Respuesta :
Answer: [tex]K_a[/tex] for the acid is [tex]2.7\times 10^{-4}[/tex]
Explanation:
[tex]C_9H_7O_4H\rightarrow H^+C_9H_7O_4^-[/tex]
c 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= 0.313 M and [tex]pH[/tex] = 2.031
[tex](\alpha)=0.030[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex][H^+]=0.500\times 0.030=0.015[/tex]
Also [tex]pH=-log[H^+][/tex]
[tex]2.031=-log[H^+][/tex]
[tex][H^+]=0.009[/tex]
[ [tex][H^+]=c\alpha[/tex]
[tex]c\alpha=0.009[/tex]
Putting in the values we get:
[tex]K_a=\frac{(0.009)^2}{(0.313-0.009)}[/tex]
[tex]K_a=\frac{(0.009)^2}{(0.304)}=2.7\times 10^{-4}[/tex]
[tex]K_a[/tex] for the acid is [tex]2.7\times 10^{-4}[/tex]
