Question

The pH of a solution of propanoic acid is measured to be . Calculate the acid dissociation constant of propanoic acid. Round your answer to significant digits.

Answer 1

The given question is incomplete. The complete question is:

The pH of a 0.98 M solution of propanoic acid is measured to be 2.43. Calculate the acid dissociation constant of propanoic acid. Round your answer to two significant digits.

Answer: The acid dissociation constant of propanoic acid is
`1.4* 10^(-5)`

Step-by-step explanation:

`CH_3CH_2COOH\rightarrow CH_3CH_2COO^-+H^+`

cM 0 0

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

So dissociation constant will be:

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

Give c= 0.98 M and pH = 2.43

`pH=-log[H^+]`

`2.43=-log[c* \alpha]`

`2.43=-log[0.98* \alpha]`

`3.72* 10^(-3)=0.98* \alpha`

`\alpha=3.79* 10^(-3)`

`K_a=((0.98* 3.79* 10^(-3))^2)/((0.98-0.98* 3.79* 10^(-3)))=1.4* 10^(-5)`

The acid dissociation constant of propanoic acid is
`1.4* 10^(-5)`