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=(|CH_(3)COO^(-)|*|H^(+)|)/(|CH_(3)COOH|)=(x^(2))/(|CH_(3)COOH|)\\\\ 1,78*10^(-5)=(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`