Question

Calculate the solubility of $Pb{{I}_{2}}({{K}_{sp}}=1.4\times {{10}^{-8}})$ in water at $25{}^\circ C$ , which is $90%$ dissociated.

Answer 1

Hint: Solubility is defined as the ability of a compound to get soluble. Solubility is expressed in terms of molarity, ${{K}_{sp}}$ is defined as the solubility product constant which is an equilibrium constant. It tells us about the level at which a solute dissolves a solution.

Formula used:
$\alpha =\dfrac{D.M}{I.M}$
where, $\alpha $ is the degree of dissociation, $D.M$ is the dissociated moles and $I.M$ is the initial moles.

Complete step by step answer:
Here, it is given that ${{K}_{sp}}=1.4\times {{10}^{-8}}$
To calculate the degree of dissociation,
$\alpha =\dfrac{D.M}{I.M}$
where, $\alpha $ is the degree of dissociation, $D.M$ is the dissociated moles and $I.M$ is the initial moles.
In this question, it is given that the solution is $90%$ dissociated, so we can say that the dissociated moles $=\,\,90$ and initial moles $=\,\,100$
Now, if we substitute the values in the above formula, we get,
$\alpha =\dfrac{90}{100}$
The degree of dissociation $=0.9$
Let us see the reaction
$Pb{{I}_{2}}_{(s)}\rightleftharpoons P{{b}^{2+}}_{(aq)}+2{{I}^{-}}_{(aq)}$

	$P{{b}^{2+}}$	$2{{I}^{-}}$
Solubility	$s\times 0.9$	$2s\times 0.9$ .

In this reaction, $Pb{{I}_{2}}$ dissociates to give one $P{{b}^{2+}}$ (cation) and two ${{I}^{-}}$ (anion).
Here, $s$ is the solubility therefore, the solubility of one $P{{b}^{2+}}$ is $s$ and of two ${{I}^{-}}$ is $2s$
In this reaction, $P{{b}^{2+}}$ and $2{{I}^{-}}$ are $90%$ dissociated. Hence, we multiply the solubility by $0.9$
${{K}_{sp}}=(0.9\times s){{(0.9\times 2s)}^{2}}$
Now, substituting the value of ${{K}_{sp}}$ , we get,
$\Rightarrow 1.4\times {{10}^{-8}}=4{{(0.9)}^{3}}{{s}^{3}}$
$\Rightarrow {{s}^{3}}=\dfrac{1.4\times {{10}^{-8}}}{4\times {{(0.9)}^{3}}}$
On further solving, we get,
$s=0.78\times {{10}^{-2}}mol/L$

Therefore, the solubility of $Pb{{I}_{2}}$ is $0.78\times {{10}^{-2}}mol/L$

Additional information:
If the ionic product is more than the saturated product, precipitation will take place.
If the ionic product is less than the saturated product, precipitation will not take place.
If the ionic product is equal to the saturated product, then the solution will be saturated and there will be no precipitation.
${{K}_{sp}}$ is defined as the solubility product which is defined as the equilibrium constant for solid to dissolve in an aqueous solution.
Solubility product tells us about the slightly ionic compound.

Note: Solubility product depends only on temperature.
Factors that affect solubility are common ion effect and simultaneous solubility.
As we increase the solubility product, there will be an increase in the solubility.