Answer: The percent dissociation of 4-chlorobutanoic acid is 21%
Explanation:
The dissociation of weak acid is shown as:
[tex]C_3H_6ClCOOH\rightarrow C_3H_6ClCOO^-+H^+[/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= 0.54 mM = [tex]5.4\times 10^{-4}M[/tex] and [tex]\alpha[/tex] = ?
[tex]K_a=3.02\times 10^{-5}[/tex]
Putting in the values we get:
[tex]3.02\times 10^{-5}=\frac{(5.4\times 10^{-4}\times \alpha)^2}{(5.4\times 10^{-4}-5.4\times 10^{-4}\times \alpha)}[/tex]
[tex](\alpha)=0.21[/tex]
[tex]\%(\alpha)=0.21\times 100=21\%[/tex]
Thus the percent dissociation of 4-chlorobutanoic acid is 21%
