Calculate the percent dissociation of crotonic acid (C, H,CO, H) in a 0.63 mM aqueous solution of the stuff.

16%

Explanation:

Crotonic acid : C₃H₅CO₂H

C₃H₅CO₂H  ⇄ C₃H₅CO₂⁻ +  H⁺

   C                      O                O

where : C = C ( 1 - ∝ ) ,   O = C∝

also:  Ka = C∝² / ( 1 - ∝ )  ---- ( 1 )

From Alexa data resource :

Pka = 4.69 ,  [  Ka = 2.04 * 10^-5 = C∝² / ( 1 - ∝ ) ]

back to equation 1

2.04 * 10^-5 = [ ( 0.63 * 10^-3 ) * ∝² / ( 1 - ∝ ) ]  ----- ( 2 )

∴ ∝² / ( 1 - ∝ ) = 3.24 * 10⁻²

Resolving equation above

∝ = 0.1645 =  16.45%

the answer is 1 i believe