Respuesta :
Answer:
Incomplete question: A student prepares a 0.54 mM aqueous solution of butanoic acid...
The answer is: The percentage of dissociation is 15.4%
Explanation:
Given:
Ka = 1.51x10⁻⁵
Concentration = 0.54 mM = 0.54x10⁻³M
The reaction is:
C₄H₈O₂ = C₄H₇O₂⁻ + H⁺
I 0.54x10⁻³ 0 0
E 0.54x10⁻³(1-x) 0.54x10⁻³x 0.54x10⁻³x
Where x is the degree of dissociation
[tex]Ka=\frac{[H^{+}][C_{4}H_{7}O_{2}^{-} }{[C_{4}H_{8}O_{2} } \\1.51x10^{-5} =\frac{0.54x10^{-3}x^{2} }{1-x} \\x^{2} +0.028x-0.028=0\\x_{1} =0.154\\x_{2} =-0.182[/tex]
We will choose the positive value
x = 0.154
The percentage of dissociation is 15.4%
