Question

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

Answer 1

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
`CH_3COOH+H_2O\rightleftharpoons H_3O^+ +CH_3COO^-`

Answer: 0.000017

Step-by-step explanation:

`CH_3COOH+H_2O\rightleftharpoons H_3O^+ +CH_3COO^-`

cM 0 M 0 M

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

So dissociation constant will be:

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

Give c= 0.10 M and
`\alpha` = dissociation constant = ?

`K_a=?`

Putting in the values we get:

`K_a=((0.10* \alpha)^2)/((0.10-0.10* \alpha))`

`pH=-log[H^+]`

`2.87=-log[H^+]`

`[H^+]=1.35* 10^(-3)`

`[H^+]=c* \alpha`

`1.35* 10^(-3)=0.10* \alpha`

`\alpha=0.013`

`K_a=((0.10* 0.013)^(2))/(0.10-0.10* 0.013)`

`K_a=0.000017`

Thus the acid dissociation constant is 0.000017