Outline:
3/7/07
Pick up CAPA 15 & 16 - outside
 4 more lectures until Exam 2…

Today:

Chapter 18
Buffers
Buffer calculations
Titrations
Lots of problems to practice:
 Chapter
17:
17.9, 17.11, 17.13, 17.15, 17.17, 17.21,
17.23, 17.25, 17.27, 17.29, 17.31, 17.33,
17.35, 17.37, 17.39, 17.41, 17.43, 17.45,
17.49, 17.51, 17.53, 17.55, 17.57, 17.59,
17.61, 17.63, 17.67, 17.69, 17.71, 17.73,
17.81, 17.83
Lots of problems to practice:
 Chapter
18:
18.1, 18.3, 18.5, 18.9, 18.11, 18.13,
18.15, 18.17, 18.19, 18.21, 18.23, 18.27,
18.29, 18.31, 18.33, 18.35, 18.37, 18.39,
18.41, 18.45, 18.47, 18.51, 18.55, 18.57,
18.59, 18.61, 18.63, 18.65, 18.67, 18.69,
18.71, 18.73, 18.75, 18.77, 18.79, 18.81,
18.83, 18.85, 18.87, 18.93, 18.101, 18.103
Summary of acid base rxns:
weak acid + weak base
Keq
weak acid + strong base
calculations
strong acid + weak base
solve for x
strong acid + strong base
pH + pOH = 14
Find Keq from Ka Kb = Kw
Acid Base reactions (remember):
 Add
0.10 mol NaOH to ….
(water) No common ions:
pH goes from 7.00 to 13.00
change = +6.00
(HOAc/OAc- mix) Common ions:
pH goes from 4.74 to 5.11
change = +0.37
Buffer calculations….
pH = pKa + log ([conj base]/[acid])
This one also exists:
pOH = pKb + log ([conj acid]/[base])
Practice!
 Definition:
A buffer is usually a mixture of
conjugate acid-base pairs in solution.
A buffer resists strong changes in pH.
pH of a buffer…
0.3 L of 0.4 M acetic acid is added to
0.7 L of 0.2 M sodium acetate, what
will be the pH of the resultant buffer?
pH = pKa + log ([conj base]/[acid])
pH = 4.75 + log [OAc-]/[HOAc]
Do the dilutions:
0.3/1.0 × 0.4 = 0.12 M HA
0.7/1.0 × 0.2 = 0.14 M A-
pH of a buffer…
0.3 L of 0.4 M acetic acid is added to
0.7 L of 0.2 M sodium acetate, what
will be the pH of the resultant buffer?
pH = pKa + log ([conj base]/[acid])
pH = 4.75 + log (0.14/0.12)
= 4.75 + 0.07
= 4.82
A buffer example:
Human blood must be maintained at a
pH of 7.40  0.05 for the rest of the body
biochemistry to function. This is
achieved with a buffer made from CO2:
CO2 (aq) + H2O (l)  H2CO3 (aq)
If the carbonic acid concentration of a
healthy human is 1.3510-2 M, what is
the normal human concentration of
HCO3-?
pH = pKa + log ([conj base]/[acid])
7.40 = 6.25 + log ([HCO3-]/ 1.3510 -2)
[HCO3-] = 1.5210-1 M
 Rather
“twisted” addition to the
problem:
Inject 10 mL of 1.0 M HCl into a
(previously healthy) human with a
total blood volume of 8.0 L. What is
the new pH?
 “Buffer capacity” problem
How many moles H+ added?
0
.010 L  1.0 mol/L = 0.010 mol
How many moles H+ there already?
= 10-7.40 = 3.1810-7 mol/L8.0 L
= 2.5410-6 mol
= negligible!
Initial [H+]?
 [H+]=(0.010 mol+2.5410-6 mol)/ 8.0 L
= 1.25  10-3 M
 [H+]
Same old procedure….
H2CO3  HCO3- + H +
1.3510-2
1.5210-1 1.2510-3 I
+1.2510-3 -1. 2510-3 -1.2510-3 D
1.4810-2
1.5110-1
0
-x
+x
+x
D
solve for x; x = 4.210-8 M; pH = 7.38
Ka = 4. 2710-7: calc in which direction?
or use:
Easier pH = pKa+log ([conj base] /[acid])
-1/1.4810-2) = 7.38
=
6.37+log(1.5110
to use
More Buffer pH…

Try another one….
A buffer solution is prepared by mixing
0.300 L of 0.400 M (CH3)3N and 0.700 L
of 0.200 M (CH3)3NHCl. What is the pH
of this buffer solution? The pKb of
(CH3)3N is 4.19.
This one also exists:
pOH = pKb + log ([conj acid]/[base])
pH= 9.74
Addition of acid/base to a buffer
 1.0
L of a buffer solution is prepared
with 0.025 M formic acid and 0.010 M
sodium formate (pKa = 3.74). If 0.005
mols (0.2 g) of NaOH is added to the
solution, what is the pH?
OHHCOOH  HCOO- + H+
pH = pKa + log ([conj base]/[acid])
pH= 3.74 + log (0.010+0.005/0.025-0.005)
or pH = 3.74 – 0.12 = 3.62
Addition of acid/base to a buffer
 What
if you exceed the buffer capacity?
If 0.050 mols (2.0 g) of NaOH is added
to the solution, what is the pH?
pH = pKa + log ([conj base]/[acid])
pH= 3.74 + log (0.010+0.050/0.025-0.050)
Simply neutralize
Not defined!
0.025 mols, and
=
0.025
M
OH
calculate the pOH of
the remainder…
Or pH = 12.4
weak acid
Buffer region
Equivalence
point
Mid-point
of OH- added