Question

Consider citric acid, H3C6H5O7:

a. How many acidic hydrogens does the acid have?

b. What is the Ka for loss of the first acidic hydrogen?

c. What is the Ka for the weakest loss of hydrogens?

d. Write reactions representing successive losses of the acidic hydrogens in water.

e. Explain using Le Chatelier's principle what happens as strong acid is added to each reaction.

Answer 1

Solution-

Citric acid C6H8O7 can be written as H3C6H5O7 i.e. H3A

a) Citric acid is a tribasic acid it means it has 3 acidic hydrogens that can be ionized as the proton.

b) An applicable equation for ionization of the first proton is

H3A (aq)   H2A-(aq) + H+.

Ka1 = [H2A-][H+] / [H3A] = 1.7 x 10-4.

c) Ionization equation for 3rd proton is, the 3rd proton is the weakest one to be lost.

HA2- (aq)   A3-(aq) + H+.

Ka3 = [A3-][H+] / [HA2-] = 6.4 x 10-6.

d) Now the reactions representing successive losses of the acidic hydrogens in water are-

1st ionization : H3A (aq) + H2O H2A-(aq) + H3O+.

2nd ionization : H2A-(aq) + H2O HA2-(aq) + H3O+.

3rd ionization : HA2-(aq) + H2O A3-(aq) + H3O+.

d) According to Le Chatelier's Principle states "any system at equilibrium when subjected to constrain like the change in volume, pressure, concentration tries to minimize constraint imposed by nullifying it by shifting the equilibrium in opposite direction.

Addition of strong acid will decrease the ionization of proton at each stage (above).

A strong electrolyte ionizes completely and gives large [H+] in solution and hence to neutralize this increased concentration weak citric acid molecule lowers ionization and corresponding conjugate bases start getting protonated back.