Question

Calculate the pH of a 0.42 M solution of sodium acetate, CH3COONa. (Ka(acetic acid) = 1.8 x 10-5)

Answer 1

this is a salt

A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

Ka can be calcualted as follows:

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10

and;

assume x = [OH-]

[H3O+] = x= [B] due to equilibrium

[B] = M-x = 0.42-x

Kb = [HA ][OH-]/[A-]

5.55*10^-10 = x*x/(0.42-x)

solve for x with quadratic equation

x = OH- =1.526*10^-5

[OH-  = 1.526*10^-5 M

pOH = -log(OH-) = -log(1.526*10^-5) = 4.81644

pH = 14-4.81644 = 9.18356

pH = 9.18356