Final answer:
The acid dissociation constant (Ka) for acetic acid, given the initial concentration of 0.200 M and the equilibrium concentration of H3O+ is 0.0019 M, is 1.82 × 10⁻µ.
Step-by-step explanation:
To calculate the acid dissociation constant (Ka) for acetic acid, we can use the information provided that the initial concentration of acetic acid is 0.200 M and the equilibrium concentration of H3O+ is 0.0019 M. Using the chemical equation for the dissociation of acetic acid (CH3COOH):
CH3COOH(aq) ⇌ CH3COO−(aq) + H3O+(aq),
we know that at equilibrium, the concentration of H3O+ ions will be the same as the concentration of CH3COO− ions. Therefore, [CH3COO−] = [H3O+] = 0.0019 M.
Because acetic acid is a weak acid, we can assume that the change in concentration of acetic acid is roughly equal to the concentration of H3O+ ions produced. Thus, the equilibrium concentration of CH3COOH will be 0.200 M - 0.0019 M = 0.1981 M. The formula for Ka is:
Ka = [CH3COO−][H3O+] / [CH3COOH]
Substituting the equilibrium concentrations into this formula gives:
Ka = (0.0019 M)(0.0019 M) / (0.1981 M) = 1.82 × 10⁻µ,
which is the Ka for acetic acid.