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

Hint: 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.