Lecture 0: Acid-Base Chemistry
• Reading: Zumdahl Chapters 7 and 8
• Outline
– Definition of Acids and Bases
– Equilibrium Chemistry and Acids/Bases
– Titrations
Arrhenius (or Classical) Acid-Base Definition
An acid is a substance that contains hydrogen and dissociates
in water to yield a hydronium ion : H3O+
A base is a substance that contains the hydroxyl group and
dissociates in water to yield : OH Neutralization is the reaction of an H+ (H3O+) ion from the
acid and the OH - ion from the base to form water, H2O.
H 3O+ (aq) + OH − (aq) 
→ 2H 2O(aq)
Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species that donates an H+ ion.
An acid must contain H in its formula; HNO3 and H2PO4- are two
examples, all Arrhenius acids are Brønsted-Lowry acids.
A base is a proton acceptor, any species that accepts an H+ ion.
A base must contain a lone pair of electrons to bind the H+ ion;
a few examples are NH3, CO32-, F -, as well as OH -.
Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius
bases contain the Brønsted-Lowry base OH-.
Therefore in the Brønsted-Lowry picture, acids donate
protons and bases accept protons.
Ka: the Acid Dissociation Constant
For the ionization of an acid, HA, in water:
+
−
→
HA(aq) + H 2O(aq) ←
H
O
(aq)
+
A
(aq)

3
The equilibrium constant for this reaction is referred to as Ka:
Ka =
+
−
O
A
H
[ 3 ][ ]
[HA]
The stronger the acid, the higher the [H3O+] at
equilibrium, and the larger Ka:
Stronger acid
higher [H3O+]
larger Ka
• Strong acid undergoes
complete dissociation.
• Weak acid undergoes
partial dissociation.
[H3O+] and pH
• The hydronium ion concentration is generally expressed in
terms of pH.
pH = −log[H 3O+ ]
• The lower the pH, the greater the hydronium ion concentration.
• Example: If the pH of a solution is 3.4, what is the hydronium ion
concentration?
[H O ]= 10
+
3
− pH
= 10−3.4 = 3.98x10−4 M
Determining Concentrations from Ka and Initial [HA]
Problem: What is the pH of a 0.125 M HClO (hypochlorous acid)
solution? Ka = 3.5 x 10-8
Plan: We need to find [H3O+]. First, write the balanced equation and
the expression for Ka and solve for the hydronium ion concentration.
Solution:
−
+
→
HClO(aq) + H 2O(aq) ←
ClO
(aq)
+
H
O
(aq)

3
+] [ClO-]
[H
O
3
Ka =
= 3.5 x 10-8
[HClO]
Example (continued)
−
+
→
HClO(aq) + H 2O(aq) ←
ClO
(aq)
+
H
O
(aq)

3
Concentration (M)
Initial
Change
Equilibrium
HClO H2O
0.125
----x
---0.125 - x
----
H3O+
0
+x
x
ClO0
+x
x
X is so important it has a name: The extent of Advancement
X
x=
V
Ka =
( x )( x )
= 3.5 ⋅10−8
assume 0.125 - x = 0.125
0.125 − x
x 2 = 4.38 ⋅10−9 or x =  H 3O +  = 6.6 ⋅10−5
pH = − log10  H 3O +  = − log ( 6.6 ⋅10−5 ) = 4.2
Autoionization of Water
+
−
→
2H 2O(aq) ←
H
O
(aq)
+
OH
(aq)

3
• For pure water the concentration of hydroxyl and hydronium ions
must be equal:
 H 3O +  OH −  = K w = 1⋅10−14
 H 3O +  = OH − 
 H 3O +  ⋅  H 3O +  = 1 ⋅10−14
 H 3O +  = 1 ⋅10−14 = 1⋅10−7
pH = 7
• The molarity of pure water is:
1000g/L
18.02 g/mol
= 55.4M
Summary: Titration Curve Calculations
Acid/Base Titration Problems
• Problems are a mix of stoichiometry and equilibrium.
• Stoichiometry: Add a strong base (acid) to a weak acid (base), and
determine the extent of neutralization using stoichiometric concepts.
• Equilibrium: The concentration of weak base/acid conjugate
species is determined, and the equilibrium expression used to
determine [H3O+], or pH.
Weak Acid Titration
Midpoint (Buffer)
Initial
Calculating the pH During a
Weak Acid-Strong Base Titration
Problem: Calculate the pH during the titration of 20.00 mL of
0.250 M nitrous acid (HNO2; Ka = 4.5 x 10-4) after adding the
following volumes of 0.250 M NaOH : (a) 0.00 mL (b) 8.00 mL
(c) 10.00 mL (d) 20.00 mL.
Plan: (a) Just calculate the pH of a weak acid.
(b)-(d) Calculate the amounts of acid remaining after the
reaction with the base in addition to the conjugate base
concentration, and plug these into the equilibrium
expression.
Initial pH
• Before addition of base, problem is exactly like a weak-acid
equilibrium problem.
HNO2 (aq) + H2O(l)
Init
0.25
Change -x
Eq.
0.25-x
[H3O+] [NO2-]
Ka =
= x (x)
[HNO2]
0.250 M
H3O+(aq) + NO2-(aq)
0
0
x
x
x
x
= 4.5 x
10-4
pH = -log(1.061 x 10-2) = 1.97
x2 = 1.125 x 10-4
x = 1.061 x 10-2
After 8 ml Addition of NaOH: Stoichiometry
HNO2 (aq) + OH-(l)
H2O(l) + NO2-(aq)
• OH- converts HNO2 to NO2-.
• 8.00 mL x 0.250 mmol/mL = 2.00 mmol OH- which will
convert (neutralize) 2.00 mmol of HNO2.
• Initially 20 mL x 0.250 mmol/ml = 5.00 mmol HNO2.
• Therefore, 5.00 mmol - 2.00 mmol = 3.00 mmol of HNO2 left,
and 2.00 mmol of NO2- are produced.
After 8 ml Addition of NaOH: Equilibrium
• Need concentrations: mmol/total mL. Total ml = 28 ml
Concentration (M)
Initial
Change
Equilibrium
HNO2 (aq) + H2O(l)
H3O+(aq) + NO2-(aq)
0.107
-x
0.107 - x
0
x
x
 H 3O +   NO2− 
HNO2 ]
[
+
Ka =
⇒  H 3O  = K a
 NO2− 
[ HNO2 ]
= ( 4.5 x10
−4
= 6.8 x10−4
pH = 3.17
( 0.107 )
) ( 0.071)
0.071
x
0.071 + x
After 10 ml Addition of NaOH: Stoichiometry
HNO2 (aq) + OH-(l)
H2O(l) + NO2-(aq)
• OH- converts HNO2 to NO2-.
• 10.00 mL x 0.250 mmol/mL = 2.5 mmol OH- which will
convert (neutralize) 2.5 mmol of HNO2.
• Initially 20 mL x 0.250 mmol/ml = 5 mmol HNO2.
• Therefore, 5mmol - 2. 5 mmol = 2.5 mmol of HNO2 left,
and 2.5 mmol of NO2- are produced.
After 10 ml Addition of NaOH: Equilibrium
• Need concentrations, given by mmol/total mL. Total ml = 30 ml
Concentration (M)
Initial
Change
Equilibrium
HNO2 (aq) + H2O(l)
0.083
-x
0.083 - x
H3O+(aq) + NO2-(aq)
0
x
x
 H 3O +   NO2− 
HNO2 ]
[
+
⇒  H 3O  = K a
Ka =
 NO2− 
[ HNO2 ]
= ( 4.5 x10
−4
= 4.5 x10−4
pH = 3.35
( 0.083)
) ( 0.083)
0.083
x
0.083 + x
20 ml of NaOH: Equivalence
HNO2 (aq) + OH-(l)
H2O(l) + NO2-(aq)
• 20.00 mL x 0.250 mmol/mL = 5 mmol OH- which will
convert (neutralize) 5 mmol of HNO2.
• Initially 20 mL x 0.250 mmol/ml = 5 mmol HNO2.
• Therefore, all of the HNO2 is converted to NO2-!
• pH is determined by NO2- equilibrium:
NO2- (aq) + H2O(l)
OH-(aq) + HNO2(aq)
Equivalence (cont.)
• Initial [NO2-] = 5mmol/40 ml = 0.125M
NO2- (aq) + H2O(l)
Initial
Change
Equilibrium
OH-(aq) + HNO2(aq)
0.125
-x
0.125 - x
K
K b = w = 2.2x10−11 =
Ka
0
x
x
−
OH
[ ][HNO2 ]
−
NO
[ 2]
0
x
x
x2
=
0.125
x = [OH − ]= 1.65x10−6 M
pOH = −log[OH − ]= 5.78
pH = 14 − pOH = 8.22
Generic Titration Curves
: Initial
: Midpoint Buffer
: Equivalence
For 0.10 M HA