Question

Calculate the percent dissociation of butanoic acid (C2H,CO,H) in a 0.87 mM aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits. 1 % x 6 ?

Answer 1

Ans :-  Percent dissociation of butanoic acid = 12 %

Explanation :-

Given,

Initial concentration of butanoic acid i.e. C₃H₇CO₂H = 0.87 mM = 8.7 x 10^-4 M

Let "α" is the degree of dissociation of C₃H₇CO₂H

Acid dissociation constant (k_a) of butanoic acid = 1.5 x 10^-5

Acid dissociation constant (ka) is ratio product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage.

ICE table of C₃H₇CO₂H is :

..........................C₃H₇CO₂H (aq)-----------------------------> C₃H₇CO₂^- (aq)....................+........................H⁺ (aq)

Initial (I)...............8.7 x 10^-4 M...........................................0.0 M.............................................................0.0 M

Change (C) ...............-8.7 x 10^-4 α.................................+8.7 x 10^-4 α................................................+8.7 x 10^-4 α

Equilibrium (E)......8.7 x 10^-4 (1-α)..................................8.7 x 10^-4 α M...............................................8.7 x 10^-4 α M

Expression of equilibrium constant (k_c) is :

k_c = [C₃H₇CO₂^- ].[H⁺] / [C₃H₇CO₂H]

1.5 x 10^-5 = (8.7 x 10^-4 α)² / 8.7 x 10^-4 (1-α)

1.5 x 10^-5 = (8.7 x 10^-4).α² / (1-α)

1.5 x 10^-5 .(1-α) = (8.7 x 10^-4).α²

α² = 0.0172(1-α)

α² + 0.0172 α - 0.0172 = 0

On solving this

α = 0.1228

So,

Percent dissociation = α x 100 % = 0.1228 x 100 % = 12.28 % = 12 % (upto two significant figures).

Hence, Percent dissociation of butanoic acid = 12 %