Question

Consider that you have two solutions of acetic acid, Ka = 1.8x10-5, one solution that is 1.59 M and another solution that is 0.186 M. Compare the percent dissociation of acetic acid in these two solutions. What is the ratio of percent dissociation of the 1.59 M solution to the 0.186 M solution? (% dissociation of 1.59 M / % dissociation of 0.186 M) Enter your answer numerically to three significant figures.

Answer 1

The ratio of percent dissociation of the 1.59 M solution to the 0.186 M solution: 0.343 : 1.

Step-by-step explanation:

`HAc\rightleftharpoons Ac^-+H^+`

Initially c

At equilibrium c-cα cα cα

Dissociation constant of an acetic acid:

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

Degree of dissociation = α

Dissociation constant of an acid is given by:

`K_a=(c\alpha * c\alpha )/(c(1-\alpha ))=(c\alpha ^2)/((1-\alpha ))`

1) Concentration of acid = c = [HAc] = 1.59 M

`1.8* 10^(-5)=(c\alpha ^2)/((1-\alpha ))`

Degree of dissociation = α

`1.8* 10^(-5)=(1.59 M\alpha ^2)/((1-\alpha ))`

`\alpha =0.003359`

Percentage of dissociation = 0.3359%

2) Concentration of acid = c' = [HAc] = 0.186 M

`1.8* 10^(-5)=(c'(\alpha ')^2)/((1-\alpha '))`

Degree of dissociation = α'

`\alpha '=0.009789`

Percentage of dissociation = 0.9789%

The ratio of percent dissociation of the 1.59 M solution to the 0.186 M solution:

`(0.3359\%)/(0.9789\%)=0.343`