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Calculate the acid dissociation constant [tex]$K_a$[/tex] of acrylic acid given the following information:

- The pH of a 1.5M solution of acrylic acid [tex](HC_3H_3CO_2)[/tex] is measured to be 2.04.

Round your answer to 2 significant digits.

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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