Ch 2 - Problems

Problems

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1. Identify the potential pot ential hydrogen bond donors and acceptors acce ptors in the following molecules: molecules:

Answer: (a) Donors: NH1, NH2 at C2, NH9; acceptors: N3, O at C6, N7. (b) Donors: NH1, NH2 at C4; acceptors: O at C2, N3.

group, OH group; acceptors: acce ptors: COO – group, OH group.

(c) Donors:

2. Occasionally, Occasionally, a C—H group can form a hydrogen h ydrogen bond. Why would such a group be more likely to be a

hydrogen bond donor group when the C is next to N? Answer:

A protonated (and the refore positively charged) nitrogen nitrogen would promote the separation separat ion of charge in the adjacent C—H bond so that the C would have a partial negative charge and the H would have a partial positive positive charge. cha rge. This This would make make the th e H more likely to be donated to a hydrogen bond acceptor acce ptor group. 3. Rank the th e water wate r solubility solubility of the following following compounds: compounds: (a) H3C—CH2 —O—CH3 (b)

(c)

(d) H3C—CH2 —CH3 (e)

Answer:

From most most soluble soluble (most polar) to least lea st soluble soluble (least polar): po lar): c, b, e, a, d. 4. Where would the following substances substances partition in water containing palmitic palmitic acid ac id micelles? micelles? (a) +

–

+

–

–

H3 N—CH2 —COO , (b) H3 N—(CH2)11 —COO , (c) H3C—(CH2)11 —COO

Answer:

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Problems


(a) Water; (b) water; (c) micelle. 5. Explain why water forms nearly spherical spherical droplets on the surface of a freshly waxed car. Why doe sn't

water bead on a clean windshield? Answer:

The waxed car is a hydrophobic surface. surface . To To minimiz minimizee its interaction interac tion with the hydrophobic molecules (wax), each water drop minimizes minimizes its its surface area are a by becomi bec oming ng a sphere (the geometrical shape shape with the lowest possible possible ratio of surface to volume). Water does not bead on o n glass, glass, because beca use the glass glass presents a hydrophilic hydrophilic surface with which the water molecules can interact. This allows allows the water to spread out. 6. Describe what happens when a dialysis dialysis bag containing pure water is suspended suspended in a beaker of seawater.

What would happen if the dialysis membrane were permeable to water but not solutes? Answer:

Water molecules move from inside the dialysis bag to the surrounding seawater by osmosis. Ions from the seawater diffuse into the dialysis dialysis bag. At equilibrium, equilibrium, the compositions compositions of the solutions solutions inside and outside out side the dialysis bag are identical. If the membrane were solute-impermeable, solute-impermeable, essentially all the water would leave the th e dialysis bag. bag. 7. Draw the structures of the t he conjugate bases of the t he following following acids:

Answer:



Problems


(a) pH 4, (b)

; pH 8,

pH 4,

; pH 11, NH 3.

; pH 8,

; pH 11,

.

9. Calculate the pH of a 200 mL solution of pure water to which has be en added adde d 50 mL of 1 mM HCl HCl.. Answer: +

–4

The increase in [H ] due to the addition a ddition of HCl is (50 mL) (l mM)/(250 mL) mL) = 0.2 mM = 2 × 10

M.

Because the [H+] of pure water, 10 –7 M, is relatively insignificant, the pH of the solution is equal to –log(2 × 10 –4) or 3.7. 10. Calculate the pH of a 1 L solution containing (a) 10 mL of 5 M NaOH, (b) 10 mL of 100 mM glycine glycine and

20 mL of 5 M HCl, and (c) 10 mL of 2 M acet ic acid and 5 g of sodium acetate acetat e (formula weight weight 82 g · –1

mol ). Answer: (a) (0.010 L)(5 mol · L –1 NaOH)/(1 L) = 0.05 M NaOH = 0.05 M OH – +

–

–14

[H ] = K w/[OH ] = (10

–13

)/(0.05) = 2 × 10

M

pH = –log [H+] = –log (2 × 10 –13) = 12.7 (b) (0.020 L)(5 mol · L –1 HCl)/(1 L) = 0.1 M HCl ≡ 0.1 M H+

Since the contribution of 0.01 L × 100 mM/(1 L) = 1 mM glycine glycine is insig insignifi nificant cant in the presenc e of 0.1 M HCl, pH = –log [H+] = –log (0.1) = 1.0 (c) pH = pK + log ([acetate]/[acetic acid])

[acetate] = (5 g)(1 mol/82 g)/(1 L) = 0.061 M –1

[acetic acid] = (0.01 L)(2 mol · L )/(1 L) = 0.02 M pH = 4.76 + log (0.061/0.02) = 4.76 + 0.48 = 5.24 11. Calculate the standard free energy change for the dissociation of HEPES. Answer:

The standard free energy change can be calculated using Eq. 1-16 and the value of K from Table 2-4.



Problems


The concentration of A – is (50 mL)(2.0 M)/(200 mL) = 0.5 M, and the concentration of HA is (25 mL)(2.0 M)/(200 mL) = 0.25 M. Substitute these values into the Henderson–Hasselbalch equation (Eq. 2-9):

– 13. What is the pK of the weak acid ac id HA if if a solution containi conta ining ng 0.1 M HA and 0.2 M A has a pH of 6.5?

Answer:

Use Use the Henderson–Hasselbalch Henderson–Hasselbalch equation equa tion (Eq. 2-9) and solve for pK :

14. How many grams of sodium succinate (formula weight 140 g · mol –1) and disodium disodium succinate (formula (formula –1

weight 162 g · mol ) must be added to 1 L of water to produce a solution with pH 6.0 and a total solute concentration of 50 mM? Answer:

Let HA = sodium succinate and A – = disodium succinate. –

–

[A ] + [HA] = 0.05 M, so [A ] = 0.05 M – [HA] From Eq. 2-9 and Table 2-4, log ([A]/[HA]) = pH – pK = 6.0 – 5.64 = 0.36 –

[A ]/[HA] = antilog 0.36 = 2.29 (0.050 M – [HA])/[HA] = 2.29 [HA] = 0.015 M –

[A ] = 0.050 M – 0.015 M = 0.035 M –1

–1

grams of sodium sodium succinate = (0.015 mol · L )(140 g · mol ) × (1 L) = 2.1 g grams of disodium succinate = (0.035 mol · L –1)(162 g · mol –1) × (1 L) = 5.7 g



Problems


form (Fig. 2-18). Therefore, the concentration of OH – required is equivalent to the –1

concentration of the acid: (0.100 mol · L phosphoric acid)(0.1 L) = 0.01 mol NaOH NaOH required = (0.01 mol)(1 mol)(1 L/5 mol NaOH) NaOH) = 0.002 0.0 02 L = 2 mL. 16. (a) Would phosphoric acid or succinic acid be a better buffer at pH 5? (b) Would ammonia or piperidine be a better buffer at pH 9? (c) Would HEPES or Tris Tris be a better be tter buffer at pH 7.5? 7 .5? Answer: (a) Succinic acid; a cid; (b) ammonia; (c) HEPES. 17. You need nee d a buffer at pH 7.5 for use in purifying purifying a protein at 4°C. You You have chosen Tris, Tris, pK 8.08, Δ H ° =

50 kJ · mol –1. You You carefull care fully y make up 0.01 M Tris buffer, buffer, pH 7.5 at 25°C, and store it in the cold c old box to equilibrate it to the temperature of the purification. When you measure the pH of the temperatureequilibrated buffer it has increased to 8.1. What is the explanation for this increase? How can you avoid this problem? Answer:

The dissociation of TrisH+ to its basic form and H+ is associated associated with a large, positive positive enthalpy e nthalpy change. Consequently, heat is taken up by the reactant on dissociation. When the temperature is lowered, there is less heat ava ilable ilable for this process, shifting the equilibrium equilibrium constant toward the t he associated a ssociated form (the effect of o f temperature on the equil e quilibri ibrium um constant of a reaction is given given by Eq. Eq . 1-18). To To avoid this problem, the buffer should be prepared at the same temperature as its planned use. – + 18. Glycine Glycine hydrochloride (Cl H3 N CH2COOH) is is a diprotic acid that contains a carboxyli ca rboxylicc acid ac id group group and

an ammonium group group and is therefore ca lled lled an a n amino acid. It is often used in biochemical buffers. (a) Which proton would you expect to dissociate at a lower pH, the proton of the carboxylic acid

group or the ammonium group? (b) Write Write the th e chemi che mical cal equations equa tions describing describing the dissociation of the first and sec ond protons of

Cl – H3 N+CH2COOH. (c) A solution containing 0.01 M Cl – H3 N+CH2COOH and 0.02 M of the monodissociated monodissociated species

has pH = 2.65. What is the pK of this dissociation? (d) In analogy with Figure Figure 2-18, sketch the titration curve c urve of this t his diprotic diprotic acid. ac id. Answer:

ammonium groups and therefore lose their protons (a) Carboxylic acid groups are stronger acids than ammonium



Problems


(c) The pK values of glycine's two ionizable ionizable groups groups are sufficiently sufficiently different so that the Henderson– Hende rson–

Hasselbalch Hasselbalch equation equa tion (Section (Section 2-2B) ade quately describes the behavior be havior of the solution of the diacid and the monodissociated monodissociated species.

(d)

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Ch 2 - Problems

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