What is the Ka value for the reaction?

Question

CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO^–(aq)

[H₃O⁺] = 4.0 × 10^–3 M

[CH₃COOH] = 0.90 M

What is the K_a value for the reaction?

Answer 1

We have to start this by assuming the concentration values given are at equilibrium. Otherwise, we cannot calculate a value for the K_a. This isn't noted in the wording above but it's a safe assumption given the information in the problem.

Equilibrium constants are calculated via the following form: K = [products]^coeff/[reactants]^coeff , where each product/reactant is raised to the power of its individual coefficient.

You might think, hey, we only have the values for two concentrations out of these three chemicals! But using the ratios of coefficients, and basic assumptions about weak acid dissociation, we can assume that the concentration of CH₃COO^- is equal to the concentration of H₃O⁺.

So, our K_a = [H₃O⁺]*[CH₃COO^-]/[CH₃COOH]

Plugging in values, we get: (4.0 * 10^-3)²/0.9 = 0.000017777, and rounding for SigFigs we get K_a = 1.8 * 10^-5.