Final answer:
The percent ionization of a 0.133 M solution of a monoprotic weak acid with a Ka of 6.68 × 10⁻³ is approximately 2.16%.
Step-by-step explanation:
The percent ionization of a weak acid can be calculated using the acid ionization constant (Ka) and the initial concentration of the acid. In this case, the Ka of the monoprotic weak acid is 6.68 × 10⁻³. To find the percent ionization, we need to determine the concentration of H₃O⁺ (hydronium ions) in the solution.
We can use an ICE table to solve this problem. Let's assume that x is the amount of acid that ionizes. This means that the change in concentration of the acid ([Hox⁻]) will be -x, and the change in concentration of hydronium ions ([H₃O⁺]) will be +x. At equilibrium, the concentration of the acid will be (0.133 M - x) and the concentration of hydronium ions will be x.
We can use the relationship between Ka and x to solve for x. The expression for Ka is Ka = [H₃O⁺][A⁻] / [HA]. Substituting the values, we get (6.68 × 10⁻³) = x * x / (0.133 - x). By solving this equation, we find that x ≈ 2.87 × 10⁻³.
Finally, we can calculate the percent ionization using the formula: (x / initial concentration) * 100%. Plugging in the values, we get (2.87 × 10⁻³ / 0.133) * 100% ≈ 2.16%.