Question

What is the pH of a 0.1 M solution of Sodium Acetate? (Hint: the acetate ion is a conjugate base of acetic acid)

Answer 1

(Assuming that the acetic acid has a Ka of 1.8 x10^-5), first it's necessary to state the ocurring dissolving process:

CH3COONa <-> CH3COO- + Na+
Then, it would be convenient to state the reaction of the acetate ion with water as follows:
CH3COO- + H2O <-> CH3COOH + OH-
The equilibrium of this reaction would be represented by Kb.
Kb=([CH3COOH][OH-]) / [CH3COO-]

With the value of Ka, we can obtain the value of Kb by dividing Kb= Kw/Ka= 1x10^-14 / 1.8x10^-5 = 5.56 x10^-5
Then, elaborating an ICE, we can conclude the values for the concentrations of each component as follows:

	CH3COO- <->	CH3COOH	OH-
Initial	0.1	0	0
Changing	-x	+x	+x
Final	0.1-x	x	x

Substituting in the original Kb equilibrium:

5.56x10^-10= (x^2)/(0.1-x) (x value can be considered as being very small because Kb is too small, therefore this equation can be simplified)

5.56x10^-10= (x^2)/(0.1)
Solving for x, x= 7x10^-6

Since x represents [OH-], we can calculate pOH with pOH= -log[OH-]
pOH= 5.128
pH= 14 - pOH = 8.872.