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

Answer:

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

Explanation:

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

Initially c  

At equilibrium c-cα  cα cα

Dissociation constant of an acetic acid:

$K_a=1.8\times 10^{-5}$

Degree of dissociation = α

Dissociation constant of an acid is given by:

$K_a=\frac{c\alpha \times c\alpha }{c(1-\alpha )}=\frac{c\alpha ^2}{(1-\alpha )}$

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

$1.8\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha )}$

Degree of dissociation = α

$1.8\times 10^{-5}=\frac{1.59 M\alpha ^2}{(1-\alpha )}$

$\alpha =0.003359$

Percentage of dissociation = 0.3359%

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

$1.8\times 10^{-5}=\frac{c'(\alpha ')^2}{(1-\alpha ')}$

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:

$\frac{0.3359\%}{0.9789\%}=0.343$