Question

Acetylsalicylic acid (aspirin) HC9H7O4, is the most widely used pain reliever and fever reducer. Find the pH of .040 M aqueous aspirin at body temperature (Ka = 3.6e-4 @ 37.0 degrees C)

Answer 1

this is an equilibrium problem, as aspirin is not a strong acid/base.

HC9H7O4 <--> C9H7O4- + H+

initial concentration is 0.040 M. let change in concentration to equilibrium equal "x". thus, equilibrium concentration of HC9H7O4 is 0.040-x, cause aspirin is a reactant so it decreases in the reaction. thus, "x" amount C9H7O4- (conjugate base) is present at equilibium. so is "x" amount of H+.

Ka for this reaction is ([C9H7O4][H+])/[HC9H7O4]=3.6*10^-4. so substitute values.

Ka=([x][x])/0.040-x=3.6*10^-4. the -x in the denominator is negligible, as the power of Ka is less than 10^-3

so you essentially have x^2/0.040=3.6*10^-4

so x^2=14.4*10^-6. take square root of both sides and x is 3.79 *10^-3

X is the concentration or molarity of H+ ions at equilibrium. pH = -log (3.79 *10^-3)=2.42