Ch 15 – Describing Buffering Capacity
(Henderson-Hasselbalch)
Acids:
pH =
pKa + log (A-/HA)
Problem 1: Calculate the pH of each of the following buffered solutions: (Ka for acetic
acid is 1.8 x 10-5
a.
b.
c.
d.
0.10 M acetic acid/0.25 M sodium acetate
0.25 M acetic acid/0.10 M sodium acetate
0.08 M acetic acid/0.20 M sodium acetate
0.20 M acetic acid/0.08 M sodium acetate
Problem 2: Calculate the pH of each of the following solutions.
a.
b.
c.
d.
0.10 M propanoic acid (HC3H5O2, Ka = 1.3 x 10-5)
0.10 M sodium propanoate (NaC3H5O2)
Pure H2O
A mixture containing 0.10 M H C3H5O2 and 0.10 M NaC3H5O2
Problem 3: Calculate the change in pH that occurs when 0.010 M HCl is added to 1.00 L
of each of the following solutions:
Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Solution B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
Ka for acetic acid is 1.8 x 10-5
Henderson-Hasselbalch
pKa = -(log 1.8 x 10-5) (1) = 4.74
H+
Initial
After
0.010 M
0
C2H3O2-
HC2H3O2
5.00 M
5.00 M
4.99 M
5.01 M
PH = pKa + log (A-/HA)
= 4.74 + log (4.99/5.01)
= 4.74
Virtually no change in pH for solution A when 0.010 M HCl is added
Try solution B
pKa = -(log 1.8 x 10-5) (1) = 4.74
H+
Initial
After
0.010 M
0
C2H3O2-
HC2H3O2
0.050 M
0.050M
0.040 M
0.060 M
pH = pKa + log (A-/HA)
= 4.74 + log (0.040/0.060)
= 4.56
Due Tuesday
p 759-760 #12, and 17 (refer to our problem #2. This is the
book’s #15)