pH of precipitation in the
United States 2001, and in
Europe as reported in 2002.

What are Acids and Bases?
acid = substance that increases the concentration of H
3
O + base = decreases the concentration of H
3 increasing the amount of OH )
O + (by

Bronsted-Lowry Acid-Base Theory acid = proton donor (H + ) base = proton acceptor
HCl + H
2
O →
HCl + NH
3
→

Conjugate Acid-Base Pairs

Relation Between [H + ], [OH ], and pH
H
2
O + H
2
O
H
2
O H + + OH -
H
3
O + + OH equivalent
K w
= [H + ][OH ] = 1.01 x 10 -14 at 25 o C

Example, p. 169: Concentration of
H + and OH in Pure Water at 25 o C
Calculate the concentrations of H + and OH in pure water at 25 o C.

As the concentration of H + increases,
OH must decrease and vica-versa
Example, p. 169: Finding [OH-] when H+ is
Known.
What is the concentration of OH if [H + ] =
1.0 x 10 -3 M at 25 o C?

pH - a measure of the aciity of a solution ("puissance d'hydrogen") pH = -log [H + ] (approximately!)
[H+] = 10 -3 M
[H+] = 10.0 M
[H+] = 10 -10 M

Strengths of Acids and Bases strong = complete (100%) dissociation
MEMORIZE these strong acids and bases - all other acids and bases are weak

weak = incomplete dissociation
HA H + + AOR HA + H
2
O
K a

[H

][A

]
HA
H
3
O + + A -
K a

[H
3
O

][A

]
HA
B + H
2
O BH + + OH -
K b

[BH

][OH

]
B

Classes of Weak Acids and Bases carboxylic acids = weak acids amines = weak bases
RNH
2
R
2
NH
R
3
N primary secondary tertiary polyprotic acids and bases
H
2
CO
3
H
3
PO
4
Ca(OH)
2
CO
PO
3
2-
4
3-

Relation Between K a and K b
HA + H
2
O
A + H
2
O
H
3
O + + A salt = conjugate base undergoes hydrolysis
HA + OH -

Example, p. 174: K a
1.75 x 10 -5 . Find K b for acetic acid is for the acetate ion.

pH of solutions of strong acids and bases
HA → H + + A -
BOH → B + + OH -
H
2
O H + + OH -
[H + ] = [OH ] = 1.0 x 10 -7 strong acids and bases completely dissociate
Case I: concentration of acid or base >> 10 -7

pH of a strong acid: Example p. 175
Find the pH of 4.2 x 10 -3 M HClO
4

Case II: concentration of acid or base  10 -7
Now the contribution of H + from water must be included -

[H

]

C
HA

2 C
HA

4K w
2
[OH

]

C
BOH

2 C
BOH

4K w
2 note that w hen 2 C
HA or 2 C
BOH

4K w
[H

]

C
HA and [OH

]

C
BOH

pH of a strong base at a low concentration:
“trick question” top of p. 176
Find the pH of 4.2 x 10 -9 M KOH

pH of solutions of weak acids and bases (Sec 8-6, 8-7) – the “ICE” table
Calculate the pH of a 0.020 M benzoic acid solution. K a
= 6.28 x 10 -5
I: Exact solution using quadratic equation

II. Approximate solution

Weak Base Equilibrium, Example p. 183
Find the pH of a 0.0372 M solution of the commonly encountered (?) weak base cocaine. K b
= 2.6 x 10 -6

Ch 9:
Buffers buffer = resists changes in pH; solution of a weak acid or base and their salts

Henderson-Hasselbach Equation derivation and assumptions:

Example, p.191: Using the H-H Equation
Sodium hypochlorite (NaOCl) was dissolved in a solution buffered to pH = 6.20. Find the ratio
[OCl ]/[HOCl]

Example, p. 192: A Buffer Solution
Find the pH of a solution prepared by dissolving 12.43 g of TRIS (FM = 121.136) plus 4.67 g TRIS hydrochloride
(FM = 157.597) in 1.00 L of water.

If add H +
A Buffer in Action
Weak Acid & Salt e.g. CH
3
COO / CH
3
COOH
Weak Base & Salt e.g. NH
4
+ / NH
3
If add OH -

How to Prepare a Buffer Solution
1.
2.
3.
4.
Consult a table of pKa's and pick the weak acid or base closest to the pH you need.
Solve for the ratio mol salt/(mol acid/base)
Choose a reasonable value for either mol salt or mol acid/base and solve for the other
After preparing the buffer, adjust the pH to the desired value (you never get exactly what you calculate because of the assumptions made in deriving the H-H equation

Example: buffer with pH = 4.8
acid/base acetic acid pK a
4.757
pK b
4.202
benzoic acid ammonia dimethylamine
4.74
3.13

Buffer Capacity: How well a solution resists changes in pH when an acid or base is added: when the pH = pKa!

Example, p.198
HA = H + + A mol A = 0.0383, mol HA = 0.9617