What is the approximate value of the equilibrium constant, k n, for the neutralization of acetic acid withsodium hydroxide, shown in the equation below? the k afor acetic acid is 1.8 × 10 –5?

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

You use the atomic number to find out how many protons and neutrons there are in the atom. you use the atomic mass to find out how many protons and electrons there are in the atom. 

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