This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 1 Equilibria IInvolving
nvolving
Weak Acids and Bases 2 3 Equilibria IInvolving
nvolving
Weak Acids and Bases
Acid
Acid
acetic, CH3CO2H
ammonium, NH 4+
bicarbonate, HCO 3 Aspirin is a good example
of a weak acid,
Ka = 3.2 x 10 4 Equilibria IInvolving
nvolving
Weak Acids and Bases Conjugate Base
CH3CO2, acetate
NH3, ammonia
CO32, carbonate Consider acetic acid
HOAc + H2 O
¸ A weak acid (or base) is one that ionizes
to a VERY small extent (< 5%). 4 0.003 M Values of Ka for acid and Kb for bases are
found in TABLE 17.4 — page 799
found TABLE
Notice the relation of TABLE 17.4 to the
table of relative acid/base strengths
(Table 17.3). 0.06 M 2.0 M Equilibria
Involving A
Weak Acid
Determining the pH
Determining the pH
of an acetic acid
of an acetic acid
solution.
solution.
See Screen 17.8.
See Screen 17.8. OAc [H3 O+ ][OAc ]
[HOAc] 1.8 x 105 (K is designated Ka for ACID)
ACID
Because [H 3O+] and [ OAc] are SMALL, Ka << 1. 5 0.0001 M + Conj. base Ka Equilibria IInvolving
nvolving
Weak Acids and Bases H3O+ Acid Equilibria Involving A Weak Acid
Involving
You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, H3O+, OAc,
and the pH.
Step 1. Define equilibrium concs.
[HOAc]
initial
change
equilib [H3O+] [OAc] 1.00
1.00 0
0 0
0 x
x +x
+x
x
x +x
+x
x
x 1.00x
1.00x Note that we neglect [H 3O+] from H 2O. a pH meter Page 1 6 Equilibria Involving A Weak Acid
Involving 7 You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc, and the pH. Ka [HOAc] Ka 1.00  x [H3O+ ][OAc ]
1.8 x 10 5 =
[HOAc] And so x = [ 1.8 x 10 5 =
=[ H3O+ Equilibria Involving A Weak Acid
Involving
Calculate the pH of a 0.0010 M solution of
formic acid, HCO2H.
HCO2H + H2O ¸
HCO2  + H3O+
10 4 = [ Ka • Exact Solution
[H3O+] = [HCO2] = 3.4 x 104 M
[H
[HCO2H] = 0.0010  3.4 x 10 4 = 0.0007 M
pH = 3.47
pH x = [H3O+] = [OAc] = 4.2 x 10 3 M Is approximate solution valid? 11 Exact or Approximate Solution?
1.8 x 10 5 = x2
1.00 pH =  log [H3O+] = log (4.2 x 10 3) = 2.37
2.37 10 Ka 1.8 x 10 5 = x = [H3O+] = [OAc] = [Ka • 1.00]1/2 1.00]1/2 [H3O+ ][OAc ]
[HOAc] Approximate solution
[H3O+] = [Ka • Co]1/2 = 4.2 x 104 M, pH = 3.37
[H
pH Ka x2
1.00 OAc x2
1.00  x Equilibria Involving A Weak Acid
Involving
Consider the approximate expression
Ka 1.8 x 10 5 = x2
1.00 x [H3O+ ] = [K a • 1 .00]1/2 For many weak acids
When do you use the exact
solution and when the
approximate solution? Ka 1.8 x 10 5 = 2 x
1.00 Page 2 9 Step 3. Solve K a approximate expression
approximate x2
1.00  x First assume x is very small because Ka is so
small.
Ka Equilibria Involving A Weak Acid
Involving
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc, and the pH. Step 3. Solve K a expression approximately
approximately
x2 This is a quadratic. Solve using quadratic
formula or method of approximations (see
Appendix A). Ka = 1.8 x 8 You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc, and the pH. Step 2. Write K a expression
[H3O+ ][OAc ]
1.8 x 10 5 = Equilibria Involving A Weak Acid
Involving [H3O+] = [conj. base] = [ Ka • Co]1/2
[H
where C0 = initial conc. of acid
The Great SUCO Rule of Thumb:
If 100 • Ka < Co, then [H3O+] = [Ka • Co]1/2
If 12 ...
View Full
Document
 Fall '08
 McQuade
 Organic chemistry, Bases

Click to edit the document details