Question

Please show all work. Thank you.

Calculate the pH of a 0.10 M solution of sodium acetate NaC_2H_3O_2.

Answer 1

Ka of CH3COOH = 1.80*10^-5

Lets find Kb for CH3COO-

use:

Kb = E-14/Ka

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

Kb = 5.556*10^-10

for simplicity lets write CH3COO- as A-

A- + H2O -----> AH + OH-

0.1 0 0

0.1-x x x

Kb = [AH][OH-]/[B-]

Kb = x*x/(c-x)

since kb is small, x will be small and it 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)*0.1) = 7.454*10^-6

pOH = -log [OH-] = -log (7.454*10^-6) = 5.13

PH = 14 - pOH = 14 - 5.13 = 8.87

Answer: 8.87