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?

Ka of aspirin = 3.69*10^-4

Explanation:

Mass of aspirin(C9H8O4) = 2.00 g

Molar mass of aspirin = 180 g/mol

Therefore:

It is given that:

pH = 2.62

Set up an ICE table corresponding to the dissociation of aspirin :

                        C9H8O4     ↔     H+         +        C9H7O4-

Initial                   0.018                    -                            -

Change                 -x                      +x                          +x

Equilib               (0.018-x)                x                           x

The acid dissociation constant Ka is given as:

since [H+] = x = 2.40*10^-3M

[C9H7O4-] = x = 2.40*10^-3M

[C9H8O4] = (0.018-x) = (0.018-2.40*10^-3)M=1.56*10^-2M

i believe the answer is true.