Chapter 1
Page 4
The periodic table in Figure 1-2 is drawn incorrectly; however, the identities of the shaded
elements are correct.
Page 5
The diphosphoric acid ester at the bottom of Table 1-1 should not be labeled
“(Phosphodiester)”. In the functional group column, the second structure from the bottom is
mislabeled as “(phosphodiester linkage)”. It should be labeled “(phosphoanhydride linkage)”.
Chapter 2
Page 36
The first line of text states that Kw, the ionization constant of water, is 1014 at 25ºC. The correct
value of Kw is 10–14.
Page 39
Sample Calculation 2-3 uses too few significant figures. At least two significant figures should
be used.
Pages 40–41
Sample Calculation 2-4
In this calculation, it is assumed that the amount of boric acid that is added (1.9 mL) does not
significantly dilute the solution (600 mL). That’s true in this case, but not in others (see
Problems 40 and 43 at the end of Chapter 2).
Page 666
The Solution to Problem 2-34 is printed incorrectly. In the correct version shown here,
the acidic protons are circled.
Page 49
Problem 40
The solution to this problem on page 667 is incorrect. A revised problem and the correct
solution are provided here:
Problem 40. What is the volume (in mL) of 6 M acetic acid that would have to be added to 500
mL of a solution of 0.20 M sodium acetate in order to achieve a pH of 5.0?
Solution 40. The concentration of acetate is 0.20 M, since sodium acetate is a salt that
dissociates completely in solution and the mole ratio of sodium acetate to acetate is 1:1.
Substitute the known quantities into the Henderson–Hasselbalch equation: the concentration of
acetate (A–), the desired pH, and the pK of acetic acid (HA; from Table 2-2 on page 40).
pH = pK + log([A–]/[HA])
5.0 = 4.76 + log((0.20 M)/[HA])
0.24 = log((0.20 M)/[HA])
1.74 = (0.20 M)/[HA]
[HA] = 0.115 M
Next, calculate the volume (V) of stock 6.0 acetic acid that must be added to the solution to
achieve a final concentration of 0.115 M.
V1M1 = V2M2
V1(6.0 M) = (500 mL)(0.115 M)
V1 = 9.6 mL
The addition of 9.6 mL to a 500 mL solution dilutes the solution by 2%, which doesn’t
introduce significant error. An alternative way to solve the problem would be to convert the
concentrations in molarities to moles.
Page 49
Problem 42
There is an error in the statement of the original problem. In parts (c) and (d), hydrochloric acid
(not sodium hydroxide) must be added to the solution of the sodium salt of HEPES in order to
achieve the desired pH. The correct version of the problem and its solution are provided here:
Problem 42. An experiment requires the buffer HEPES, pH = 8.0 (see Table 2-2 on page 40).
(a) Write the equation for the dissociation of HEPES in water. Identify the weak acid and the
conjugate base.
(b) What is the effective buffer range for HEPES?
(c) The buffer will be prepared by making 1.0 L of a 0.10 M solution of HEPES. Hydrochloric
acid will be added until the desired pH is achieved. Describe how you will make 1.0 L of 0.10
M HEPES. (HEPES is supplied by the chemical company as a sodium salt with a molecular
weight of 260.3 g/mol.)
(d) What is the volume (in mL) of a stock solution of 6.0 M HCl that must be added to the 0.1
M HEPES to achieve the desired pH of 8.0? Describe how you will make the buffer.
Solution 42
(a) The correct structures are shown on page 667.
(b) The pK for HEPES is 7.55; therefore its effective buffering range is 6.55–8.55. In
practice, note that the buffering range might be smaller than given by the pH = pK ± 1
rule. On their web site, Sigma-Aldrich lists the effective buffering range of HEPES as
6.8–8.2.
(c) 1.0 L × 0.10 mol/L × 260.3 g/mol = 26 g
Weigh 26 g of the HEPES salt and add to a beaker. Dissolve in slightly less than 1.0 liter
of water (leave “room” for the HCl solution that will be added in the next step).
(d) Adding HCl to the solution will convert some of the conjugate base (A–) to the weak
acid form of HEPES. The number of moles of H+ added to the solution will be equal to
the number of moles of weak acid produced ([H+] = [HA]). The final ratio of A– to HA
can be calculated:
pH = pK + log([A–]/[HA])
8.0 = 7.55 + log([A–]/[HA])
0.45 = log([A–]/[HA])
100.45 = [A–]/[HA]
2.82 = [A–]/[HA]
Since the final concentration of the buffer is 0.10 M, [HA] + [A–] = 0.10 M. We can use
this information to figure out the absolute concentrations of the weak acid and the
conjugate base, given that the ratio of conjugate base to weak acid is 2.82.
Adding HCl to the HEPES conjugate base will convert some of the conjugate base to the
weak acid. So the original concentration of 0.10 M for the conjugate base will decrease
and the concentration of the HEPES weak acid will increase by the same amount, x.
2.82 = [A–]/[HA]
2.82 = (0.10 M – x)/x
2.82x = 0.10 M – x
3.82x = 0.10 M
x = 0.026 M
[A–] = 0.10 M – x = 0.074 M
[HA] = [H+] = x = 0.026 M
Next, calculate the volume of stock 6.0 M HCl that must be added to the solution to
achieve a final concentration of 0.026 M:
V1M1 = V2M2
V1(6.0 M) = (1000 mL)(0.026 M)
V1 = 4.3 mL
Add 4.3 mL of 6.0 M HCl to the 1.0 liter of 0.1 M HEPES prepared. Verify the pH with
a pH meter and adjust if necessary. Dilute to a final volume of 1.0 liter.
Page 49
Problem 43
The solution to this problem on page 668 does not take into account the dilution of the
buffer. Although the dilution of the buffer is small and does not affect the answer
significantly, it is easy enough to include a dilution factor in the calculations. The
problem and its solution—incorporating the dilution factors—are provided here:
Problem 43. One (1.0) liter of a 0.1 M Tris buffer (see Table 2-2) is prepared and
adjusted to a pH of 8.2.
(a) Draw the structures of the weak acid and the conjugate base forms of Tris.
(b) What is the effective buffering range of Tris?
(c) What are the concentrations of the conjugate acid and weak base at pH 8.2?
(d) What is the ratio of conjugate base to weak acid if 1.5 mL of 3.0 M HCl is added to
1.0 L of the buffer? What is the new pH? Has the buffer functioned effectively?
Compare the pH change to that of Problem 29a in which the same amount of acid was
added to the same volume of pure water.
(e) What is the ratio of conjugate base to weak acid if 1.5 mL of 3.0 M NaOH is added to
1.0 L of the buffer? What is the new pH? Has the buffer functioned effectively?
Compare the pH change to that of Problem 29b in which the same amount of base was
added to the same volume of pure water.
Solution 43
(a) The correct structures are shown on page 668.
(b) The pK of Tris is 8.30; therefore its effective buffering range is 7.30–9.30.
(c) Use the Henderson-Hasselbalch equation to determine the ratio of conjugate base to
weak acid:
pH = pK + log([A–]/[HA])
8.20 = 8.30 + log([A–]/[HA])
–0.10 = log([A–]/[HA])
0.79 = [A–]/[HA]
Since [A–] = 0.79[HA], and [A–] + [HA] = 0.1 M,
0.79[HA] = 0.1 M – [HA]
1.79[HA] = 0.1 M
[HA] = 0.056 M
[A–] = 0.1 – 0.056 M = 0.044 M
(d) First, determine the [H+] contributed by the HCl (which dissociates completely), so
[HCl] = [H+]:
V1M1 = V2M2
(1.5 mL)(3.0 M) = (1001.5 mL)M2
M2 = 0.0045 M
The acid will convert some of the conjugate base to weak acid. (Also, the addition of the
acid will dilute the buffer somewhat, so the 1000/1001.5 factor accounts for this dilution.)
Therefore the new concentrations are:
[A–] = (0.044 M)(1000/1001.5) – 0.0045 M = 0.0394 M
[HA] = (0.056 M)(1000/1001.5) + 0.0045 M = 0.0604 M
The new pH is determined by substituting the new concentrations of A– and HA into the
Henderson–Hasselbalch equation:
pH = pK + log([A–]/[HA])
pH = 8.30 + log(0.0394/0.0604)
pH = 8.30 + (–0.186)
pH = 8.11
The buffer has been effective—the pH has declined about 0.1 unit (from pH = 8.2 to pH
= 8.11) with the addition of the strong acid. In comparison, the addition of the same
amount of acid to water, which is not buffered, resulted in a pH change from
approximately 7.0 to 2.35 (see Problem 29a).
(e) First, determine the [OH–] contributed by the NaOH (which dissociates completely):
V1M1 = V2M2
(1.5 mL)(3.0 M) = (1001.5 mL)M2
M2 = 0.0045 M
The base will convert some of the weak acid to conjugate base. (Also, the addition of the
base will dilute the buffer somewhat, so the 1000/1001.5 factor accounts for this
dilution.) Therefore the new concentrations are:
[A–] = (0.044 M)(1000/1001.5) + 0.0045 M = 0.0484 M
[HA] = (0.056 M)(1000/1001.5) – 0.0045 M = 0.0514 M
The new pH is determined by substituting the new concentrations of A– and HA into the
Henderson–Hasselbalch equation:
pH = pK + log([A–]/[HA])
pH = 8.30 + log(0.0484/0.0514)
pH = 8.30 + (–0.026)
pH = 8.27
The buffer has been effective—the pH has increased only 0.07 unit (from pH = 8.2 to pH
= 8.27) with the addition of the strong base. In comparison, the addition of the same
amount of base to water, which is not buffered, resulted in a pH change from
approximately 7.0 to 11.6 (see Problem 29b).
Chapter 3
Page 57
There is an error in the drawing of the GC base pair so that the carbonyl carbon on the guanine
has a carbon with five bonds. The double bond in the ring should be eliminated. The correct
structure is shown below:
Page 671
The answer to Problem 5 should read as follows (correction in boldface):
5. The organism must also contain 19% A (since [A] = [T] according to Chargaff’s
rules) and 62% C + G (or 31% C and 31% G, since [C] = [G]). Each cell is a diploid,
containing 60,000 kb or 6 × 107 bases. Therefore,
[A] = [T] = (0.19)(6 × 107 bases) = 1.14 × 107 bases
[C] = [G] = (0.31)(6 × 107 bases) = 1.86 × 107 bases
Page 672
In the figure shown as part of the Solution to Problem 16, both of the DNA strands at the top
should be orange (both contain 15N).
Chapter 4
Page 110
In the section titled “Proteins Have Hydrophobic Cores,” the second sentence of the third
paragraph reads “This is because the formation of α helices and β sheets, which are
internally hydrogen-bonded, minimizes the hydrophobicity of the polar backbone
groups.” The sentence should state that these secondary structures minimize the
hydrophilicity of the backbone groups.
Page 131
Problem 44 is incorrect. The problem should read “Explain why hemoglobin from the
llama, a species native to the Andes of South America, has a higher oxygen affinity than
human hemoglobin.” The solution (page 677) should read “At high altitude, less oxygen
is available to bind to hemoglobin in the lungs. Llama hemoglobin has higher oxygen
affinity than human hemoglobin, so it can more easily bind oxygen in order to deliver it
to the tissues.”
Chapter 6
Page 180
Some of the labels in Figure 6-9 are misleading. The two tetrahedral intermediates shown
(the results of Steps 1 and 4) are not identical to the transition states of these reaction
steps, although they are believed to be very similar to the transition states.
Page 197
Problem 42(a) should read “The compound shown below also reacts to form pnitrophenolate. Draw the reaction mechanism for this reaction.”
Chapter 8
Page 262
The cone-shaped phospholipid on the right is drawn incorrectly. Carbon 2 of its glycerol
backbone should bear an H and an OH group.
Page 264
Problem 1 should read as follows (corrections in boldface):
1. Fatty acids are often referred to in a shorthand form which consists of two numbers
separated by a colon. The first number is the number of carbons; the second number
is the number of double bonds. Therefore, palmitate would be represented by the
shorthand 16:0. An unsaturated fatty acid is indicated by a Δ sign, with the positions
of the double bonds represented by superscripts (the carboxylate carbon is #1). For
example, oleate would be represented by the shorthand 18:1 Δ9 and linoleate as 18:2
Δ9,12. Draw the structures of the following fatty acids:
(a) Myristate (14:0)
(b) Palmitoleate (16:1 Δ9)
(c) Linolenate (18:3 Δ9,12,15)
Chapter 9
Page 285
Three lines from the end, the text states that NAD+ is strictly a one-electron carrier. In
fact, NAD+ is strictly a two-electron carrier.
Chapter 12
Page 712
In the solution for Problem 8, the number of electrons is 2, not 1. The correct equations
are:
!G °" = #nF !E °"
(
)
!G °" = # ( 2 ) 96, 485 J $ V#1 $ mol#1 (1.13 V)
!G °" = #218 kJ $ mol#1
Chapter 16
Page 512
In Figure 16-16, the green arrow from phosphorylase kinase to glycogen synthase should
be a red arrow, indicating an inactivating event. Thus, the legend should state that all but
two of the arrows in the diagram represent activation events.
Chapter 20
Page 618
In the explanation of the wobble hypothesis, the text reads “…the third anticodon
position and the 5´ anticodon position experience some flexibility or wobble…” Wobble
pairing occurs between the third codon position and the 5´ anticodon position, as shown
in the figure on the same page.