Answer:
1.33%
Explanation:
In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:
CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )
Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:
Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]
<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>
<em />
When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.
The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:
[CH₃CO₂H] = 0.100M - X
[H₃O⁺] = X
[CH₃CO₂⁻] = X
<em>Where X is reaction coordinate.</em>
Replacing in Ka expression:
1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]
1.8x10⁻⁵ = [X] [X] / [0.100M - X]
1.8x10⁻⁶ - 1.8x10⁻⁵X = X²
1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0
Solving for X:
X = -0.00135 → False solution. There is no negative concentrations.
X = 0.00133 → Right solution.
That means concentration of acetate ion is:
[CH₃CO₂⁻] = 0.00133M.
Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:
% Ionization = 0.00133M / 0.100M × 100 =
<h3>1.33%</h3>
<em />
<em />
<em />
