - 5.12 = 8.87
As expected, the [OH-] in solution is > 1.0 x 10-7; the solution is
basic.
 What about the dissolution of NH4Cl in water.
NH4Cl (aq)  NH4+ (aq) + Cl- (aq)
But we know that the ammonium ion, NH4+ (aq), is the conjugate
acid of the weak base NH3 (aq). Therefore, it will hydrolyze in
aqueous solution according to the following reaction.
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+(aq)
Ka (NH4+)
 Therefore, the dissolution of the salt of a conjugate acid base in
water will give an acidic solution.
 Similarly for the dissolution of CH3NH3Br in water.
CH3NH3Br (aq)  CH3NH3+ (aq) + Br - (aq)
Hydrolysis of the methylammonium ion, CH3NH3+(aq), will occur.
CH3NH3+ (aq) + H2O (l) ⇌ CH3NH2 (aq) + H3O+(aq)
Ka (CH3NH3+)
22
Also gives an acidic solution
Example
Calculate the pH of a solution of 0.010 M NH4Cl.
NH4Cl (aq)  Cl- (aq) + NH4+ (aq)
NH4+(aq) + H2O  NH3 (aq) + H3O+ (aq)
This is the acid hydrolysis reaction for the ammonium ion.
[NH 3 ][H 3O  ]
Ka 
 5.6 x10 10

[NH 4 ]
Data Table
[NH4+]
[NH3]
[H+]
Start
0.010
0
0
Change
-x
+x
+x
m
0.10 - x
x
x
[x ][x ]
 5.6  10 10
[0.010  x ]
assume 0.010 - x  0.010
Ka 
x2
Ka 
 5.6x 1010
[0.010]
x 2 = 5.6x 1012
x = 5.6x 10 12  2.37x 106
23
pH   log(2.37x 10 6 )
pH  5.63
As expected, the [H+] in solution is > 1.0 x 10-7; the solution is
acidic.
 What about salts in which both the cation and anion hydrolyze?
The pH of the solution will depend on the magnitude of the Ka
and the Kb values of the respective acidic and basic ions.
Ka vs. Kb
Ka > Kb
Ka < Kb
Ka = Kb
Type of Solution
Acidic Solution
Basic Solution
Neutral Solution