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
