Answer: [tex]K_a[/tex] for the acid is [tex]3.34\times 10^{-4}[/tex]
Explanation:
[tex]HC_9H_7O_4\rightarrow H^+C_9H_7O_4^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
Give c = 0.328 M and [tex]pH=1.987[/tex]
[tex]1.987=-log[H^+][/tex]
[tex][H^+]=0.0103[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex]0.0103=0.328\times \alpha[/tex]
[tex]\alpha=0.0314[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Putting in the values we get:
[tex]K_a=\frac{(0.328\times 0.0314)^2}{(0.328-0.328\times 0.0314)}[/tex]
[tex]K_a=3.34\times 10^{-4}[/tex]
