Question

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?

Answer 1

Answer:

Ka of aspirin = 3.69*10^-4

Explanation:

Mass of aspirin(C9H8O4) = 2.00 g

Molar mass of aspirin = 180 g/mol

Therefore:

$Moles\ of\ C9H8O4 = \frac{Mass}{Molar\ Mass} =\frac{2.00g}{180g/mol} =0.011\ mol$

$Molarity\ C9H8O4 = \frac{Moles}{Liter} = \frac{0.011}{0.600} =0.018M$

It is given that:

pH = 2.62

$Since\ pH = -log[H+]\\\\the \ [H+] = 10^{-pH } = 10^{-2.62} = 2.40*10^{-3} M$

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:

$Ka = \frac{[H+][C9H7O4-]}{[C9H8O4]} = \frac{x^{2} }{0.018-x}$

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

$Ka = \frac{2.40*10^{-3}*2.40*10^{-3}  }{1.56*10^{-2} } =3.69*10^{-4}$