Q19.94P - Chemistry: Molecular Nature Of Matter And Change

Question

Consider the dissolution of $\mathbf{\text{PbS}}$ in water:

 $\mathbf{PbS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{HS}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$

Adding aqueous $\mathbf{\text{NaOH}}$  causes more $\mathbf{\text{PbS}}$ to dissolve. Does this violate Le Chatelier’s principle? Explain.

Step-by-Step Solution

Answer

In the reaction ${\text{PbS(s) + H}}_{\text{2}}\text{O}\iff {\text{Pb}}^{\text{2 + }}{\text{(aq) + HS}}^{\text{ - }}{\text{(aq) + OH}}^{\text{ - }}\text{(aq)}$;  when aqueous $\mathrm{NaOH}$  is added   is formed which allows more  $\mathrm{PbS}$ to dissolve. This does not violate the Le Chatelier’s principle.

Concept Introduction

According to Le Chatelier's principle, if a system in equilibrium is disturbed, it moves in a direction to nullify the effect of the disturbance and re-establish a new equilibrium.

Equation fo r PbS

First, the dissolution equation for $\mathrm{PbS}$ is given below.

$\mathrm{PbS}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\iff {\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{HS}}^{-}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

According to Le Chatelier's principle, when the concentration of reactants is increased or the concentration of products is decreased, the reaction goes toward the products (to the right). It is similar the opposite way, if the concentration of reactants is decreased or the concentration of products is increased, the reaction goes toward the reactants (to the left).

Addition of NaOH

If we add aqueous $\mathrm{NaOH}$  to this solution, we would expect the reaction to shift to the left, because of the increased concentration of  ${\text{OH}}^{\text{ - }}$ ions, but this doesn't happen and  $\mathrm{PbS}$ becomes easier to dissolve. This happens because dissolved   ion forms a complex with  ${\text{OH}}^{\text{ - }}$ ions –

 ${\text{Pb}}^{\text{2 + }}{\text{(aq) + 2OH}}^{\text{ - }}\text{(aq)}\iff {\text{Pb(OH)}}_{\text{2}}$

This complex can precipitate and leave the solution, allowing more   $\mathrm{PbS}$ to dissolve, so this does not violates the Le Chatelier's principle.

 

Therefore, violation of Le Chatelier's principle does not take place.