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

Idk the answer imthink snail cant mate they dont lol

(no3)⁻n has oxidation number x,o has oxidation number -2.(x*1 -2*3) charge of the nitrate ion, and we know that charge of the nitrate ion =-1.so,(x*1 -2*3) =-1x-6=-1x=+5answer a. +5.