Calculate the [H+] and pH of a 0.0045 M acetic acid solution. The 𝐾a of acetic acid is 1.76×10−5

Answer 1

Acetic acid = CH3COOH

CH3COOH <==> H⁺ + CH3COO^-

Ka = [H⁺][CH3COO^-] / [CH3COOH]

1.76x10^-5 = (x)(x) / 0.0045 - x

x² + 1.76x10^-5x - 7.92x10^-8 = 0

x = [H⁺] = 0.000273 M

pH = -log 0.000273

pH = 3.56