Respuesta :
The question is incomplete , complete question is ;
A student prepares a 0.30 mM aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. The [tex]pK_a=4.756[/tex]
Answer:
21.43% is the percentage of acetic acid that is in the dissociated form in its solution.
Explanation:
The [tex]pK_a[/tex] value of acetic acid = 4.756
The dissociation constant of acetic acid = [tex] K_a[/tex]
[tex]pK_a=-\log[K_a][/tex]
[tex]4.756=-\log[K_a][/tex]
[tex]K_a=10^{-4.756}=1.754\times 10^{-5}[/tex]
The initial concentration of acetic acid = c = 0.30 mM = [tex]0.30\times 10^{-3} M[/tex]
[tex]1 mM =m 10^{-3} M[/tex]
Degree of dissociation of acetic acid = [tex]\alpha [/tex]
[tex]HAc\rightleftharpoons Ac^-+H^+[/tex]
initially
c 0 0
At equilibrium
(c-cα) cα cα
The expression of dissocation constant :
[tex]K_a=\frac{[Ac^-][H^+]}{[HAc]}[/tex]
[tex]1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha }{c-c\alpha}[/tex]
[tex]1.754\times 10^{-5}=\frac{c\times (\alpha )^2}{(1-\alpha)}[/tex]
[tex]1.754\times 10^{-5}=\frac{0.30\times 10^{-3}\times (\alpha )^2}{(1-\alpha)}[/tex]
[tex]\alpha =0.2143=21.43\%[/tex]
21.43% is the percentage of acetic acid that is in the dissociated form in its solution.
