Answer:
1.68% is ionized
Step-by-step explanation:
The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:
C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)
Ka = 6.46x10⁻⁵ = [C₇H₅O₂⁻] [H₃O⁺] / [C₇H₆O₂]
Where [] are concentrations in equilibrium
In equilibrium, some 0.225M of the acid will react producing both C₇H₅O₂⁻ and H₃O⁺, the equilibrium concentrations are:
[C₇H₆O₂] = 0.225-X
[C₇H₅O₂⁻] = X
[H₃O⁺] = X
Replacing:
6.46x10⁻⁵ = [X] [X] / [0.225-X]
1.4535x10⁻⁵ - 6.46x10⁻⁵X = X²
1.4535x10⁻⁵ - 6.46x10⁻⁵X - X² = 0
Solving for X:
X = -0.0038. False solution, there is no negative concentrations.
X = 0.00378M. Right solution.
That means percent ionization (100 times Amount of benzoic acid ionized over the initial concentration of the acid) is:
0.00378M / 0.225M * 100 =
1.68% is ionized