Question

The molar solubility of $Pb{{I}_{2}}$ in 0.2M $Pb{{(N{{O}_{3}})}_{2}}$ solution in terms of solubility product, ${{K}_{sp}}$is-
(A) ${{({{K}_{sp}}/0.2)}^{1/2}}$
(B) ${{({{K}_{sp}}/0.4)}^{1/2}}$
(C) ${{({{K}_{sp}}/0.8)}^{1/2}}$
(D) ${{({{K}_{sp}}/0.8)}^{1/3}}$

Answer 1

Hint: product of equilibrium concentrations of constituent ions raised to the power of their respective coefficients in the balanced equilibrium expression at a given temperature is the Solubility product. The relation would be obtained as ${{K}_{sp}}=4{{S}^{3}}$ . Apply the principle of common ion effect as well.

Complete step by step solution:
We know that for a general electrolyte ${{A}_{x}}{{B}_{y}}$ solubility at equilibrium,
\[{{A}_{x}}{{B}_{y}}\rightleftarrows x{{A}^{y+}}+y{{B}^{x-}}\]
The solubility product is
\[{{K}_{sp}}={{[{{A}^{y+}}]}^{x}}{{[{{B}^{x-}}]}^{y}}\]
That is a product of equilibrium concentrations of constituent ions raised to the power of their respective coefficients in the balanced equilibrium expression at a given temperature is the Solubility product.
If S is the molar solubility of the compound given, the equilibrium concentrations of the ions in the saturated solution can be written as -
\[\begin{align}
& [{{A}^{y+}}]=xS \\
& [{{B}^{x-}}]=yS \\
\end{align}\]
\[{{K}_{sp}}={{[xS]}^{x}}{{[yS]}^{y}}\] …(i)
Dissociation of $Pb{{I}_{2}}$ can be expressed as:
\[Pb{{I}_{2}}\rightleftarrows P{{b}^{2+}}+2{{I}^{-}}\] …(ii)
Hence from equation (i), we get ,
\[\begin{align}
& {{K}_{sp}}=S\times {{(2S)}^{2}} \\
& {{K}_{sp}}=4{{S}^{3}} \\
\end{align}\]
In presence of $Pb{{(N{{O}_{3}})}_{2}}$ which dissociates completely as-
$Pb{{(N{{O}_{3}})}_{2}}\rightleftarrows P{{b}^{2+}}+2N{{O}_{3}}^{-}$
As \[[Pb{{(N{{O}_{3}})}_{2}}]=0.2M\] ,
$[P{{b}^{2+}}]=0.2M$
Let the new solubility be s’ of $Pb{{I}_{2}}$ . At equilibrium , concentration of $P{{b}^{2+}}$would be 0.2+s’.
The common ion effect, a consequence of Le Chatlier’s principle or the Equilibrium Law, is an effect that suppresses the ionization of an electrolyte when another electrolyte that contains an ion which is also present in the first electrolyte, that is, a common ion, is added.
So Due to the common ion effect, equilibrium (ii) will shift to the left and $[P{{b}^{2+}}]\sim 0.2M$.
\[{{K}_{sp}}=0.2\times {{(2s')}^{2}}=0.2\times 4s{{'}^{2}}=0.8s{{'}^{2}}\]
\[s'={{({{K}_{sp}}/0.8)}^{1/2}}\]

Therefore, the correct answer is option (C)\[{{({{K}_{sp}}/0.8)}^{1/2}}\].

Note: The ionic product of an electrolyte is defined in the same way as solubility product but the difference is that the ionic product consists of concentration of ions under any condition, whereas expression of solubility product consists of only equilibrium concentrations.