Question

In a 1.2mM aqueous solution of butanoic acid C3H7CO2H, what is the percentage of butanoic acid that is dissociated? You can find some data that is useful for solving this problem in the ALEKS Data resource. Round your answer to 2 significant digits.

Answer 1

Answer:

11% is the percentage of butanoic acid that is dissociated.

Explanation:

$HBu\rightleftharpoons H^++Bu^-$

Initially

c        0         0

At equilibrium

(c-cα)   cα     cα

Concentration of acid = c = [HBu] = 1.2 mM = 0.0012 M

Dissociation constant of butanoic acid = $K_a=1.51\times 10^{-5}$

Degree of dissociation =$\alpha$

Dissociation constant of an acid is given by:

$K_a=\frac{[Bu^-][H^+]}{[HBu]}$

$1.51\times 10^{-5}=\frac{(c\alpha )^2\times x}{(c -c\alpha )}$

$1.51\times 10^{-5}=\frac{c(\alpha )^2\times x}{(1-\alpha )}$

$1.51\times 10^{-5}=\frac{0.0012 M\time (\alpha )^2}{(1-\alpha )}$

$\alpha = 0.1060\approx 0.11 =11\%$

11% is the percentage of butanoic acid that is dissociated.