Acids & Bases Chemistry Math Question 9

Question

In a 0.22 M solution of sodium acetate, what concentration (M) of hydronium ions comes from the self-ionization of water? pK_b for acetate anion is 9.246

Answer 1

Sodium acetate = CH3COONa

In water, you have Na⁺ and CH3COO^-

The CH3COO- will hydrolyze as follows:

CH3COO^- + H2O ==>CH3COOH + OH^- (note this produces a weak acid and strong base, so pH will be>7)

pKb = 9.246 so Kb = 1x10^-9.246 = 5.675x10^-10

Kb = 5.675x10^-10 = (CH3COOH][OH^-]/[CH3COO^-] = (x)(x)/0.22-x

x² = 5.675x10^-10x - 1.249x10^-10

x² - 5.675x10^-10 + 1.249x10^-10 = 0

Using quadratic to solve, we have

x = 1.1x10^-5 M = [OH-]

[H⁺] [OH^-] = Kw = 1x10^-14

[H⁺] = 1x10^-14/1.1x10^-5 = 9.1x10^-10 M = [H3O⁺] = concentration of hydronium ion