Question

1. 3. Consider the dissociation of acetic acid: CH3COOH(aq) + H2O(l) ⎯→ ←⎯ CH3COO-(aq) + H3O+(aq) Ka = 1.8 x 10-5    pKa = 4.74

a. Write the equilibrium expression.

b. Would the addition of an aqueous solution of sodium hydroxide favor the dissociation of acetic acid? Explain your answer.

c. Would the addition of an aqueous solution of sodium acetate favor the dissociation of acetic acid? Explain your answer.

Answer 1

The dissociation equation is

CH₃COOH (aq) + H₂O (l) ------> CH₃COO^- (aq) + H₃O⁺ (aq)

K_a = 1.8*10^-5, pK_a = 4.74

a) The expression for equilibrium constant is

K = [CH₃COO^-][H₃O⁺]/[CH₃COOH] (H₂O, l is present in large excess and hence not included in the expression for K).

b) Write out the equation that would take place:

CH₃COOH (aq) + NaOH (aq) ------> CH₃COO^-Na⁺ (aq) + H₂O (l)

H⁺ (aq) + OH^- (aq) -----> H₂O (l)

As aqueous NaOH is added to acetic acid, neutralization of acetic acid to form sodium acetate (acetate is the conjugate base of acetic acid) and water takes place. As more NaOH is added, more acetate and water are formed. Water is removed from the system and hence H⁺ (or H₃O⁺) concentration decreases. To keep K constant, therefore, acetic acid should undergo further dissociation and furnish more H₃O⁺.

b) Sodium acetate reacts with water as

CH₃COO^- (aq) + H₂O (l) -------> CH₃COOH (aq) + OH^-(aq)

The reaction produces more CH₃COOH. K must remain constant as the temperature is kept constant. Hence, to keep K constant, the numerator must increase and hence acetic acid will undergo further dissociation.