Question

$Pb{Cl}_2$ is insoluble in cold water. Addition of $HCl$ increases its solubility due to:
a) formation of soluble complex anions like ${[Pb{Cl}_3]}_{-}$
b) oxidation of Pb(II) to Pb(IV)
c) formation of ${[Pb{(H_{2}O)}_6]}_{2+}$
d) formation of polymeric lead complexes

Answer 1

Hint: $Pb{Cl}_2$ is a compound that is insoluble in cold water but the addition of $HCl$ provides the ${Cl}^{-}$ ions that helps in overcoming the attraction forces and hence increases the solubility of $Pb{Cl}_2$.

Complete step by step answer:
Lead(II) chloride $(Pb{Cl}_2)$ is an inorganic compound which is a white solid under ambient conditions. It is one of the most important lead-based reagents. It is poorly soluble in water. And it is insoluble in cold water.

This is due to the following reason:
We know that, for a solid to dissolve in a solvent sufficient energy must be available to overcome the force of attraction between the ions in the lattice. This is termed the lattice energy.Much of the energy required to overcome this lattice energy comes from the solvent molecules. In order to extract this energy from the solvent there must be an interaction between the solvent (water in this case) and the ions ${Pb}^{2+}$ and ${Cl}^{-}$for $Pb{Cl}_2$.

The water molecule is polar with the H atoms being positive and thus are attracted to the ${Cl}^{-}$ ions and the O atom is negative and so it is attracted to the ${Pb}^{2+}$ ions. The Chloride anion due to its high electronegativity and high charge/radius ratio or e/r ratio has a strong attraction to water.

As a result, although water molecules are attracted and “trapped”, the interaction is insufficient to release sufficient energy to overcome the lattice energy to any great extent.
As a result, the solid $Pb{Cl}_2$ does dissolve but only slightly at room temperature (about $11g/{dm}^3$ at 20degC).

But, when $HCl$ is added to $Pb{Cl}_2$, its solubility increases. It is due to the formation of soluble complex anions like ${[Pb{Cl}_3]}_{-}$ which are formed due to the reaction between $Pb{Cl}_2$ and $HCl$. The reaction may be given as follows:

${Pb{Cl}_2}_{(s)}+2{Cl}^{-}\underset{Cool}{\overset{}{\to }}{{[Pb{Cl}_3]}^{-}}_{(aq)}$

${Pb{Cl}_2}_{(s)}+2{Cl}^{-}\underset{ExcessofHCl}{\overset{}{\to }}{{[Pb{Cl}_4]}^{2-}}_{(aq)}$
Thus, the addition of excess amount of ${Cl}^{-}$ ions changes the $Pb{Cl}_2$ as a soluble complex of ${[Pb{Cl}_4]}^{2-}$. Hence, it becomes soluble.

Therefore, option (a) is the correct answer.

Note: Although Pb changes its oxidation state from +2 to +4 due to the addition of $HCl$, it is not the reason that Pb becomes soluble in water. The main reason is the formation of a soluble complex anion ${[Pb{Cl}_3]}^{-}$.