Respuesta :
The given question is incomplete. The complete question is:
Determine the acid dissociation constant for a 0.10 M acetic acid solution that has a pH of 2.87. Acetic acid is a weak monoprotic acid and the equilibrium equation of interest is [tex]CH_3COOH+H_2O\rightleftharpoons H_3O^+ +CH_3COO^-[/tex]
Answer: 0.000017
Explanation:
[tex]CH_3COOH+H_2O\rightleftharpoons H_3O^+ +CH_3COO^-[/tex]
cM 0 M 0 M
[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= 0.10 M and [tex]\alpha[/tex] = dissociation constant = ?
[tex]K_a=?[/tex]
Putting in the values we get:
[tex]K_a=\frac{(0.10\times \alpha)^2}{(0.10-0.10\times \alpha)}[/tex]
[tex]pH=-log[H^+][/tex]
[tex]2.87=-log[H^+][/tex]
[tex][H^+]=1.35\times 10^{-3}[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex]1.35\times 10^{-3}=0.10\times \alpha[/tex]
[tex]\alpha=0.013[/tex]
[tex]K_a=\frac{(0.10\times 0.013)^{2}}{0.10-0.10\times 0.013}[/tex]
[tex]K_a=0.000017[/tex]
Thus the acid dissociation constant is 0.000017
