A laboratory technician wants to determine the aspirin content of a headache pill by acid-base titration. Aspirin has a Ka of 3.0 x 10-4. If the pill is dissolved in water to give a solution about 0.0050 M, what is the pH of this solution? (Neglect dilution effects.)

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⁺]

pH = 2.97