S-14
1. Solubility of Ethanol in Water Explain why ethanol (CH
3
CH
2
OH) is more soluble in water than is
ethane (CH
3
CH
3
).
2. Calculation of pH from Hydrogen Ion Concentration What is the pH of a solution that has an
H
concentration of (a) 1.75 10
5
mol/L; (b) 6.50 10
10
mol/L; (c) 1.0 10
4
mol/L;
(d) 1.50 10
5
mol/L?
3. Calculation of Hydrogen Ion Concentration from pH What is the H
concentration of a solution
with pH of (a) 3.82; (b) 6.52; (c) 11.11?
4. Acidity of Gastric HCl In a hospital laboratory, a 10.0 mL sample of gastric juice, obtained several
hours after a meal, was titrated with 0.1
M
NaOH to neutrality; 7.2 mL of NaOH was required. The pa-
tient’s stomach contained no ingested food or drink; thus assume that no buffers were present. What
was the pH of the gastric juice?
Answer Multiplying volume (L) by molar concentration (mol/L) gives the number of moles in
5. Calculation of the pH of a Strong Acid or Base (a) Write out the acid dissociation reaction for
hydrochloric acid. (b) Calculate the pH of a solution of 5.0 10
4
M
HCl. (c) Write out the acid
dissociation reaction for sodium hydroxide. (d) Calculate the pH of a solution of 7.0 10
5
M
NaOH.
Water
chapter 2
c02Water.qxd 12/6/12 4:12 PM Page S-14
Chapter 2 Water S-15
Answer
6. Calculation of pH from Concentration of Strong Acid Calculate the pH of a solution prepared
by diluting 3.0 mL of 2.5
M
HCl to a final volume of 100 mL with H
2
O.
7. Measurement of Acetylcholine Levels by pH Changes The concentration of acetylcholine (a
neurotransmitter) in a sample can be determined from the pH changes that accompany its hydrolysis.
When the sample is incubated with the enzyme acetylcholinesterase, acetylcholine is converted to
choline and acetic acid, which dissociates to yield acetate and a hydrogen ion:
In a typical analysis, 15 mL of an aqueous solution containing an unknown amount of acetylcholine had
a pH of 7.65. When incubated with acetylcholinesterase, the pH of the solution decreased to 6.87. As-
suming there was no buffer in the assay mixture, determine the number of moles of acetylcholine in
the 15 mL sample.
8. Physical Meaning of pK
a
Which of the following aqueous solutions has the lowest pH: 0.1
M
HCl; 0.1
M
acetic acid (pK
a
4.86); 0.1
M
formic acid (pK
a
3.75)?
9. Meanings of K
a
and pK
a
(a) Does a strong acid have a greater or lesser tendency to lose its proton
than a weak acid? (b) Does the strong acid have a higher or lower K
a
than the weak acid? (c) Does
the strong acid have a higher or lower pK
a
than the weak acid?
Acetylcholine
O
Choline Acetate
NH
C
O
CH
3
HO
CH
3
CH
3
CH
3
CH
2
CH
2
NH
2
O
CH
3
CH
2
CO
O
CH
2
CH
3
CH
3
CH
3
c02Water.qxd 12/6/12 4:12 PM Page S-15
S-16 Chapter 2 Water
10. Simulated Vinegar One way to make vinegar (not the preferred way) is to prepare a solution of
acetic acid, the sole acid component of vinegar, at the proper pH (see Fig. 2–15) and add appropriate
flavoring agents. Acetic acid (M
r
60) is a liquid at 25 C, with a density of 1.049 g/mL. Calculate the
volume that must be added to distilled water to make 1 L of simulated vinegar (see Fig. 2–16).
Answer From Figure 2–16, the pK
a
of acetic acid is 4.76. From Figure 2–15, the pH of vinegar
is ~3; we will calculate for a solution of pH 3.0. Using the Henderson-Hasselbalch equation
11. Identifying the Conjugate Base Which is the conjugate base in each of the pairs below?
(a) RCOOH, RCOO
(b) RNH
2
, RNH
3
(c) H
2
PO
4
, H
3
PO
4
(d) H
2
CO
3
, HCO
3
12. Calculation of the pH of a Mixture of a Weak Acid and Its Conjugate Base Calculate the pH
of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pK
a
4.76) of
(a) 2:1; (b) 1:3; (c) 5:1; (d) 1:1; (e) 1:10.
Answer Using the Henderson-Hasselbalch equation,
pH pK
a
log
[A
]
[HA]
13. Effect of pH on Solubility The strongly polar hydrogen-bonding properties of water make it an
excellent solvent for ionic (charged) species. By contrast, nonionized, nonpolar organic molecules, such
as benzene, are relatively insoluble in water. In principle, the aqueous solubility of any organic acid or
base can be increased by converting the molecules to charged species. For example, the solubility of
benzoic acid in water is low. The addition of sodium bicarbonate to a mixture of water and benzoic acid
raises the pH and deprotonates the benzoic acid to form benzoate ion, which is quite soluble in water.
Are the following compounds more soluble in an aqueous solution of 0.1
M
NaOH or 0.1
M
HCl? (The dis-
sociable proton are shown in bold.)
Answer
14. Treatment of Poison Ivy Rash The components of poison ivy and poison oak that produce the
characteristic itchy rash are catechols substituted with long-chain alkyl groups.
If you were exposed to poison ivy, which of the treatments below would you apply to the affected area?
Justify your choice.
(a) Wash the area with cold water.
(b) Wash the area with dilute vinegar or lemon juice.
C
O
OHC O⫺
Benzoic acid Benzoate ion
pK
a
≈5
O
N⫹
H
Pyridine ion
pK
a
≈5
(b) (c)
(a)
-Naphthol
pK
a
≈10
C
HN
H
C
H
C
O O CH
3
CH
2
N-Acetyltyrosine methyl ester
pK
a
≈10
CH
3
OH
O
O
OH
(CH
2
)
n
CH
3
pK
a
≈8
OH
Chapter 2 Water S-17
c02Water.qxd 12/6/12 4:12 PM Page S-17
S-18 Chapter 2 Water
(c) Wash the area with soap and water.
(d) Wash the area with soap, water, and baking soda (sodium bicarbonate).
15. pH and Drug Absorption Aspirin is a weak acid with a pK
a
of 3.5 (the ionizable H is shown in bold):
It is absorbed into the blood through the cells lining the stomach and the small intestine. Absorption
requires passage through the plasma membrane, the rate of which is determined by the polarity of the
molecule: charged and highly polar molecules pass slowly, whereas neutral hydrophobic ones pass
rapidly. The pH of the stomach contents is about 1.5, and the pH of the contents of the small intestine
is about 6. Is more aspirin absorbed into the bloodstream from the stomach or from the small
intestine? Clearly justify your choice.
16. Calculation of pH from Molar Concentrations What is the pH of a solution containing 0.12 mol/L
of NH
4
Cl and 0.03 mol/L of NaOH (pK
a
of NH
4
/NH
3
is 9.25)?
17. Calculation of pH after Titration of Weak Acid A compound has a pK
a
of 7.4. To 100 mL of a 1.0
M
solution of this compound at pH 8.0 is added 30 mL of 1.0
M
hydrochloric acid. What is the pH of the
resulting solution?
Answer Begin by calculating the ratio of conjugate base to acid in the starting solution, using
the Henderson-Hasselbalch equation:
C
O
O
C
O
OH
CH
3
18. Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer in
biochemical experiments. The amino group of glycine, which has a pK
a
of 9.6, can exist either in the
protonated form (ONH
3
) or as the free base (ONH
2
), because of the reversible equilibrium
RONH
3
RONH
2
H
(a) In what pH range can glycine be used as an effective buffer due to its amino group?
(b) In a 0.1
M
solution of glycine at pH 9.0, what fraction of glycine has its amino group in the
ONH
3
form?
(c) How much 5
M
KOH must be added to 1.0 L of 0.1
M
glycine at pH 9.0 to bring its pH to
exactly 10.0?
(d) When 99% of the glycine is in its ONH
3
form, what is the numerical relation between the pH of
the solution and the pK
a
of the amino group?
Answer
z
y
c02Water.qxd 12/6/12 4:12 PM Page S-19
S-20 Chapter 2 Water
19. Calculation of the pK
a
of an Ionizable Group by Titration The pK
a
values of a compound with
two ionizable groups are pK
1
4.10 and pK
2
between 7 and 10. A biochemist has 10 mL of a 1.0
M
solution of this compound at a pH of 8.00. She adds 10.0 mL of 1.00
M
HCl, which changes the PH to
3.20. What is pK
2
?
20. Calculation of the pH of a Solution of a Polyprotic Acid Histidine has ionizable groups with pK
a
values of 1.8, 6.0, and 9.2, as shown below (His imidazole group). A biochemist makes up 100 mL of
a 0.100
M
solution of histidine at a pH of 5.40. She then adds 40 mL of 0.10
M
HCl. What is the pH of
the resulting solution?
Chapter 2 Water S-21
H3NCH
COOH
CH2
C
H
N
CH
C
HN
H
1.8
pK1
6.0
pK2
9.2
pK3
H3N
CH
COO
CH2
C
H
N
CH
C
HN
H
H3N
CH
COO
CH2
C
H
N
CH
C
HN
H2NCH
COO
CH2
C
H
N
CH
C
HN
COOH COO
HisHHis
NH3NH2
Ionizable
group
Answer The initial pH of 5.40 is so far below pK
(that of the amino group of histidine), that
21. Calculation of the Orginal pH from the Final pH after Titration A biochemist has 100 mL of
a .10
M
solution of a weak acid with a pK
a
of 6.3. She adds 6.0 mL of 1.0
M
HCl, which changes the pH
to 5.7. What was the pH of the original solution?
Answer First calculate the ratio of acid to conjugate base in the final solution:
c02Water.qxd 12/7/12 8:19 PM Page S-21
S-22 Chapter 2 Water
22. Preparation of a Phosphate Buffer What molar ratio of HPO
4
2
to H
2
PO
4
in solution would
produce a pH of 7.0? Phosphoric acid (H
3
PO
4
), a triprotic acid, has 3 pK
a
values: 2.14, 6.86, and
12.4. Hint: Only one of the pK
a
values is relevant here.
23. Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in com-
mercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. To convert
these responses to a pH reading, the electrode must be calibrated against standard solutions of known
H
concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH
2
PO
4
ⴢH
2
O;
FW 138) and disodium hydrogen phosphate (Na
2
HPO
4
; FW 142) needed to prepare 1 L of a standard
buffer at pH 7.00 with a total phosphate concentration of 0.100
M
(see Fig. 2–16). See Problem 22 for
the pK
a
values of phosphoric acid.
24. Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a
pK
a
of 6.0, calculate the ratio of conjugate base to acid at a pH of 5.0.
Answer Using the Henderson-Hasselbalch equation,
25. Preparation of Buffer of Known pH and Strength Given 0.10
M
solutions of acetic acid (pK
a
4.76) and sodium acetate, describe how you would go about preparing 1.0 L of 0.10
M
acetate buffer
of pH 4.00.
Answer Use the Henderson-Hasselbalch equation to calculate the ratio [Ac
]/[HAc] in the
final buffer.
26. Choice of Weak Acid for a Buffer Which of these compounds would be the best buffer at pH 5.0:
formic acid (pK
a
3.8), acetic acid (pK
a
4.76), or ethylamine (pK
a
9.0)? Briefly justify your answer.
27. Working with Buffers A buffer contains 0.010 mol of lactic acid (pK
a
3.86) and 0.050 mol of
sodium lactate per liter. (a) Calculate the pH of the buffer. (b) Calculate the change in pH when 5 mL
of 0.5
M
HCl is added to 1 L of the buffer. (c) What pH change would you expect if you added the
same quantity of HCl to 1 L of pure water?
Chapter 2 Water S-23
c02Water.qxd 12/6/12 4:12 PM Page S-23
S-24 Chapter 2 Water
Answer Using the Henderson-Hasselbalch equation,
28. Use of Molar Concentrations to Calculate pH What is the pH of a solution that contains 0.20
M
sodium acetate and 0.60
M
acetic acid (pK
a
4.76)?
29. Preparation of an Acetate Buffer Calculate the concentrations of acetic acid (pK
a
4.76) and
sodium acetate necessary to prepare a 0.2
M
buffer solution at pH 5.0.
30. pH of Insect Defensive Secretion You have been observing an insect that defends itself from ene-
mies by secreting a caustic liquid. Analysis of the liquid shows it to have a total concentration of for-
mate plus formic acid (K
a
1.8 10
4
) of 1.45
M
; the concentration of formate ion is 0.015
M
. What
is the pH of the secretion?
31. Calculation of pK
a
An unknown compound, X, is thought to have a carboxyl group with a pK
a
of 2.0
and another ionizable group with a pK
a
between 5 and 8. When 75 mL of 0.1
M
NaOH is added
to 100 mL of a 0.1
M
solution of X at pH 2.0, the pH increases to 6.72. Calculate the pK
a
of the second
ionizable group of X.
32. Ionic Forms of Alanine Alanine is a diprotic acid that can undergo two dissociation reactions (see
Table 3–1 for pK
a
values). (a) Given the structure of the partially protonated form (or zwitterion; see
Fig. 3–9) below, draw the chemical structures of the other two forms of alanine that predominate in
aqueous solution: the fully protonated form and the fully deprotonated form.
Of the three possible forms of alanine, which would be present at the highest concentration in solutions
of the following pH: (b) 1.0; (c) 6.2; (d) 8.02; (e) 11.9. Explain your answers in terms of pH relative to
the two pK
a
values.
Answer
Alanine
H
3
NCH
COO
CH
3
Chapter 2 Water S-25
c02Water.qxd 12/6/12 4:12 PM Page S-25
S-26 Chapter 2 Water
33. Control of Blood pH by Respiratory Rate
(a) The partial pressure of CO
2
in the lungs can be varied rapidly by the rate and depth of breathing.
For example, a common remedy to alleviate hiccups is to increase the concentration of CO
2
in
the lungs. This can be achieved by holding one’s breath, by very slow and shallow breathing
(hypoventilation), or by breathing in and out of a paper bag. Under such conditions, pCO
2
in the
air space of the lungs rises above normal. Qualitatively explain the effect of these procedures on
the blood pH.
(b) A common practice of competitive short-distance runners is to breathe rapidly and deeply
(hyperventilate) for about half a minute to remove CO
2
from their lungs just before the race
begins. Blood pH may rise to 7.60. Explain why the blood pH increases.
(c) During a short-distance run, the muscles produce a large amount of lactic acid
(CH
3
CH(OH)COOH, K
a
1.38 10
4
M
) from their glucose stores. In view of this fact, why
might hyperventilation before a dash be useful?
Answer
34. Calculation of Blood pH from CO
2
and Bicarbonate Levels Calculate the pH of a blood plasma
sample with a total CO
2
concentration of 26.9 m
M
and bicarbonate concentration of 25.6 m
M
. Recall
from page 67 that the relevant pK
a
of carbonic acid is 6.1.
Answer Use the Henderson-Hasselbalch equation:
35. Effect of Holding One’s Breath on Blood pH The pH of the extracellular fluid is buffered by the
bicarbonate/carbonic acid system. Holding your breath can increase the concentration of CO
2
(g) in the
blood. What effect might this have on the pH of the extracellular fluid? Explain by showing the rele-
vant equilibrium equation(s) for this buffer system.
Data Analysis Problem
36. “Switchable” Surfactants Hydrophobic molecules do not dissolve well in water. Given that water is
a very commonly used solvent, this makes certain processes very difficult: washing oily food residue
off dishes, cleaning up spilled oil, keeping the oil and water phases of salad dressings well mixed, and
carrying out chemical reactions that involve both hydrophobic and hydrophilic components.
Surfactants are a class of amphipathic compounds that includes soaps, detergents, and emulsifiers.
With the use of surfactants, hydrophobic compounds can be suspended in aqueous solution by forming
micelles (see Fig. 2–7). A micelle has a hydrophobic core consisting of the hydrophobic compound and
the hydrophobic “tails” of the surfactant; the hydrophilic “heads” of the surfactant cover the surface of
the micelle. A suspension of micelles is called an emulsion. The more hydrophilic the head group of the
surfactant, the more powerful it is—that is, the greater its capacity to emulsify hydrophobic material.
When you use soap to remove grease from dirty dishes, the soap forms an emulsion with the grease
that is easily removed by water through interaction with the hydrophilic head of the soap molecules.
Likewise, a detergent can be used to emulsify spilled oil for removal by water. And emulsifiers in com-
mercial salad dressings keep the oil suspended evenly throughout the water-based mixture.
There are some situations in which it would be very useful to have a “switchable” surfactant: a mol-
ecule that could be reversibly converted between a surfactant and a nonsurfactant.
(a) Imagine such a “switchable” surfactant existed. How would you use it to clean up and then
recover the oil from an oil spill?
Liu et al. describe a prototypical switchable surfactant in their 2006 article “Switchable
Surfactants.” The switching is based on the following reaction:
(b) Given that the pK
a
of a typical amidinium ion is 12.4, in which direction (left or right) would you
expect the equilibrium of the above reaction to lie? (See Fig. 2–17 for relevant pK
a
values.) Jus-
tify your answer. Hint: Remember the reaction H
2
O CO
2
H
2
CO
3
.
Liu and colleagues produced a switchable surfactant for which R C
16
H
33
. They do not name the
molecule in their article; for brevity, we’ll call it s-surf.
(c) The amidinium form of s-surf is a powerful surfactant; the amidine form is not. Explain this ob-
servation.
Liu and colleagues found that they could switch between the two forms of s-surf by changing the
gas that they bubbled through a solution of the surfactant. They demonstrated this switch by measur-
ing the electrical conductivity of the s-surf solution; aqueous solutions of ionic compounds have
higher conductivity than solutions of nonionic compounds. They started with a solution of the ami-
dine form of s-surf in water. Their results are shown below; dotted lines indicate the switch from one
gas to another.
z
y
Amidine form Amidinium form
H
2
O
CO
2
⫹⫹ HCO
3
⫹
NCH
3
CH
3
N
H
R
EN
CH
3
CC
A
ⴙNCH
3
CH
3
N
H
H
R
E
E
CH
3
A
⫺
S-28 Chapter 2 Water
(d) In which form is the majority of s-surf at point A? At point B?
(e) Why does the electrical conductivity rise from time 0 to point A?
(f) Why does the electrical conductivity fall from point A to point B?
(g) Explain how you would use s-surf to clean up and recover the oil from an oil spill.
Answer
Reference
A
B
Time (min)
Gas bubbled in: CO2Ar CO2Ar
Electrical
conductivity
0
0100 200