Question

Calculate the pH of a 0.391 M aqueous solution of acetylsalicylic acid (aspirin) (HC9H7O4, Ka = 3.0×10-4).

pH =

Answer 1

Given [HC9H7O4]=0.391 M.

Ka=3x10^-4.

Since asprin is a weak acid then the ICE table is

HC9H7O4 + H2O <----------> H3O^+ + C9H7O4^-

Initial. 0.391. 0 . 0

Change . - x . +x . +x

Equilibrium. 0.391 - x. x. x

Ka=[H3O^+][C9H7O4^-]/[HC9H7O4]

(3x10^-4)=x^2/(0.391-x)

x^2 + (3x10^-4)x - (1.173x10^-3)=0

After solving this equation, x=0.013.

Therefore [H3O^+]=0.0107 M.

pH=- log[H3O^+]

pH=- log(0.0107)=1.97

pH=1.97.

Please let me know if you have any doubt. Thanks.