Question

Measuring the conductivity of an aqueous solution in which 0.0200 mol CH3COOH has been dissolved in 1.00 L of solution shows that 2.96 % of acetic acid molecules have ionized to CH3COO- ions and H3O+ ions. Calculate the equilibrium constant for ionization of acetic acid and compare your result with a known value of 1.8 × 10-5.

Answer 1

CH3COOH = HAc
.......HAc --> H^+ + Ac^-
Ka = (H^+)(Ac^-)/(HAc)
(H^+) = 0.02 x 0.0296 = ?\
(Ac^-) = 0.02 x 0.0296
(HAc) = 0.02-(0.02 x 0.0296) = ?
Substitute, calculate and compare.