acetylsalicylic acid (aspirin) HC9H7O4 is the most widely used pain reliever and fever reducer. Find the PH of 0.040 M aqueous aspirin at body temperature (Ka at 37 C = 3.6 x 10-4)

Question

is this correct?
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 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

thanks.

Answer 1

I agree with this.