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.
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.
To calculate the equilibrium constant for the ionization of acetic acid, we need to first determine the concentrations of the ions in the solution.
Given that 2.96% of acetic acid molecules have ionized, we can calculate the concentrations of CH3COO- ions and H3O+ ions.
Since the concentration of acetic acid (CH3COOH) is 0.0200 mol/L, the concentration of CH3COO- ions and H3O+ ions can be calculated as follows:
Concentration of CH3COO- ions = 0.0200 mol/L × 0.0296 = 0.000592 mol/L
Concentration of H3O+ ions = 0.0200 mol/L × 0.0296 = 0.000592 mol/L
Now, let's calculate the equilibrium constant (K) using the formula:
K = [CH3COO-] × [H3O+] / [CH3COOH]
K = (0.000592 mol/L) × (0.000592 mol/L) / (0.0200 mol/L)
K = 1.75 × 10^-5
Finally, we can compare this result with the known value of 1.8 × 10^-5. The calculated equilibrium constant is close to the known value, indicating that our calculation is consistent with the known value.