Question

What is the pH of a 1.288 M solution of sodium acetate? Ka of acetic acid is 1.8*10^-5. (2 decimal places)

Answer 1

The dissociation reaction

C2H3NaO2 = CH3COO- + Na+

Base dissociation constant

Kb = Kw/Ka = (1*10^-14)/(1.8*10^-5)

= 5.55*10^-10

The reaction with ICE TABLE

CH3COO- + H2O = CH3COOH + OH-

I 1.288

C - x +x +x

E (1.288-x) x x

Equilibrium constant expression of the reaction

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

5.55*10^-10 = [x] [x] /[1.288-x]

1.288 >> x

5.55*10^-10 = [x] [x] /[1.288]

x² = 5.55*10^-10*1.288

x = 2.6749*10^-5 = [OH-]

pOH = - log [OH-]

= - log (2.6749*10^-5)

= 4.5726

pH = 14 - pOH = 14 - 4.5726

= 9.43