Answer :
The acid dissociation constant (Ka) of acrylic acid is approximately 2.3 x 10^(-5).
To calculate the acid dissociation constant (Ka) of acrylic acid (HC3H3CO2) from the given pH and molarity, you can use the equation for the ionization of a weak acid:
HC3H3CO2 ⇌ H+ + C3H3CO2-
The equilibrium expression for this reaction is:
Ka = [H+][C3H3CO2-] / [HC3H3CO2]
Given that the pH of the solution is 2.04, we can determine the concentration of H+ ions by taking the negative logarithm of the hydrogen ion concentration:
[H+] = 10^(-pH)
[H+] = 10^(-2.04) = 0.00748 M
Since acrylic acid is a monoprotic acid, the concentration of HC3H3CO2 is equal to its initial concentration, which is 1.5 M.
Now, let's substitute the values into the equation for Ka:
Ka = [H+][C3H3CO2-] / [HC3H3CO2]
Ka = (0.00748)(0.00748) / 1.5
Ka = 0.000035 / 1.5
Ka = 2.3 x 10^(-5) (rounded to 2 significant digits)
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