Answer:
[tex]Ka=2.5x10^{-6}[/tex]
Explanation:
Hello,
In this case, by knowing that in terms of the change [tex]x[/tex] due to the reaction extent, the percent dissociation is:
[tex]\% Dissociation:\frac{x}{[acid]_0}[/tex]
Thus, we compute [tex]x[/tex] as:
[tex]x=\% Dissociation*[acid]_0=0.89\%*0.031M=2.759x10^{-4}M[/tex]
With that [tex]x[/tex], we could assume the law of mass action for a typical dissociation:
[tex]Acid\rightleftharpoons H^++HA^-[/tex]
As:
[tex]Ka=\frac{x*x}{[acid]_0-x}=\frac{2.759x10^{-4}M*2.759x10^{-4}M}{0.031M-2.759x10^{-4}M} \\\\Ka=2.5x10^{-6}[/tex]
Best regards.
