Ch 8:
Introducing Acids and Bases
pH of precipitation in the
United States 2001, and in
Europe as reported in 2002.
1
What are Acids and Bases?
acid = substance that increases the concentration of H3O+
base = decreases the concentration of H3O+ (by
increasing the amount of OH-)
Bronsted-Lowry Acid-Base Theory
acid = proton donor (H+)
base = proton acceptor
HCl + H2O →
HCl + NH3 →
2
Conjugate Acid-Base Pairs
Relation Between [H+], [OH-], and pH
H3O+ + OH-
H2 O + H 2 O
H2 O
H+
+
equivalent
OH-
Kw = [H+][OH-] = 1.01 x 10-14 at 25 oC
3
Example, p. 169: Concentration of
H+ and OH- in Pure Water at 25 oC
Calculate the concentrations of H+ and OH- in pure
water at 25 oC.
4
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 oC?
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
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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
H+ + A-
HA
Ka 
[H ][A  ]
HA
B + H2 O
Kb 
OR
HA + H2O
Ka 
H3O+ + A-
[H3O  ][A  ]
HA
BH+ + OH-
[BH ][OH ]
B
6
Classes of Weak Acids and Bases
carboxylic acids = weak acids
:
amines = weak bases
primary
:
RNH2
secondary
:
R2NH
R3N
tertiary
polyprotic acids and bases
H2CO3
CO32-
H3PO4
PO43-
Ca(OH)2
Relation Between Ka and Kb
HA + H2O
H3O+ + A-
A- + H2O
HA + OH-
salt = conjugate base
undergoes hydrolysis
7
Example, p. 174: Ka for acetic acid is
1.75 x 10-5. Find Kb for the acetate ion.
8
pH of solutions of strong acids and
bases
HA → H+ + ABOH →
H2O
B+
+
OH-
strong acids and bases
completely dissociate
H+ + OH-
[H+] = [OH-] = 1.0 x 10-7
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 HClO4
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Case II: concentration of acid or base  10-7
Now the contribution of H+ from water must be included -

[H ] 

2
CHA  CHA
 4K w
[OH ] 
2
2
CBOH  CBOH
 4K w
2
2
2
note that when CHA
or CBOH
 4K w
[H ]  CHA and [OH ]  CBOH
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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. Ka = 6.28 x 10-5
I: Exact solution using quadratic equation
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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. Kb = 2.6 x 10-6
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Ch 9:
Buffers
buffer = resists changes in pH; solution of a weak
acid or base and their salts
Henderson-Hasselbach Equation
derivation and assumptions:
13
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.
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A Buffer in Action
Weak Acid & Salt
Weak Base & Salt
e.g. CH3COO- / CH3COOH
e.g. NH4+ / NH3
If add H+
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
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Example: buffer with pH = 4.8
acid/base
pKa
acetic acid
4.757
benzoic acid
4.202
pKb
ammonia
4.74
dimethylamine
3.13
Buffer Capacity: How well a solution resists changes in
pH when an acid or base is added: when the pH = pKa!
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Example, p.198
HA = H+ + A-
mol A- = 0.0383, mol HA = 0.9617
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