Question

2. What is the pH of a 0.36 M solution of sodium propionate, NaC3H302, at 25°C? (For propionic acid, HC2H502, Ka=1.3 10-5 at 25°C.) a. 9.22 b. 6.14 c. 4.78 d. 7.86 e. 11.10

Answer 1

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1.3*10^-5

Kb = 7.692*10^-10

C3H3O2- dissociates as

C3H3O2- + H2O -----> HC3H3O2 + OH-

0.36 0 0

0.36-x x x

Kb = [HC3H3O2][OH-]/[C3H3O2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((7.692*10^-10)*0.36) = 1.664*10^-5

since c is much greater than x, our assumption is correct

so, x = 1.664*10^-5 M

use:

pOH = -log [OH-]

= -log (1.664*10^-5)

= 4.7788

use:

PH = 14 - pOH

= 14 - 4.7788

= 9.2212

Answer: a