Question

If the solubility of $PbC{{l}_{2}}$ in water is 0.01M at ${{25}^{\circ }}C$ , its maximum concentration in 0.2 M NaCl will be
(A)- $2\times {{10}^{-3}}M$
(B)- $1\times {{10}^{-4}}M$
(C)- $1.6\times {{10}^{-2}}M$
(D)- $4\times {{10}^{-4}}M$

Answer 1

Hint: The relation between the concentration and the solubility of the compound is given by its solubility product constant, as the ions are present in maximum equilibrium with the solid molecule. Also, the common ion effect will reduce the solubility of the compound.


Complete step by step answer:
The given compound of lead chloride is an insoluble ionic compound, which is present in equilibrium with its dissolved ions in water, as it does not dissociate completely. Thus, having the equation as follows:
$PbC{{l}_{2}}\rightleftarrows P{{b}^{2+}}+2C{{l}^{-}}$ ------------ (a)

Then, the solubility product will be ${{K}_{sp}}=\left[ P{{b}^{2+}} \right]{{\left[ C{{l}^{-}} \right]}^{2}}$ , where ${{K}_{sp}}$ is the solubility product constant.
Given the concentration of the solid compound to be 0.01 M, from the stoichiometry the concentration of $\left[ P{{b}^{2+}} \right]$ will also be 0.01 M and $\left[ C{{l}^{-}} \right]$ will be $(2\times 0.01)=0.02M$.
So, we have the solubility product ${{K}_{sp}}=0.01\times {{(0.02)}^{2}}=4\times {{10}^{-6}}$ .

When the solid compound is added to the NaCl solution, having the chloride ion as the common ion between both the solid and the solution. Through the common ion effect, the equilibrium will shift to the left and the solubility of the solid compound is reduced.
In the solution of NaCl, the concentration of $\left[ C{{l}^{-}} \right]=0.2M$.
Substituting this value in the solubility product of $PbC{{l}_{2}}$, we will get the concentration of $\left[ P{{b}^{2+}} \right]$ ion as follows:
$\left[ P{{b}^{2+}} \right]=\dfrac{{{K}_{sp}}}{{{\left[ C{{l}^{-}} \right]}^{2}}}=\dfrac{4\times {{10}^{-6}}}{{{(0.2)}^{2}}}=4\times {{10}^{-4}}M$
As seen in equation (a), the concentration of $\left[ P{{b}^{2+}} \right]$ comes from $PbC{{l}_{2}}$. So, we now have the maximum concentration of the solid in NaCl solution to be option (D)- $4\times {{10}^{-4}}M$ , that is equal to the concentration of $\left[ P{{b}^{2+}} \right]$ as calculated above.
So, the correct answer is “Option D”.

Note: The solubility product constant is the equilibrium constant of the ionic compound, as it is slightly soluble in water. It depends on temperature.
The shift in the direction of equilibrium is given by Le Chatelier’s principle, as the number of ions vary in the solution due to various factors.