You are given the neutralization of acetic acid with sodium hydroxide. Also, you are given the k for acetic acid, which is 1.8 x 10⁻⁵. You are asked to find the approximate value of the equilibrium constant, kn, for the neutralization. We will have a reaction of both acetic acid and sodium hydroxide.
CH₃COOH + NaOH → CH₃COONa + H₂O
which comes from
CH₃COOH → CH₃COO⁻ + H⁺
H⁺ + OH⁻ → H₂O
The k for water is always 1.0 x 10¹⁴. The Ksp for the reaction will be
Ksp = [CH₃COOH][H₂O]
Ksp = (1.8 x 10⁻⁵)(1.0 x 10¹⁴)
Ksp = 1.8 x 10⁹
