Answer:
11% is the percentage of butanoic acid that is dissociated.
Explanation:
[tex]HBu\rightleftharpoons H^++Bu^-[/tex]
Initially
c 0 0
At equilibrium
(c-cα) cα cα
Concentration of acid = c = [HBu] = 1.2 mM = 0.0012 M
Dissociation constant of butanoic acid = [tex]K_a=1.51\times 10^{-5}[/tex]
Degree of dissociation =[tex]\alpha [/tex]
Dissociation constant of an acid is given by:
[tex]K_a=\frac{[Bu^-][H^+]}{[HBu]}[/tex]
[tex]1.51\times 10^{-5}=\frac{(c\alpha )^2\times x}{(c -c\alpha )}[/tex]
[tex]1.51\times 10^{-5}=\frac{c(\alpha )^2\times x}{(1-\alpha )}[/tex]
[tex]1.51\times 10^{-5}=\frac{0.0012 M\time (\alpha )^2}{(1-\alpha )}[/tex]
[tex]\alpha = 0.1060\approx 0.11 =11\%[/tex]
11% is the percentage of butanoic acid that is dissociated.
