3.10 Common ion effect :
Can you recall?
Which reagents are used to
precipitate (i) group II, (ii) group III
B, (iii) group III A of basic radicals/
cations?Consider a solution of weak acid
CH 3 COOH and its soluble ionic salt
CH 3 COONa.
CH 3 COOH is weak acid, dissociates only
slightly in solution
CH 3 COOH CH 3 COO(aq) + H⊕ (aq)
CH 3 COONa being a strong electrolyte
dissociates almost completely in solution.
CH 3 COONa(aq) CH 3 COO + Na⊕
Both the acid and the salt produce
CH 3 COO ions in solution. CH 3 COONa
dissociates completely. Therefore it provides
high concentration of CH 3 COO^ ions.
According to Le-Chatelier principle, the
addition of CH 3 COO^ from CH 3 COONa to
the solution of CH 3 COOH, shifts equilibrium
of dissociation of CH 3 COOH to left. Thus
reverse reaction is favoured in which
CH 3 COO combines with H⊕ to form unionised
CH 3 COOH. Hence dissociation of CH 3 COOH
is supressed due to presence of CH 3 COONa
containing a common CH 3 COO ion.
Therefore [Ba^2 ⊕] = 0.025 M and
[F] = 0.010M
Hence ionic product of BaF 2 is
IP = [Ba^2 ⊕][F]^2
= 0.025 × (0.01)^2
= 2.5 × 10-6
Ksp (BaF 2 ) = 1.7 × 10-6 Thus, Ksp < IP
Ionic product in the solution is greater than
Ksp. Hence BaF 2 will precipitate from the
solution.The common ion effect states that the
ionisation of a weak electrolyte is supressed
in presence of a strong electrolyte containing
an ion common to the weak electrolyte.Can you tell?
How does the ionization of
NH 4 OH suppressed by addition of
NH 4 Cl to the solution of NH 4 OH?Remember...
Common ion effect is a special
case of Le-Chatelier's principle in
which the stress applied to an equilibrium
system is an increase in the concentration
of one of the product (ions). The effect
of this stress is reduced by shifting the
equilibrium to the reactant side.Do you know?
The hardness of water is due to
presence of Ca^2 ⊕ ions. It is surprising
to know that Ca^2 ⊕ ions can be removed by
adding more Ca^2 ⊕ ions in the form of lime
Ca(OH) 2 , to the hard water. The OH ions
of lime react with HCO 3 ions present in the
hard water to form CO 32 ions.
OH(aq) + HCO 3 (aq)
CO 32 (aq) + H 2 O(l)
Solubility product of CaCO 3 is very low
(Ksp = 4.5 × 10-9). Addition of lime makes
IP >>Ksp which results in the precipitation
of CaCO 3 and thereby removal of hardness.The presence of a common ion also
affects the solubility of a sparingly soluble
salt. Consider, the solubility equilibrium of
AgCl,
AgCl(s) Ag⊕(aq) + Cl(aq)
The solubility product of AgCl is
Ksp = [Ag⊕][Cl]3.10.1 Common ion effect and solubility