Question

What is the solubility (in M) of $PbC{l_2}$ in a $0.15{\text{ M}}$ solution of $HCl$ ? The ${K_{sp}}$ of $PbC{l_2}$ is $1.6{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 5}}$ .

Answer 1

Here we have to find the solubility of $PbC{l_2}$ in a $0.15{\text{ M}}$ solution of $HCl$ . Initially we have chlorine ions from $HCl$ . Thus we will find the total concentration of chlorine atoms at equilibrium position. Then with the help of ${K_{sp}}$ of $PbC{l_2}$ we can easily find the solubility of $PbC{l_2}$ .

Complete answer:
We are given initially with the concentration of $HCl$ as $0.15{\text{ M}}$ . It can be represented as:
$HCl{\text{ }} \to {\text{ }}{H^ + }{\text{ }} + {\text{ }}C{l^ - }$
The concentration of hydrochloric acid will be equal to the concentration of hydrogen and chlorine ions. Therefore we can say that initial concentration of chlorine will be:
$\left[ {HCl} \right]{\text{ = }}\left[ {{H^ + }} \right]{\text{ = }}\left[ {C{l^ - }} \right]{\text{ = 0}}{\text{.15 M}}$
Now when we add $PbC{l_2}$ it will be dissociated as:
$PbC{l_2}{\text{(s) }} \rightleftharpoons P{b^{2 + }}(aq.){\text{ }} + {\text{ }}2C{l^ - }(aq.)$
Since lead chloride is in solid state, therefore ${K_{sp}}$ for above reaction can be written as:
$\Rightarrow {K_{sp}}{\text{ = }}\left[ {P{b^{2 + }}} \right]{\text{ }}{\left[ {C{l^ - }} \right]^2}$
Let say $s$ be the solubility of the $PbC{l_2}$ then at equilibrium it can be represented as:

Time	$\left[ {PbC{l_2}} \right]$	$\left[ {P{b^{2 + }}} \right]$	$2\left[ {C{l^ - }} \right]$
$t = 0$	$1$	$0$	$0.15$
$t = {t_{eq.}}$	$1 - s$	$s$	$0.15{\text{ }} + {\text{ }}2s$

The ${K_{sp}}$ for above reaction will be :
$\Rightarrow {K_{sp}}{\text{ = }}\left[ s \right]{\text{ }}{\left[ {0.15 + 2s} \right]^2}$
Let us assume that $2s \ll 0.15$ then it can be reduced as,
$\Rightarrow {\text{ }}0.15{\text{ }} + {\text{ }}2s{\text{ }} \approx {\text{ }}0.15$
Thus the equation reduced as:
$\Rightarrow {K_{sp}}{\text{ = }}\left[ s \right]{\text{ }}{\left[ {0.15} \right]^2}$
$\Rightarrow \left[ s \right]{\text{ = }}\dfrac{{{K_{sp}}}}{{{{\left[ {0.15} \right]}^2}}}$
On substituting the values we get the result as,
$\Rightarrow \left[ s \right]{\text{ = }}\dfrac{{1.6{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 5}}}}{{0.15{\text{ }} \times {\text{ 0}}{\text{.15}}}}$
$\Rightarrow \left[ s \right]{\text{ = 7}}{\text{.11 }} \times {\text{ 1}}{{\text{0}}^{ - 4}}$
Thus the solubility of $PbC{l_2}$ is ${\text{7}}{\text{.11 }} \times {\text{ 1}}{{\text{0}}^{ - 4}}{\text{ M}}$ . Thus we can say that the condition $2s \ll 0.15$ holds true.

Note:
It must be noted that when lead chloride is being added to a solution of hydrochloric acid, the chlorine s atoms are initially present inside the solution. Therefore the initial concentration of chlorine atoms is $0.15{\text{ M}}$ . Also for finding the solubility product the concentration of solid is not included because solid substance is not soluble. We have assumed $2s \ll 0.15$ to make our calculations easier. If we do not make this assumption then our calculations will be lengthy.