Answer :
Answer: Concentration of acetic acid in vinegar is 0.06 M
Explanation:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
cM 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]
[tex]pH=-log[H^+][/tex]
[tex]3.0=-log[H^+][/tex]
[tex][H^+]=c\times \alpha=10^{-3}[/tex]
[tex]K_a=1.8\times 10^{-5}[/tex]
Putting in the values we get:
[tex]1.8\times 10^{-5}=\frac{(10^{-3})^2}{(c-10^{-3})}[/tex]
[tex]c=0.06M[/tex]
Thus concentration of acetic acid in vinegar is 0.06 M