At equilibrium an acetic acid solution has 0.0787 M acid and 0.00118 M each of hydronium and acetate. What is Ka?

1. The expression puts hydronium times acetate over the intact acetic acid. 2. Multiply the two ion concentrations: 0.00118 x 0.00118 = 1.39 x 10^-6. 3. Divide by the acid concentration: 1.39 x 10^-6 divided by 0.0787. 4. That gives 1.77 x 10^-5, the acid ionisation constant of acetic acid. 5. Stopping at the numerator gives 1.39 x 10^-6, which is the common slip here. 6. Inverting the ratio gives a huge value, which would describe the reverse reaction. _Source: OpenStax Chemistry (CC BY 4.0), Ch 14 "Acid-Base Equilibria", section 14.3 Relative Strengths of Acids and Bases_