Final answer:
To calculate the percent ionization of a 0.120 M weak acid with a Ka of 0.00380, we determine the equilibrium concentration of H+ and then use it to find the percent ionization. The calculated percent ionization of the acid in solution is approximately 5.975%.
Step-by-step explanation:
To determine the percent ionization of a 0.120 M solution of a weak acid with a given Ka of 0.00380, we use the acid ionization expression:
For a weak acid HA dissociating into H+ and A-, the ionization is represented as: HA → H+ + A-.
The equilibrium expression is written as: Ka = [H+][A-] / [HA].
Since the initial concentration of A- is 0, we let x be the concentration of H+ and A- at equilibrium and (0.120 - x) be the concentration of HA. The equilibrium expression becomes:
Ka = x² / (0.120 - x). Assuming x is small compared to the initial concentration, we can simplify the equilibrium expression: Ka ≈ x² / 0.120.
Solving for x, we get:
x = √(Ka × 0.120)
Upon calculation, x ≈ √(0.00380 × 0.120) ≈ 0.00717 M as the concentration of H+ and A- at equilibrium.
Now we can calculate the percent ionization:
Percent Ionization = (x / initial concentration of HA) × 100%
Percent Ionization = (0.00717 / 0.120) × 100% ≈ 5.975%
The percent ionization of the 0.120 M weak acid with a Ka of 0.00380 is approximately 5.975%.