Cliffs AP Chemistry, 3rd Edition

Step 3: Substitute into the equilibrium expression known (and unknown) information. Let x
equal the amount of H+that ionizes from HA. Because the molar ratio of [H+]:[A–] is 1:1, [A–]
also equals x, and we can approximate 0.50 −x as 0.50 (5% rule).

174 10.^5050 .x

2

# - =

Step 4: Solve for x.

x^2 = (1.74 × 10 –5)(0.50)

= 8.7 × 10 –6

x = 2.9 × 10 –3M = [H+]

(b) Restatement: pH and pOH in 0.50 M HA.

pH = −log[H+] = −log (2.9 × 10 –3) = 2.47

pH + pOH = 14

pOH = 14.0 −2.47 = 11.53

Note: Significant figures for logarithms is equal to the number of significant figures in the

mantissa.

(c) Restatement: % HA that does ionize.

%%%WholePart

HA

H

## 100 = 100

+

6

7

@

A

.

. %.%
050
==29 10# -^3 # 100 0 58

However, 0.58% represents the percentage of the HA molecules that doionize. Therefore,
100.00 −0.58 = 99.42% of the HA molecules do notionize.

(d) Given: Buffer pH = 6.50

Restatement: [HA]/[A–]?

Step 1: Recognize the need to use the Henderson-Hasselbalch equation.

log

acid

base

pH p= Ka+ (^106)
6
@
@
Step 2: Substitute into the equation the known (and unknown) information.
pKa= −log Ka= 4.76

.. log
HA

A

65 476=+

• 
  

6

7

@

A

log ...

HA

A

=-=6 50 4 76 1 74

• 
  

6

7

@

A

.

HA

A

=55 10#^1

• 
  

6

7

@

A

Because the question is asking for [HA] / [A–], you need to take the reciprocal: 0.018 155

Equilibrium