Question

Calculate the pH of a 0.51 M CH3COONa solution. (Ka for acetic acid = 1.8x10^-5)

Answer 1

Solution-

Given-

Ka for acetic acid = 1.8x10^-5

[CH3COONa ]= 0.51 M

water dissociation constant (Kw ) = 1.0 x 10^-14

Dissociation of CH3COONa is as follows

Sodium acetate is strong salt
CH3COONa(aq) + H2O(l) → CH3COO^-(aq) + Na⁺(aq)

CH3COO^-(aq) + H2O(l) → CH3COO- + OH-

We know the equation for base dissociation constant (Kb)

Kb = Kw/Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.55 x 10^-10

Kb = 5.55 x 10^-10

Let’s assume

[CH3COO-] = [OH-] = X

[CH3COO-] = 0.51-X)     Concentration at equilibrium

Then dissociation constant (Kb)is

Kb = [CH3COO-] [OH-]/ [CH3COO-] = X² /(0.51-X)

By using 5% approximation we neglect x at the denominator.

Kb = X² /0.51

(5.55 x 10^-10) * 0.51 = X²

X² = 2.83 X 10^-10

X = [OH-] = 1.68 X 10^-5 M

We know the equation of POH

pOH = - log [OH-] = -log[1.68 x 10^-5] = 4.77

pH + pOH = 14
pH = 14 - pOH = 14 - 4.77

pH = 9.23

Answer- pH of a 0.51 M CH3COONa solution = 9.23