Final answer:
To find the percent ionization of a 0.384 M solution of hydrofluoric acid (HF), we set up an ICE table, solve for x in the equilibrium expression assuming x << 0.384, and then calculate percent ionization as (x / 0.384) × 100%, resulting in 3.54%.
Step-by-step explanation:
To calculate the percent ionization of a 0.384 M solution of hydrofluoric acid, HF, with a Kᴇ (HF)=7.2×10⁻⁴, we first need to write the ionization equation of HF in water:
HF (aq) → H⁺ (aq) + F⁻ (aq)
Next, we express the equilibrium concentrations in an ICE table (Initial, Change, Equilibrium). Let 'x' represent the concentration of H⁺ and F⁻ formed in the reaction:
- Initial concentrations: [HF] = 0.384 M, [H⁺] = 0, [F⁻] = 0
- Change in concentrations: [HF] decreases by x, [H⁺] increases by x, [F⁻] increases by x
- Equilibrium concentrations: [HF] = 0.384 - x, [H⁺] = x, [F⁻] = x
Substitute the equilibrium concentrations into the expression for Kᴇ:
Kᴇ = [H⁺][F⁻]/[HF] = (x)(x)/(0.384 - x)
Assuming that x << 0.384, which is reasonable for a weak acid, the equation simplifies to:
Kᴇ = x²/0.384
Now, solve for x given that Kᴇ = 7.2×10⁻⁴:
7.2×10⁻⁴ = x²/0.384
x² = (7.2×10⁻⁴)(0.384)
x = sqrt((7.2×10⁻⁴)(0.384))
Percent ionization = (x / 0.384) × 100%
Calculating these values, x = 0.0136 M, and percent ionization = 3.54%.