Question

consider an unknown weak acid, HA. A 0.5M solution of HA was found to have a pH of 3.90. what is delta Gnot for the dissociation of HA at 298K? answers (in kJ/mol) a) 1.7 b) 42.8 c) 2.45 d) -17.3 e) 56.3

Answer 1

First, we will find Ka.

pH = -log[H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-3.90

[H+] = 1.259 x 10^-4

HA dissociates as

HA H⁺ + A^-

so [A^-] = [H⁺] =  1.259 x 10^-4

Initially HA is 0.5 M and at equilibrium it is

0.5 M -  1.259 x 10^-4 = 0.49987 M

so Ka for the dissociation HA H⁺ + A^-

can be calculated as

Ka = [H⁺] [A^-]/[HA]

Ka = 1.259 x 10^-4 x 1.259 x 10^-4/0.49987

Ka = 3.171 x 10^-8

From this Ka we can calculate G^o as

G^o =  -RT ln K

G^o = -8.314 x 298 ln 3.171 x 10^-8

G^o = 42779 J

G^o = 42.78 kJ/mol

correct answer b)42.8