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

Ka of aspirin = 3.69*10^-4

Step-by-step explanation:

Mass of aspirin(C9H8O4) = 2.00 g

Molar mass of aspirin = 180 g/mol

Therefore:

`Moles\ of\ C9H8O4 = (Mass)/(Molar\ Mass) =(2.00g)/(180g/mol) =0.011\ mol`

`Molarity\ C9H8O4 = (Moles)/(Liter) = (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 = ([H+][C9H7O4-])/([C9H8O4]) = (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 = (2.40*10^(-3)*2.40*10^(-3)  )/(1.56*10^(-2) ) =3.69*10^(-4)`