Question

What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5

Answer 1

CH₃COOH ⇔ CH₃COO⁻ + H⁺

[CH₃COO⁻] = [H⁺] = x

$K=\frac{|CH_{3}COO^{-}|*|H^{+}|}{|CH_{3}COOH|}=\frac{x^{2}}{|CH_{3}COOH|}\\\\ 1,78*10^{-5}=\frac{x^{2}}{0,06} \ \ |*0,06\\\\ 0,1068*10^{-5}=x^{2}\\\\ x_{1}\approx0,001 \ \land \ \ x_{2}=\approx-0,001\\\\ pH=-log|H^{+}|=-log0,001=3$