Question

Part A

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the Ka value calculated by the student if the pH of the solution was 2.62?

Express your answer numerically using two significant figures.

Answer I got: 3.16 x 10^-4 (incorrect) this is the reason why: You used the initial concentration of aspirin in the Ka expression. Instead, you need to use the equilibrium concentration. Consider that the amount of aspirin that reacted is equal to the amount of H3O+ produced. How much is left over?

Answer 1

Mass of aspirin in the solution = 2.00 g

Moles of aspirin = ( Mass / Molar mass) = ( 2.00 g / 180.157 g/ mol) = 0.011 mol

Volume of the solution = 0.600 L

Molarity of the solution = ( Moles / Volume) = ( 0.011/ 0.600 ) M = 0.018 M

pH = 2.62

-log [H+] = pH

-log [H+] = 2.62

[H+] = 10^-2.62

[ H+] = 0.0024

[H+] = [H3O+] = 0.0024 M

                                       C9H8O4(aq)        +      H2O(l) ---------> H3O+ (aq)       +           C9H7O4-(aq)

            Initial                     0.018 M                                                  0                                      0

            Change                  - 0.0024                                                + 0.0024                            0.0024

          Equilibrium              0.0156 M                                              0.0024 M                            0.0024 M

Ka   =    [ H3O+] x [C9H7O4-] / [C9H8O4]

Ka = [0.0024] x [ 0.0024] / [ 0.0156] = 3.7 x 10^-4