Question

What is the molar solubility of $PbC{{l}_{2}}$ with a ${{K}_{sp}}$ of $1.6\times {{10}^{-5}}$ ?

Answer 1

Hint: The ${{K}_{sp}}$ is the solubility product constant also expressed as solubility product which is defined as the product of the concentration of the products of a partially soluble salt. For the partially soluble salts, the concentration of the products depends on each other, and hence they can be expressed with the same variable and some constant.

Complete step by step answer:
Here, we are given the solution of Lead Chloride $PbC{{l}_{2}}$ , which is an ionic compound and a partially soluble salt.
Hence, when lead chloride is dissolved in water, it does not dissociate into ions completely when it achieves equilibrium.
Now, the dissociation of the insoluble salt when dissolved in water can be expressed as
$PbC{{l}_{2(s)}}\rightleftharpoons Pb_{(aq)}^{2+}+2Cl_{(aq)}^{-}$
Now, suppose that the salt taken has a concentration of $1\;M$ , but as the ionization is not complete, the concentration of the ions will be less than $1\;M$ , suppose equal to $s$
Now, from the equilibrium reaction, we can say that the moles or molar concentration of chloride ions will always be twice the concentration of the lead ions.
Hence, the concentration can be expressed through the ICE (Initial Change Equilibrium) table

	$PbC{{l}_{2(s)}}$	$Pb_{(aq)}^{2+}$	$2Cl_{(aq)}^{-}$
Initial	$1\;M$ (suppose)	$0$	$0$
Change	$-s$	$+s$	$+2s$
Equilibrium	$1-s$ (suppose)	$s$	$2\;s$

Now, the solubility product can be expressed as
${{K}_{sp}}=\left[ P{{b}^{2+}} \right]{{\left[ C{{l}^{-}} \right]}^{2}}$
Substituting the value of ions from the table,
$\therefore {{K}_{sp}}=\left[ s \right]{{\left[ 2s \right]}^{2}}$
$\therefore {{K}_{sp}}=4{{s}^{3}}$
Now, we are given the solubility product,
$\therefore 1.6\times {{10}^{-5}}=4{{s}^{3}}$
$\therefore 0.4\times {{10}^{-5}}={{s}^{3}}$
Shifting the decimal point to get a suitable power for cube root,
$\therefore 4\times {{10}^{-6}}={{s}^{3}}$

$\therefore s=1.587\times {{10}^{^{-2}}}M$

Note:
Here the solubility product obtained in the form of variable $s$ can be considered a standard value. Thus whenever the ratio of the products of a partially soluble salt is $\;1:2$ or $2:1\;$ then the solubility product will be $4s^3$ . Similarly, if the products are in the ratio $\;1:1$ then the solubility product will be $s^2$ and for the products in the ratio $\;3:1$ or $1:3\;$ , solubility product is $27s^4$ .