Answer:
The concentration of [H₃O^+] in a 0.300 M solution of CH₂CHCOOH is 3.16 × 10^-5 M.
Step-by-step explanation:
Acrylic acid (CH₂CHCOOH) is a weak acid with a dissociation constant (Ka) of 3.16 × 10^-5.
The dissociation of acrylic acid in water is:
CH₂CHCOOH + H₂O ⇌ CH₂CHCOO^- + H₃O^+
The equilibrium expression for this reaction is:
Ka = [CH₂CHCOO^-][H₃O^+] / [CH₂CHCOOH]
At equilibrium, the concentration of CH₂CHCOOH that has dissociated is equal to the concentration of CH₂CHCOO^-:
[CH₂CHCOOH] = [CH₂CHCOO^-]
Let x be the concentration of [H₃O^+] at equilibrium. Then, the equilibrium concentration of [CH₂CHCOO^-] is 0.300 - x, and the equilibrium concentration of [CH₂CHCOOH] is also 0.300 - x.
Substituting these concentrations into the equilibrium expression gives:
Ka = (0.300 - x) * x / (0.300 - x)
Simplifying this expression:
Ka = x
Solving for x:
x = [H₃O^+] = Ka = 3.16 × 10^-5 M
Therefore, the concentration of [H₃O^+] in a 0.300 M solution of CH₂CHCOOH is 3.16 × 10^-5 M.