Respuesta :
Answer : The acid dissociation constant Ka of the acid is, [tex]8.7\times 10^{-5}[/tex]
Explanation :
First we have to calculate the concentration of hydrogen ion.
[tex]pH=-\log [H^+][/tex]
Given: pH = 4.06
[tex]4.06=-\log [H^+][/tex]
[tex][H^+]=8.71\times 10^{-5}M[/tex]
The dissociation of acid reaction is:
[tex]HA\rightarrow H^++A^-[/tex]
Initial conc. c 0 0
At eqm. c-cα cα cα
Given:
Degree of dissociation = α = 0.10 % = 0.001
[tex][H^+]=c\alpha[/tex]
[tex]8.71\times 10^{-5}=c\times 0.001[/tex]
[tex]c=0.0871M[/tex]
The expression of dissociation constant of acid is:
[tex]K_a=\frac{[H^+][A^-]}{[HA]}[/tex]
[tex]K_a=\frac{(c\times \alpha)\times (c\times \alpha)}{(c-c\alpha)}[/tex]
Now put all the given values in this expression, we get:
[tex]K_a=\frac{(0.0871\times 0.001)\times (0.0871\times 0.001)}{(0.0871-0.0871\times 0.001)}[/tex]
[tex]K_a=8.7\times 10^{-5}[/tex]
Thus, the acid dissociation constant Ka of the acid is, [tex]8.7\times 10^{-5}[/tex]
