Answer: The percent dissociation of butanoic acid is 9.8%
Explanation:
[tex]C_3H_2CO_2H\rightarrow H^+C_3H_2CO_2^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 1.4 mM = [tex]1.4\times 10^{-3}[/tex] and [tex]\alpha[/tex] = dissociation constant
[tex]K_a=1.5\times 10^{-5}[/tex]
Putting in the values we get:
[tex]1.5\times 10^{-5}=\frac{(1.4\times 10^{-3}\times \alpha)^2}{(1.4\times 10^{-3}-1.4\times 10^{-3}\times \alpha)}[/tex]
[tex](\alpha)=0.098=9.8\%[/tex]
Thus percent dissociation of butanoic acid is 9.8%
