Buffer and Noncovalent Interactions Practice Problems
1. What is the final pH of a solution made by mixing 100 ml of 0.05 M acetic acid and 100 ml of 0.1
M sodium acetate? Assume the pK a
for acetic acid is 4.76.
2. Describe the preparation of 1000 ml of a 0.2 M acetate buffer pH 5. Assume the pK a
of acetic acid is 4.76.
3. Describe the preparation of 1000 ml of a 0.1 M phosphate buffer pH 7.2 starting with 0.5 M
H
3
PO
4
and 0.5 M NaOH. Assume the following pK a s: H
3
PO
4

H
2
PO
4
-
= 2.2;
H
2
PO
4

HPO
4
-2
= 6.8; HPO
4
-2 
PO
4
-3
= 12.4.
4. For a solution that is 0.08 M K
2
HPO
4
and 0.12 M KH
2
PO
4
, calculate the concentrations of
H
3
PO
4
, H
2
PO
4
-
, HPO
4
-2
, PO
4
-3
; H
+
, OH
-
, and K
+
. Use the pK a s given above.
5. Match the letter of the interaction ( indicated by ::::: ) to the number giving the correct description of the interaction:
A. CH
3
CH
2
NH
3
+ :::::
-
OC(O)CH
2
CH
3
______ 1. Ion-Dipole
B. CH
3
CH
2
NH
3
+ :::::
O=(CH
2
)
2
C. H-O-H ::::: 
O=(CH
2
)
2
______ 2. Hydrogen Bond
______ 3. Ion-Ion
6. At an air-water interface amphipathic molecules are aligned
A. With the hydrophobic tail in the water and the polar head group in the air
B. With the polar head group in the water and the hydrocarbon tail in the air.
C. Amphipathic molecules are freely soluble in water and do not accumulate at the air-water interface.
5. The pK a for a typical long chain fatty acid is about 5.0. Explain why long chain fatty acids can form micelles in solutions whose pH is greater than 7, but are insoluble in solutions whose pH is below about 4.
6. Do you expect the strength of a hydrogen bond to be greater in a solvent of high dielectric constant like water or a solvent of low dielectric constant like ethanol?
Buffer and Noncovalent Interactions Problems, Page 1

Buffer and Noncovalent Interactions Practice Problems – ANSWERS
1. What is the final pH of a solution made by mixing 100 ml of 0.05 M acetic acid and 100 ml of 0.1
M sodium acetate? Assume the pK a
for acetic acid is 4.76.
Answer:
 100 ml of 0.05 M acetic acid = 5 mmol.
 100 ml of 0.1 M sodium acetate = 10 mmol.
Using the Henderson-Hasselbalch equation
 pH = pK a
+ log{[acetate]/[acetic acid]}.
 pH = 4.76 + log{[10]/[5]}.
 pH = 5.06
2. Describe the preparation of 1000 ml of a 0.2 M acetate buffer pH 5. Assume the pK a
of acetic acid is 4.76.
Answer:
 Acetic acid + acetate = 0.2 M
 Using the Henderson-Hasselbalch equation, pH = pK a
+ log{[acetate]/[acetic acid]}.
 5.0 = 4.76 + log{[acetate]/[acetic acid]}.
 [acetate]/[acetic acid] = 1.74 / 1, so [acetate] as a fraction of total (acetic acid + acetate) =
1.74 / 2.74 = 0.635.
 Therefore, [acetate] = 0.635 x 0.2 M total = 0.127 M; and [acetic acid] = 0.073 M.
To prepare the buffer:
 Take 800 ml of water and add 0.2 mole of acetic acid.
 Add 127 ml of 1.0 M NaOH to produce 0.127 M sodium acetate.
 Add water to 1000 ml.
3. Describe the preparation of 1000 ml of a 0.1 M phosphate buffer pH 7.2 starting with 0.5 M
H
H
3
2
PO
PO
4
4
-
and 0.5 M NaOH. Assume the following pK a s: H
3
PO
4

HPO
4
-2
= 6.8; HPO
4
-2 
PO
4
-3
= 12.4

H
2
PO
4
-
= 2.2;
Answer:

 At pH 7.2, the two predominant forms of phosphate will be
H
2


The total phosphate is 100 mM = 200 ml of 0.5 M
The ratio of
The sum of [
HPO
7.2 = 6.8 + log {[
HPO
 Therefore, [
HPO
4
-2
4
-2
4
-2
 To prepare the buffer to
HPO
4
-2 which means that [
HPO
H
] + [
4
-2
 Take 200 ml of 0.5 M H
3
2
]/[
H
PO
4
-
2
H
2
]/[
H
PO
2
4
-
PO
H
3
PO
4
.
PO
4
is calculated from the Henderson-Hasselbalch equation:
PO
4
]},
4
] = 2.5 / 1.
] = 100 mM and [ HPO
4
-2 ]/[ H
2
] = (2.5/3.5) x 100 mM = 71.4 mM and [
H
2
PO
PO
PO
4
.
 Add 200 ml of 0.5 M NaOH to convert all the
H
3
PO
4
to
H
 Add 142.8 ml 0f 0.5 M NaOH to make 71.4 mM
HPO
 Add water to 1000 ml.
4
-2 .
2
PO
4
.
4
-
4
and
HPO
4
-2 .
] = 2.5 / 1.
] = 28.6 mM.
Buffer and Noncovalent Interactions Problems, Page 2

4. For a solution that is 0.08 M K
2
HPO
4
and 0.12 M KH
2
PO
4
, calculate the concentrations of
H
3
PO
4
, H
2
PO
4
-
, HPO
4
-2
, PO
4
-3
; H
+
, OH
-
, and K
+
. Use the pK a s given above.
Answer:
 [
H
2
PO
 [
HPO
 pH = 6.8 + log {[0.08]/[0.12]} = 6.62
 [H +
 [OH


[
For
H
For
[
PO
 K
2
] = 2.4 x 10 -7
-
H
4
4
-2
-
4
] = 0.12 M (the initial value)
] = 0.08 M (the initial value)
]/[
4
-3
M
] = (1 x 10 -14
3
PO
PO
4
-3
PO
4
: pH = pK a
+ log{[
H
2
PO
4
]/[
H
3
-
]/[
H
3
HPO
PO
4
-2
4
M 2 /2.4 x 10
] = 2.63 x 10 4
: pH = pK a
+ log{[
PO
] = 1.66 x 10 -6
-7 M) = 4.17 x 10
PO
4
]}; 6.62 = 2.2 + log{[
H
2
PO
4
-
; Therefore, [
H
3
4
-3 ]/[
HPO
4
-2
4
] = 4.56 x 10
]}; 6.62 = 12.4 + log{[
; therefore, [
PO
+ = 0.12 M + 2 x 0.08 M = 0.28 M
-8 M
PO
4
-3 ] = 1.32 x 10 -7
-6
M.
M.
PO
4
-3
]/[
H
3
PO
4
]};
]/[
HPO
4
-2 ]};
5. Match the letter of the interaction ( indicated by
::::: )
to the number giving the correct description of the interaction:
A. CH
3
CH
2
NH
3
+ :::::
-
OC(O)CH
2
CH
3
______ 1. Ion-Dipole
B. CH
3
CH
2
NH
3
+ :::::
O=(CH
2
)
2
C. H-O-H
::::: 
O=(CH
2
)
2
Answer:
1 = B; 2 = C, 3 = A
______ 2. Hydrogen Bond
______ 3. Ion-Ion
6. At an air-water interface amphipathic molecules are aligned
A. With the hydrophobic tail in the water and the polar head group in the air
B. With the polar head group in the water and the hydrocarbon tail in the air.
C. Amphipathic molecules are freely soluble in water and do not accumulate at the air-water interface.
Answer: B
7. The pK a for a typical long chain fatty acid is about 5.0. Explain why long chain fatty acids can form micelles in solutions whose pH is greater than 7, but are insoluble in solutions whose pH is below about 4.
Answer:
For the fatty acid to form a micelle it must have a charged polar head group. This will only occur when the carboxyl group has ionized, i.e., COOH

COO
-
+ H
+
, and this will only occur at pH values higher than the pK a
.
8. Do you expect the strength of a hydrogen bond to be greater in a solvent of high dielectric constant like water or a solvent of lower dielectric constant like ethanol?
Answer:
Buffer and Noncovalent Interactions Problems, Page 3

The hydrogen bond is electrostatic in nature and should therefore be stronger in a solvent of low dielectric constant. Coulomb’s Law states that the strength of electrostatic interactions varies inversely with the dielectric constant.
Buffer and Noncovalent Interactions Problems, Page 4