Question

The pH of a 0.23M solution of acrylic acid HC3H3CO2 is measured to be 2.44. Calculate the acid dissociation constant Ka of acrylic acid. Round your answer to 2 significant digits. Step by step would be helpful...

Answer 1

Answer:

Kₐ =  5.7 x 10⁻⁵

Explanation:

The equilbrium for this acid is

HC₃H₃CO₂ + H2O    ⇄ H₃O⁺  +   C₃H₃CO₂ ⁻  ,

and the equilibrium constant for acrylic acid is given by the expression:

Kₐ = [ H₃O⁺][ C₃H₃CO₂⁻ ] / [ HC₃H₃CO₂ ]

Since  the pH of the 0.23 M solution is known , we can calculate [ H₃O⁺].

The ][ C₃H₃CO₂⁻ ]  is equal to  [ H₃O⁺] from the above equilibria (1:1)

Finally [ HC₃H₃CO₂ ] is known.

pH = - log  [ H₃O⁺]

taking antilog to both sides of the equation

10^-pH =  [ H₃O⁺]

Substituting

10^-2.44 =  [ H₃O⁺]  = 3.6 x 10⁻³

[ C₃H₃CO₂⁻ ] = 3.6 x 10⁻³

Kₐ = ( 3.6 x 10⁻³ ) /0 .23 = 5.7 x 10⁻⁵