What is the pH of a 0.052 M aqueous solution of sodium acetate? K_a (acetic acid)= 1.8 × 10^-5 A)10.0 B)5.3 C)11.0 D)8.7 E)3.0

Question

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Sodium acetate is the conjugate base of acetic acid. When sodium acetate is dissolved in water, it undergoes hydrolysis, which releases hydroxide ions, causing the pH of the solution to increase. The equilibrium expression for the hydrolysis reaction is as follows:
CH3COO- + H2O ⇌ CH3COOH + OH-
The equilibrium constant, Kb, for this reaction is:
Kb = [CH3COOH][OH-] / [CH3COO-]
However, since the concentration of water remains constant, we can simplify the expression to:
Kb = [CH3COOH][OH-]
The Kb value can be calculated from the Ka value of acetic acid, as Kb = Kw / Ka, where Kw is the ion-product constant of water (1.0 x 10^-14 at 25°C). Therefore, Kb = 5.6 × 10^-10. 
Using the Kb expression, we can solve for the hydroxide ion concentration:
Kb = [CH3COOH][OH-]
5.6 × 10^-10 = (1.8 × 10^-5)[OH-]
[OH-] = 3.11 × 10^-6 M
Since sodium acetate is a salt, it does not contribute to the hydronium ion concentration. Therefore, the pH of the solution can be calculated using the hydroxide ion concentration:
pH + pOH = 14
pH = 14 - pOH = 14 - (-log[OH-]) = 14 - (-log(3.11 × 10^-6)) = 8.7
Therefore, the answer is D) 8.7.

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