CHM 103 GENERAL CHEMISTRY
SPRING QUARTER 2004
Lesson Plan – 4/15/2004
1. Quiz at 12:00.
2. Brønsted-Lowry Acid-Base Model
a. an acid is a proton (H+ ion) donor.
b. a base is a proton (H+ ion) acceptor.
c. in an acid-base reaction, a proton is transfered from an acid to a base.
HB(aq) + A-(aq)
B-(aq) + HA(aq)
(1)
d. Conjugate acid-base pairs in Eq. (1):
i. HB is the conjugate acid of B-.
ii. B- is the conjugate base of HB.
iii. HB and B- are a conjugate acid-base pair.
iv. A- is the conjugate base of HA.
v. HA is the conjugate acid of A-.
vi. A- and HA are a conjugate acid-base pair.
e. An amphoteric substance can act either as a Brønsted-Lowry acid or as a
Brønsted-Lowry base. (It depends on what else is there to react with.)
3. Ion Product of Water
a. Water is amphoteric:
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq)
(2)
H+(aq) + OH-(aq)
(3)
b. This can also be written:
H2O(l)
(This works because both H+(aq) and H3O+(aq) are acceptable ways of
designating a hydrated proton.)
c. The equilibrium expression for Eq. (3) is:
KW = [H+][OH-] = 1.0 x 10-14
(4)
KW is called the ion product constant of water. The numerical value, 1.0 x
10-14, holds at 25°C.
d. In a neutral solution:
[H+] = [OH-] = (KW)1/2= 1.0 x 10-7
(5)
e. In an acidic solution:
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[H+] > [OH-] ,
(6)
[H+] > 1.0 x 10-7
(7)
[OH-] < 1.0 x 10-7
(8)
f. In a basic solution:
[H+] < [OH-] ,
(9)
[H+] < 1.0 x 10-7
(10)
[OH-] > 1.0 x 10-7
(11)
4. pH and pOH
a. Definition of pH:
pH = -log10[H+]
(12)
[H+] = 10-pH
(13)
b. Definition of pOH:
pOH = -log10[OH+]
(14)
[OH-] = 10-pOH
(15)
c. In a neutral solution:
pH = pOH = 7
(16)
d. In an acidic solution:
pH < 7
(17)
pOH > 7
(18)
e. In a basic solution:
pH > 7
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(19)
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pOH < 7
(20)
f. In any (aqueous) solution:
pH + pOH = 14
(21)
5. pH of Strong Acids
a. A strong acid is strong by virtue of being totally ionized in aqueous
solution. For example, a 0.1 M solution of HCl is 0.1 M in H+ and in Cl-.
Thus:
[H+] = 1.0 x 10-1
(22)
pH = - log10(10-1) = 1.0
(23)
And a 0.05 M solution of HNO3 has:
[H+] = 5.0 x 10-2
(24)
pH = - log10(5.0 x 10-2) =
- log10(5.0) - log10(10-2) =
-0.7 + 2.0 = 1.3
(25)
6. pH of Weak Acids
a. A weak acid is weak by virtue of being incompletely ionized in aqueous
solution. Weak acids can be either:
i. Molecules containing an ionizable hydrogen atom.
HA(aq) + H2O(l)
HA(aq)
H3O+(aq) + A-(aq)
(26a)
H+(aq) + A-(aq)
(26b)
(Eq. (26a) follows the Brønsted-Lowry Acid-Base model, showing
proton transfer from HA to H2O, while Eq. (26b) is the more
conventional representation of the ionization of a weak acid.)
ii. Ions containing an ionizable hydrogen atom.
NH4+(aq) + H2O(aq)
NH4+(aq)
H3O+(aq) + NH3(aq)
(27a)
H+(aq) + NH3(aq)
(27b)
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(Eq. (27a) follows the Brønsted-Lowry Acid-Base model, showing
proton transfer from NH4+ to H2O, while Eq. (27b) is the more
conventional representation of the ionization of a cation acting as a weak
acid.)
7. Equilibrium Constant for the Dissociation of a Weak Acid
a. We can write the reaction for the dissociation of acetic acid (HC2H3O2)
as:
H+(aq) + C2H3O2-(aq)
HC2H3O2(aq)
(28)
And we can write the equilibrium expression for Eq. (28) as:
K28 =
[H+][C2H3O2-]
[HC2H3O2]
= 1.8 x 10-5
(29)
Suppose we have a 0.05 M aqueous solution of acetic acid, and we want to
know [H+] and pH for the solution. We can set up an equilibrium table:
HC2H3O2
H+
C2H3O2-
[ ]0
0.05
0
0
[ ]eq
0.05 - x
x
x
= 1.8 x 10-5
(30)
Plugging numbers into Eq. (29) we get:
x2
0.05 -x
We “guess” that x << 0.05 and simplify Eq. (30) to get:
x2
0.05
= 1.8 x 10-5
(30)
x2 =
0.05 x 1.8 x 10-5
(31)
x2 =
0.9 x 10-6
(32)
x=
0.3 x 10-3
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Now we can add the final line to the equilibrium table:
HC2H3O2
H+
C2H3O2-
[ ]0
0.05
0
0
[ ]eq
0.05 - x
x
x
[ ]eq
0.05 - 0.0003
3.0 x 10-4
3.0 x 10-4
[ ]eq
0.05
3.0 x 10-4
3.0 x 10-4
Now we can check our approximation that 0.05 - x = 0.05. We found that
x was 0.0003, which would make 0.05 - x = 0.0497. But in fact, x = 0.05
is close enough.
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