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 2.60?

Answer 1

The value of the dissociation constant will be:
`3.94* 10^(-4)`.

Step-by-step explanation:

`AsH\rightleftharpoons As^++H^+`

At ,t= 0 c 0 0

At eq'm (c-x) x x

Concentration of aspirin = c

`c=(2.00 g)/(180 g/mol* 0.600 L)=0.01851 M`

Expression for dissociation constant will be given as:

`K_a=([H^+][As^+])/([AsH])=(x* c)/((c-x))=(x^2)/((c-x))`..(1)

The pH of the solution = 2.60

The pH of the solution is due to free hydrogen ions whcih come into solution after partial dissociation of aspirin.

`pH=2.60=\log[H^+]=-\log[x]`

`x=0.002511 M`

Putting value of x in (1).

`K_a=(x^2)/((c-x))=((0.002511 M)^2)/((0.01851 M-0.002511 M))`

`K_a=3.94* 10^(-4)`

The value of the dissociation constant will be:
`3.94* 10^(-4)`.