When 0.040 mol of propionic acid, c2h5co2h, is dissolved in 750 ml of water, the equilibrium concentration of h3o+ ions is measured to be 1.84 x 10-3 m. what is ka for this acid?

Answer is: Ka for propinoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [C₂H₅COO⁻] · [H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.

pie chart because it is in percents ( % )