Find pH of solution

Question

The active ingredient in aspirin is acetylsalicylic acid (HC9H7O4), a monoprotic acid with Ka = 3.3x10-4 at 25oC. What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 500 mg of acetylsalicylic acid each, in 250 mL of water?

Answer 1

molar mass acetylsalicylic acid (HC₉H₇O₄) = 180.2 g / mol

First, we'll find the moles of HC₉H₇O₄

500 mg HC₉H₇O₄ x 1 g / 1000 mg = 0.500 g HC₉H₇O₄

0.500 g HC₉H₇O₄ x 1 mol / 180.2 g = 0.002775 moles HC₉H₇O₄

Next, we'll calculate the concentration of HC₉H₇O₄

0.002775 mols HC₉H₇O₄ / 250 ml x 1000 ml / L = 0.0111 mols / L = 0.0111 M HC₉H₇O₄

Finally, we calculate pH using the Ka expression and the fact that pH = -log [H+]

HC₉H₇O₄ ==> H⁺ + C₉H₇O₄^-

Ka = [ H⁺] [C₉H₇O₄^-] / [HC₉H₇O₄]

3.3x10^-4 = (x)(x) / 0.0111 - x

x² = 3.66x10^-6 - 3.3x10^-4x

x² + 3.3x10^-4x - 3.66x10^-6 = 0 .. use quadratic equation to solve for x

x = 0.00175 = [H⁺]

pH = -log 0.00175

pH = 2.76