BioPHYSICAL chemistry

5.5 There is a relatively small entropic component compared to the

enthalpic component.

5.6 CH 3 COO−is acting as a base, H 2 O is acting as an acid, CH 3 COOH is

acting as an acid, and OH−is acting as a base.

5.7 (a) pH =−log(10−^5 ) = 5

(b) pH =−log(10−^9 ) = 9

5.8 The standard reaction Gibbs energy is greater than zero and equal

to −RTlnK.

5.9 At equilibrium the Gibbs energy of a reaction is always zero.

5.10 The total volume is 1 L. The 0.01 mol of NaOH will convert 0.01 mol

of acetic acid to acetate, yielding final concentrations of 0.02 and

0.01 M for the acid and base forms, respectively. The pH is then

equal to:

5.11 After adding the 0.03 mol of NaOH to 0.1 mol of acetic acid the

solution has 0.07 mol of acetic acid and 0.03 mol of acetate ions.

Since the total volume is 1 L, the molarities are 0.03 and 0.07 for

the acetic acid and acetate ions, respectively. The pH is then given

by the Henderson–Hasselback equation:

5.12 Adding 0.01 M of HCl to distilled water poises the H+concentration

at 0.01 or 10−^2 M (compared to the initial concentration of 10−^7 M).

The pH of this solution is log(10−^2 ) or 2.

5.13 Glycine is an amino acid with the neutral structure: NH 2 CH 2 COOH.

The amine group on one side of the molecule is a weak base and

can be protonated to form a quaternary amine cation R-NH 3 +and

the carboxyl group on the other side can deprotonate to form the

carboxylic anion R-COO−. The pKAof the amine group is about 9.8

(this is for dissociation of the conjugate base; the protonated amine).

The pKAfor the carboxylic acid is about 4.5. The isoelectric pH is

just the average of the two pKAvalues.

5.14 The initial pH value is approximately 14.

5.15 Tris would have equal molar concentrations of Tris and its con-

jugate base at a pH equal to the pKA, or 8.3, which is the optimal

pH for buffering action.

5.16The coupling allows the overall process to proceed in a neutral fashion.

5.17 The formation is determined by the proton-transfer step rather

than the electron-transfer step.

pH p

base

acid

log. log

.

.

=+KA = + 475

003

007

= 438.

pH p

base

acid

M

M

log. log

.

.

=+KA = + 475

01

02

= 445.

ANSWERS TO PROBLEMS 447

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