Chemistry 121
Lecture 17: Acids and Bases
Sections 9.1-9.7 in Blei & Odian 2nd edition
12/01/08
Learning Objectives
1. Define the hydronium and hydroxide ions in terms of the dissociation
of water
2. Define strong and weak acids, strong and weak bases
3. Rationalize why a compound is an acid (strong or weak) or base (strong
or weak)
4. Identify the basic properties of the pH scale
5. Make simple pH calculations
6. Describe acids and bases using the BrØnsted-Lowry definition
7. Know which acid-base reactions go to completion (better: which case
must be considered on an individual basis)
8. Define the concept of conjugate acids and bases
9. Utilize conjugate acid-base theory to interpret the acid-base
behavior of a given compound
10. Describe the behavior of polyprotic acids
11. Appreciate the ability of some salts to influence pH
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9.1: Water Reacting with Water
The autoionization of water produces the hydronium ion and the hydroxide
ion:
The autoionization of water gives rise to the pH scale. Bear in mind

The autoionization of water is a pretty crummy reaction; water is very
stable, and the reverse reaction – the mixing of a strong acid and
strong base can be highly hazardous because it is so favorable

Because of the crummy extent of reaction, the pH scale is a –log
scale; that is, low pH values mean a greater molar H3O+ concentration
and there is a 10x change in concentration between pH units

The autoionization of water is an equilibrium process; as such pH + pOH =
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9.2: Strong Acids and Strong Bases
By definition, a strong acid completely dissociates in water to give an equal
concentration of H3O+. Our primary strong acids are H2SO4, HNO3, and all
the haloacids barring HF
By definition, a strong base completely dissociates in water to give an equal
concentration of OH-. Our primary strong bases are the hydroxides of
group 1 and 2, excepting Mg(OH)2 and Ca(OH)2 (which are used as antacids1)
1
Tums is much more pleasant to ingest than Drano
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Insight into acid strength – the stability of the conjugate base
Let’s look at the structure of our strong acids and compare it to the weak
acids acetic acid and phosphoric acid, as well as ethanol (which is neither
acidic or basic):
Insight into base strength – again, electronegativity and concentration of
charge
OH- vs NH3
OH- vs. NH2-
OH- vs. H:-
9.3: The pH scale and pH Calculations
Please indicate the pH of the following solutions

3.65 g HCl dissolved in 1 L water

3.65 mg HCl dissolved in 10 L water
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
3.65 g HCl dissolved in 10 L water

4.0 g NaOH dissolved in 1L water
9.5: The BrØnsted-Lowry Definition of Acids and Bases

An acid is a proton donor

A base is a proton acceptor
o A base must have a pair of electrons with which to accept a
proton

The acid-base reaction when HCl gas is bubbled into water

The acid-base reaction when HCl gas is mixed with ammonia gas
9.4: Weak Acids and Weak Bases
By definition, a weak acid dissociates incompletely in water but does produce
some H3O+. Our primary weak acids are CH3CO2H (or other carboxylic acids),
H3PO4, and H2CO3
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By definition, a weak base dissociates incompletely in water but does produce
some OH-. Our primary weak bases are NH3 (or the organic derivatives the
amines), Mg(OH)2, and Ca(OH)2
Equations for Acid-Base Reactions:
The single most important reaction between acids and bases is their reaction
together to form water (if the base is hydroxide) and a salt:
Strong Acid + Strong Base  Completion Warning!
HCl + NaOH 
Strong Acids + Weak Bases  Completion
HNO3 + NH3 
Weak Acids + Strong bases  Completion
CH3CO2H + LiOH 
Weak acids + Weak Bases
Equilibrium
CH3CO2H + CH3NH2 
As a reference for the equilibrium between a weak acid and weak base, note
that amino acids exist as zwitterions:
Problem: Write the conjugate acid-base pairs for the reactions above:
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9.6: Dissociation of Polyprotic Acids and pH of Salts
H2SO4 + 2NaOH  Na2SO4 + H2O
H2SO4 is a very strong acid; Na2SO4 is a very weak (conjugate) base
NaOH is a very strong base; H2O is a very weak conjugate acid
One may view the above reaction in water as a reaction between sulfuric acid
and water to form the hydronium ion, followed by hydroxide and the
hydronium ion forming water:
Notice that sulfuric acid can react twice; that is, donate 2 “H+”. HSO4- is a
considerably weaker than H2SO4 (why?) but the reaction is driven by the
strong base OH-.
The same holds true for phosphoric acid, which will consume 3 hydroxide
anions on a molar basis:
Notice that PO4-3 is significantly basic and will generate some hydroxide if
placed in water – it is a weak base
Question: Na3PO4 produces a mildly basic solution and NH4Cl produces a
mildly acidic solution. Please explain
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Chemistry 121
Lecture 18: Acids and Bases, Part II: Titration; Buffers & Buffer Capacity;
the Relation between Ionization and Solubility
Sections 9.8-9.10 in Blei & Odian 2nd edition
12/02/08
Learning Objectives:
1. Develop the Henderson-Hasselbach equation
2. Define buffer and buffer capacity
3. Calculate the pH of a simple buffer system
4. Prepare a simple buffer system of given pH
5. Define titration, end point, and equivalence point
6. Describe the features of the curve for titrating a strong acid with a
strong base
7. Describe the features of the curve for titrating a strong base with a
strong acid
8. Calculate the amount of strong acid in an unknown sample by titrating
with a given volume of a base solution of known concentration
9. Calculate the amount of strong base in an unknown sample by titrating
with a given volume of an acid solution of known concentration
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9.7: Buffers and Blood Buffer System
Question: What is a buffer?
Answer:
Question: How do buffers work?
Answer:
Question: What is the pH of a buffer system?
Answer: The Henderson-Hasselbach equation allows us to write
log([A-]/[AH]) = pH – pKa or [A-]/[AH] = 10pH – pKa
Notice that a 10x shift in the ratio of acid to base causes a 1 pH unit shift
from pKa
Question: What ratio of acetic acid (pKa = 4.76) to sodium acetate would be
required to make a buffer of pH 7.76?
Buffer capacity
Question: What is the difference between neutralization and buffering?
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Question: Why is it important to maintain the buffer close to the pKa of the
acidic component of the buffer in order to maintain high buffer capacity?
Common conjugate acid-base pairs for buffer systems

Acetic acid(pKa = 4.76)/sodium acetate

Dihydrogen phosphate(pKa = 7.20)/hydrogen phosphate

Carbonic acid(pKa = 6.35)/bicarbonate buffer
9.9: The Blood Buffering System
While there are several systems at play including phosphate and protein, our
primary means of regulating blood pH is via the carbonic acid/bicarbonate
buffer
Problem: Blood pH =7.4 and little variation is tolerated, and carbonic acid has
a pKa = 6.35
Fortunately, the carbonic acid/bicarbonate buffer system is effective at
buffering our blood since

We produce acidic byproducts in about a 10:1 ratio to basic products

Breathing regulates the amount of CO2 available to form H2CO3
Question: Does rapid breathing cause an increase or decrease in blood pH?
How about holding one’s breath?
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Question: Would it be appropriate to give a 1M pH = 7.4 phosphate buffer to
an individual to prevent acidosis or alkalosis?2
8.10: Titration
What do titration curves look like?
Strong Acids Titrated with Strong Bases
Strong Bases Titrated with Strong Acids
What is mass of acetic acid in 500 mL of an unknown vinegar solution if 20
mL of 6 M NaOH is required to reach the equivalence point of a 50 mL
sample?
2
NaH2PO4 = 120, 8g NaH2PO4(mole/120g) = 67 mmole
Na2HPO4 = 142, 9.47g Na2HPO4(mole/142g) = 67 mmole
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The Relationship between pH, Ionization, and Solubility
 Or “The Revenge of the Henderson-Hasselbach Equation”
o [A-]/[AH] = 10pH – pKa
 For compounds that are largely non-polar ionization can have a
tremendous influence on solubility
o Recall that an ion-dipole interaction, such as that between Na+
or Cl- in water, is highly favored energetically
Question: Why does aspirin crystallize out of solution in the stomach?
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Appendix: pKa’s, pH, and the Henderson-Hasselbach Equation
Any compound with hydrogen attached, regardless of charge status (i.e. AH
or BH+) may be treated as an acid:
AH  A- + H+
And an equilibrium ratio established:
Ka = [A-][H+]/[AH] (1)
Reorganizing (1) gives
Ka = [H+]([A-]/[AH])
Taking the –log of both sides gives
-logKa = -log[H+] + -log([A-]/[AH])
Applying the definition of “p” gives
pKa = pH - log([A-]/[AH])
log([A-]/[AH]) = pH – pKa
An easier to interpret form, obtained by exponentiating, is
[A-]/[AH] = 10pH – pKa
Major Results:
When pH = pKa, the “acid” is 50% ionized – 100 =1, [A-] = [AH]
For each unit pH and pKa differ, ionization changes by an order of
magnitude; e.g., pH = 10 & pKa = 9, [A-] = 10[AH]; pH = 11 & pKa = 9, [A-] =
100[AH]
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