Question

Calculate the percent dissociation of butanoic acid (C3H2CO2H) in a 1.4 mM aqueous solution of the stuff.

Answer 1

Answer: The percent dissociation of butanoic acid is 9.8%

Step-by-step explanation:

`C_3H_2CO_2H\rightarrow H^+C_3H_2CO_2^-`

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= 1.4 mM =
`1.4* 10^(-3)` and
`\alpha` = dissociation constant

`K_a=1.5* 10^(-5)`

Putting in the values we get:

`1.5* 10^(-5)=((1.4* 10^(-3)* \alpha)^2)/((1.4* 10^(-3)-1.4* 10^(-3)* \alpha))`

`(\alpha)=0.098=9.8\%`

Thus percent dissociation of butanoic acid is 9.8%