Question

A typical sample of vinegar has a pH of 3.0. Assuming that vinegar is only an aqueous solution of acetic acid (Ka=1.8x10^-5), calculate the concentration of acetic acid in vinegar.

Answer 1

Answer: Concentration of acetic acid in vinegar is 0.06 M

Step-by-step explanation:

`CH_3COOH\rightleftharpoons CH_3COO^-+H^+`

cM 0 0

`c-c\alpha`
`c\alpha`
`c\alpha`

So dissociation constant will be:

`K_a=((c\alpha)^(2))/(c-c\alpha)`

`pH=-log[H^+]`

`3.0=-log[H^+]`

`[H^+]=c* \alpha=10^(-3)`

`K_a=1.8* 10^(-5)`

Putting in the values we get:

`1.8* 10^(-5)=((10^(-3))^2)/((c-10^(-3)))`

`c=0.06M`

Thus concentration of acetic acid in vinegar is 0.06 M