Answer:
pH = 2.97
Explanation:
Aspirin, (HC₉H₇O₄), is in equilibrium with water, thus:
HC₉H₇O₄(aq) + H₂O(l) ⇄ C₉H₇O₄⁻(aq) + H₃O⁺(aq)
Ka = 3.0x10⁻⁴ = [C₉H₇O₄⁻][H₃O⁺] / [HC₉H₇O₄]
When you add 0.0050M of aspirine, the solution reach equilibrium when concentrations are:
[HC₉H₇O₄] = 0.0050M - x
[C₉H₇O₄⁻] = x
[H₃O⁺] = x
Replacing in Ka formula:
3.0x10⁻⁴ = [x][x] / [0.0050-x]
1.5x10⁻⁶ - 3.0x10⁻⁴X = X²
X² + 3.0x10⁻⁴X - 1.5x10⁻⁶ = 0
Solving for x:
x = - 0.00138 → False answer. There is no negative concentrations
x = 0.00108 → Right answer
As [H₃O⁺] = x; [H₃O⁺] = 0.00108M.
pH = -log[H₃O⁺]
<em>pH = 2.97</em>
