Question

What is the pH of a 0.360 M solution of sodium acetate, CH3CO2Na ? (Ka for CH3CO2H = 1.8 x 10-5)

Answer 1

use:

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

Kb = (1.0*10^-14)/1.8*10^-5

Kb = 5.556*10^-10

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.36 0 0

0.36-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

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 ((5.556*10^-10)*0.36) = 1.414*10^-5

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

so, x = 1.414*10^-5 M

use:

pOH = -log [OH-]

= -log (1.414*10^-5)

= 4.85

use:

PH = 14 - pOH

= 14 - 4.85

= 9.15

Answer: 9.15