Question

A typical aspirin tablet contains 324 mg of aspirin (acetylsalicyclic acid, C₉H₈O₄), a monoprotic acid having

K_a = 3.0*10^-4.

A) If you dissolve two aspirin tablets in a 270 mL glass of water, what is the pH of the solution?

B) What is the percent dissociation of the solution?

I got part A right but kept getting wrong answer for percent dissociation. Please show steps thank you

Answer 1

Initial concentration,

Mass of 2 tablets = 324 * 2 = 648 mg = 0.648 grams

Moles of acid = mass/molar mass = 0.648/180 = 0.0036

Molarity of acid = moles/volume in liters = 0.0036/0.270 = 0.01333 M

C9H8O4 -----> C9H7O4 - + H+

0.01333 0 0 (at t = 0)

0.01333-x x x (at equilibrium)

Ka = x * x / (0.01333 - x)

0.0003 = x^2 / (0.01333 - x)

0.000004 - 0.0003x = x^2

x^2 + 0.0003x - 0.000004 = 0

After solving above equation,

x = 0.00185562

Hence at equilibrium,

[H+].= x = 0.00185562 M

pH = - log (H+) = - log (0.00185562) = 2.73 ... Answer

B) % dissociation = 0.00185562*100 / 0.013333 = 13.92 % ....Answer

Let me know in case of any doubts.