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 (ka) 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 (kc) is :
kc = [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 %
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