CH 306 Chapter 2

Question 1

All weak interactions are said to be fundamentally electrostatic interactions. Explain.

Answer 1

Ionic bonds, hydrogen bonds, and van der Waals interactions all depend on the unequal distribution of electrons, resulting in an unequal distribution of charge.

Question 2

Explain how the following statement applies to

biochemistry: Order can be generated by an increase in randomness.

Answer 2

The statement essentially describes the hydrophobic effect. Specific complicated biochemical structures can form, powered by the increase in entropy that results when hydrophobic groups are removed from aqueous solution.

Question 3

Define Brownian motion.

Answer 3

Brownian motion is the random movement of molecules in a fluid or gas powered by the background thermal energy.

Question 4

Water is said to be polar but uncharged.

How is it possible?

Answer 4

Water is polar in that the hydrogen atoms bear a partial positive charge, whereas the oxygen atom has a partial negative charge owing to the electronegative nature of the oxygen atom. However, the total charge on the molecule is zero; that is, the positive charges are equal to the negative charges.

Question 5

Why are weak bonds important in

biochemistry?

Answer 5

Many weak bonds allow for highly specific yet transient interactions.

Question 6

What are the common types of weak bonds important in biochemistry? How does water affect these bonds?

Answer 6

Ionic bonds, hydrogen bonds, and van der Waals interactions.
Water disrupts ionic bonds and hydrogen bonds.
Because van der Waals interactions are most common between hydrophobic groups, water can be said to strengthen these bonds by facilitating their formation through the hydrophobic effect.

Question 7

In liquid water, each molecule is hydrogen bonded to approximately 3.4 molecules of water. What effect would freezing water have on the number of hydrogen bonds? Heating water?

Answer 7

Lowering the temperature would reduce the motion of the water molecules and allow the formation of more hydrogen bonds, which
is indeed the case: each molecule of water in ice is hydrogen bonded to approximately 3.7 molecules of water. The opposite takes place as the water is heated, and fewer hydrogen bonds would be expected to form. At 100°C, a molecule of water is hydrogen bonded to 3.2 water molecules.

Question 8

What would be the effect of an organic solvent on electrostatic interactions?

Answer 8

Electrostatic interactions would be stronger in an organic solvent relative to a polar solvent because there would be no competition from the solvent for the components of the electrostatic interaction.

Question 9

What is an electronegative atom, and why are such atoms important in biochemistry?

Answer 9

An electronegative atom is one that has a high affinity for electrons. Consequently, when bonded to a hydrogen atom, the electronegative atom assumes a partial negative charge and the hydrogen atom assumes a partial positive charge. Such polarity allows the formation of hydrogen bonds.

Question 10

Define the hydrophobic effect.

Answer 10

The hydrophobic effect is the tendency of nonpolar molecules to interact with one another in the presence of water. The interaction is powered by the increase in entropy of water molecules when the nonpolar molecules are removed from the watery environment.

Question 11

How does the Second Law of Thermodynamics allow for the formation of biochemical order?

Answer 11

The Second Law of Thermodynamics states that the entropy of a system and its surroundings always increases in a spontaneous process. When hydrophobic molecules are sequestered away from water, the entropy of the water increases. Such sequestration, called the hydrophobic effect, also leads to the formation of biochemical structures.

Question 12

If an aqueous solution has a hydrogen ion concentration of 10^-5 M, what is the concentration of hydroxyl ion?

Answer 12

10^-9 M

Question 13

If an aqueous solution has a hydroxyl ion concentration of 10^-2 M, what is the concentration of hydrogen ion?

Answer 13

10^-12 M

Question 14

Using the Henderson–Hasselbalch equation, show that, for a weak acid, the pKa is the pH at which the concentration of the acid equals the concentration of the conjugate base.

Answer 14

The Henderson–Hasselbalch equation is pH = pKa + log[A−]/[HA]. If [A−] = [HA], then the equation becomes
pH = pKa + log 1. But log 1 = 0. Thus, pH = pKa under the conditions stated.

Question 15

What is the relation between the pKa of an acid and the strength of the acid?

Answer 15

The lower the pKa, the greater the Ka. The greater the Ka, the stronger the acid.

Question 16

The pKa of acetic acid is 4.76 and the pKa of trichloroacetic acid, which is used to remove warts, is 0.7. Calculate the dissociation constant of each acid. Which is the stronger acid?

Answer 16

Recall that pKa = log 1/Ka. The antilog of 4.76, the pKa of acetic acid, is 57,543, and the Ka is the inverse. Thus, the Ka for acetic
acid = 1.7 × 10−5.
The pKa for trichloroacetic acid is 0.7. Calculating as above, the Ka is 2.0 × 10−1.
Trichloroacetic acid is a stronger acid; that is, it is more completely dissociated.

Question 17

Many important biochemicals are organic acids, such as pyruvic acid (pKa = 2.50) and lactic acid (pKa = 3.86). The conjugate bases are pyruvate and lactate, respectively. Determine, for each acid, which form—the acid or the conjugate base— predominates at pH 7.4.

Answer 17

As with many problems in life, this one can be solved with the Henderson–Hasselbalch equation:

(LOOK IN BOOK FOR ANSWER)

Thus, these organic acids, like most organic acids that we will encounter in our study of biochemistry, are extensively ionized at physiological pH.

Question 18

For an acid HA, the concentrations of HA and A− are 0.075 and 0.025, respectively, at pH 6.0. What is the pKa value for HA?

Answer 18

6.48

Question 19

A dye that is an acid and that appears as different colors in its protonated and deprotonated forms can be used as a pH indicator. Suppose that you have a 0.001 M solution of a dye with a pKa of 7.2. From the color, the concentration of the protonated form is found to be 0.0002 M. Assume that the remainder of the dye is in the deprotonated form. What is the pH of the solution?

Answer 19

7.8

Question 20

An acid with a pKa of 8.0 is present in a solution with a pH of 6.0. What is the ratio of the protonated to the deprotonated form of the acid?

Answer 20

100

Question 21

Two solutions of sodium acetate are prepared, one having a concentration of 0.1 M and the other having a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: 0.0025 M, 0.005 M, 0.01 M, and 0.05 M.

Answer 21

Initial acetate (M)
0.1   0.1   0.1   0.1   0.01   0.01   0.01   0.01

Initial HCl (M)
0.0025 0.005 0.01 0.05 0.0025 0.005 0.01 0.05

pH
6.3 6.0 5.7 4.8 5.2 4.8 3.4 1.4

Question 22

Calculate the concentration of acetic acid and acetate ion in a 0.2 M acetate buffer at pH 5. The pKa of acetic acid is 4.76.

Answer 22

Acetate ion = 0.128 M

acetic acid = 0.072 M

Question 23

You are working in a high-powered clinical biochemistry lab. The chief scientist rushes in and announces, “I need 500 ml of 0.2 M acetate, pH 5.0. STAT! Who is the best and brightest in this room?” All eyes turn toward you. You have solid anhydrous sodium acetate (MW = 82 g mol−1) and a solution of 1 M acetic acid. De- scribe how you would make the buffer.

Answer 23

LOOK IN BOOK FOR ANSWER

Question 24

Following a bout of intense exercise, the pH of the exerciser’s blood was found to be 7.1. If the HCO3− concentration is 8 mM, and the pKa for HCO3− is 6.1, what is the concentration of CO2 in the blood?

Answer 24

LOOK IN BOOK FOR ANSWER

Question 25

As we will see later in the course, pathological conditions arise where the blood pH falls because of excess acid production, a condition called metabolic acidosis. Excess protons in the blood decrease the amount of HCO3− and thus reduce the buffering capacity of blood. A rapid drop in pH could lead to death. Normal values for blood are: pH = 7.4, [HCO3−] = 24.0 mM, [CO2] = 1.20 mM.

(a) If a patient has a blood pH = 7.03 and [CO2] =
1. 2 mM, what is the [HCO3−] in the patient’s blood? The pKa of HCO3− = 6.1.

(b) Suggest a possible treatment for metabolic acidosis.
(c) Why might the suggestion for part (b) be of benefit to middle-distance runners?

Answer 25

(a) LOOK IN BOOK FOR ANSWER
(b) Intravenous administration of sodium bicarbonate is a common treatment.
(c) Middle-distance running involves aspects of endurance running as well as sprinting. Some studies have shown that drinking a sodium bicarbonate solution prior to the run mitigates a drop in pH and improves performance.