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

Answer:

The value of the dissociation constant will be:$3.94\times 10^{-4}$.

Explanation:

                $AsH\rightleftharpoons As^++H^+$

At ,t= 0      c              0     0

At eq'm   (c-x)            x     x

Concentration of aspirin = c

$c=\frac{2.00 g}{180 g/mol\times 0.600 L}=0.01851 M$

Expression for dissociation constant will be given as:

$K_a=\frac{[H^+][As^+]}{[AsH]}=\frac{x\times c}{(c-x)}=\frac{x^2}{(c-x)}$..(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).

$K_a=\frac{x^2}{(c-x)}=\frac{(0.002511 M)^2}{(0.01851 M-0.002511 M)}$

$K_a=3.94\times 10^{-4}$

The value of the dissociation constant will be:$3.94\times 10^{-4}$.