Answer:
pH = 3.23
Explanation:
Before the addition of any NaOH, the only you have is a 0.020M acetic acid solution. That is in equilibrium with water as follows:
HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)
The Ka of this reaction is:
Ka = 1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]
Where [] are concentrations in equilibrium of each species
As you have in solution just HC₂H₃O₂, the equilibrium concentrations will be:
[HC₂H₃O₂] = 0.020M - X
[C₂H₃O₂⁻] = X
[H₃O⁺] = X
Where X is reaction coordinate.
Repalcing in Ka expression:
1.8x10⁻⁵ = [C₂H₃O₂⁻] [H₃O⁺] / [HC₂H₃O₂]
1.8x10⁻⁵ = [X] [X] / [0.020M - X]
3.6x10⁻⁷ - 1.8x10⁻⁵X = X²
3.6x10⁻⁷ - 1.8x10⁻⁵X - X² = 0
Solving for X:
X = -0.0006M → False solution. There is no negative concentrations
X = 0.000591M → Right solution.
As:
[H₃O⁺] = X
[H₃O⁺] = 0.000591M
As pH = -log[H₃O⁺]
pH = 3.23
