Strengths of Conjugate Acid-Base Pairs
Acid strength increases
strong
HCl H2SO4
medium
weak
very weak
HNO3 H3O+ HSO4- H3PO4 HC2H3O2 H2CO3 H2S H2PO4- NH4+ HCO3- HPO42- H2O
Cl- HSO4-
negligible
NO3
H2O
SO42-
H2PO4-
very weak
C2H3O2- HCO3- HS- HPO42-
weak
Base strength increases
NH3
medium
CO32- PO43- OH-
strong
Conjugate
Acid Strength
Relative
acid
strength
Relative
conjugate
base
strength
Very
strong
Very
weak
Strong
HA
H+ + A-
Weak
[H+] [A-]
pKa =
[HA]
Weak
Strong
Very
weak
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 508
Very
strong
Solutions of Acids and Bases: The Leveling Effect
• No acid stronger than H3O+ and no base stronger than OH– can exist
in aqueous solution, leading to the phenomenon known as the
leveling effect.
• Any species that is a stronger acid than the conjugate acid of water
(H3O+) is leveled to the strength of H3O+ in aqueous solution
because H3O+ is the strongest acid that can exist in equilibrium with
water.
• In aqueous solution, any base stronger than OH– is leveled to the
strength of OH– because OH– is the strongest base that can exist in
equilibrium with water
• Any substance whose anion is the conjugate base of a compound
that is a weaker acid than water is a strong base that reacts
quantitatively with water to form hydroxide ion
Weak Acids (pKa)
Weak Acids – dissociate incompletely (~20%)
Strong Acids – dissociate completely (~100%)
A(g) + 2 B(g)
3 C(g) + D(g)
Equilibrium constant (Keq) =
Keq =
[Products]
[Reactants]
[C ]3 [D]
[ A][B ]2
LeChatelier’s Principle
(lu-SHAT-el-YAY’s)
H+(aq) + C2H3O21-(aq)
HC2H3O2(aq)
1
Equilibrium constant
Keq =
[H  ][C2H3O2 ]
[HC2H3O2 ]
= Ka = Acid dissociation constant
Ka = 1.8 x 10-5 @ 25 oC for acetic acid
1
[H  ][C2H3O2 ]
Ka =
[HC2H3O2 ]
1
1.8 x 10-5
[H  ][C2H3O2 ]
=
[HC2H3O2 ]
Ionization Constants for Acids
Ka
HCl
H+ + Cl1-
very large
HNO3
H+ + NO31-
very large
H2SO4
H+ + HSO41-
HC2H3O2
H2S
large
H+ + C2H3O21-
1.8 x 10-5
H+
9.5 x 10-8
+
HS1-
Ionization of Acids
Acid
Hydrochloric
Ionization Equation
HCl
Sulfuric
H2SO4
Acetic
HC2H3O2
Ionization Constant, pKa
H1+ + Cl1H1+ + HSO41H1+ + C2H3O21-
very large
large
1.8 x 10-5
Formula
Name
Value of Ka*
HSO4HClO2
HC2H2ClO2
HF
HNO2
HC2H3O2
HOCl
HCN
NH4+
HOC6H5
hydrogen sulfate ion
chlorous acid
monochloracetic acid
hydrofluoric acid
nitrous acid
acetic acid
hypochlorous acid
hydrocyanic acid
ammonium ion
phenol
1.2 x 10-2
1.2 x 10-2
1.35 x 10-3
7.2 x 10-4
4.0 x 10-4
1.8 x 10-5
3.5 x 10-8
6.2 x 10-10
5.6 x 10-10
1.6 x 10-10
*The units of Ka are mol/L but are customarily omitted.
Increasing acid strength
Values of Ka for Some Common Monoprotic Acids
Sample 1)
One gram of concentrated sulfuric acid (H2SO4) is diluted to a 1.0 dm3 volume
with water. What is the molar concentration of the hydrogen ion in this solution?
What is the pH?
Solution)
First determine the number of moles of H2SO4
x mol H2SO4 = 1 g H2SO4
H2SO4
H+ + HSO41-
1 mol H2SO4
98 g H2SO4
&
= 0.010 mol H2SO4
HSO41-
H+ + SO42-
OVERALL:
H2SO4
0.010 M
2 H+ + SO42-
in dilute solutions...occurs ~100%
0.020 M
pH = - log [H+]
substitute into equation
pH = - log [0.020 M]
pH = 1.69
A volume of 5.71 cm3 of pure acetic acid, HC2H3O2, is diluted with water at
25 oC to form a solution with a volume of 1.0 dm3.
What is the molar concentration of the hydrogen ion, H+, in this solution?
(The density of pure acetic acid is 1.05 g/cm3.)
Step 1) Find the mass of the acid
Mass of acid = density of acid x volume of acid
= 1.05 g/cm3 x 5.71 cm3
= 6.00 g
Step 2) Find the number of moles of acid. (From the formula of acetic acid,
you can calculate that the molar mass of acetic acid is 60 g / mol).
 1 mol HC 2H3O 2 
= 0.10 mol acetic acid (in 1 L)
x mol acetic acid = 6.00 g HC2H3O2 
60
g
HC
H
O

2 3 2 
Molarity: M = mol / L
Substitute into equation
M = 0.10 mol / 1 L
M = 0.1 molar HC2H3O2
Step 3) Find the [H+]
Ka =
HC2H3O2
Step 3) Find the
0.1 M
[H+]
weak acid
H+ + C2H3O21-
0.1? M
Ka = 1.8 x 10-5 @ 25 oC for acetic acid
1
[H  ][C2H3O2 ]
Ka =
[HC2H3O2 ]
1
1.8 x
10-5
[H  ][C2H3O2 ]
=
[HC2H3O2 ]
How do the concentrations of
H+ and C2H3O21- compare?
[x][x]
[HC2H3O2 ]
Substitute into equation:
1.8 x 10-5 
1.8 x 10
-5
x2

[0.10 M]
pH = - log[H+]
x2 = 1.8 x 10-6 M
x = 1.3 x 10-3 molar
pH = - log [1.3 x10-3 M]
= [H+]
pH = 2.9
H+ Concentrations
…Strong vs. Weak Acid
Moles of Acid used to make
1 L of solution
H+
pH
0.010 mol H2SO4
0.0200 M
1.7
Strong acid
0.100 mol HC2H3O2
0.0013 M
2.9
Weak acid
Note: although the sulfuric acid is 10x less
concentrated than the acetic acid...
…it produces > 10x more H+
pH = - log[H+]
Practice Problems:
1a) What is the molar hydrogen ion concentration in a 2.00 dm3 solution
of hydrogen chloride in which 3.65 g of HCl is dissolved?
1b) pH
2a) What is the molar concentration of hydrogen ions in a solution
containing 3.20 g of HNO3 in 250 cm3 of solution?
2b) pH
3a) An acetic acid solution is 0.25 M. What is its molar concentration of
hydrogen ions?
3b) pH
4) A solution of acetic acid contains 12.0 g of HC2H3O2 in 500 cm3
of solution. What is the molar concentration of hydrogen ions?
1a) 0.0500 M
1b) pH = 1.3
2a) 0.203 M
2b) pH = 0.7
3a) 2.1 x 10-3 M
3b) pH = 2.7
4) 2.7 x 10-3 M
Weak Acids
Cyanic acid is a weak monoprotic acid. If the initial concentration of cyanic
+ is 4.8 x 10-3 M,
acid is 0.150 M and the equilibrium concentrationHof3O
H3+O(aq)
calculate Ka for cyanic acid.
4.8 x 10-3 M
0.150 M
Ka =
+ CN1-(aq)
H+(aq)
HCN(aq)
[Products]
[Reactants]
Ka =
4.8 x 10-3 M
[H3O+] [CN1-]
[HCN]
[4.8 x 10-3 M][4.8
[CNx1-10
] -3 M]
Ka =
[0.150 M]
Ka = 1.54 x 10-4
How is [H3O+] determined?
Measure pH of solution and work backwards