## Chemistry: The Molecular Science (5th Edition)

The equilibrium would favor the formation of $PbCl_2$, due to the common ion effect.
This is the equilibrium reaction: $PbCl_2(s) \lt -- \gt Pb^{2+} (aq) + 2Cl^-(aq)$ When we add $NaCl$, the common ion effect (because it has $Cl^{-}$ ions) will act, and the reaction will be more reactant-favored. That follows the Le Chatelier's principle. Since we are raising the concentration of one of the products, the equilibrium will try to reduce this concentration, to reduce the change.