• # question_answer 52) When solid lead iodide is added to water, the  equilibrium concentration of ${{\text{I}}^{-}}$becomes $2.6\times {{10}^{-3}}\,M.$ What is the ${{K}_{sp}}$for $Pb{{I}_{2}}?$ A) $2.2\times {{10}^{-9}}$                B) $8.8\times {{10}^{-9}}$C) $1.8\times {{10}^{-8}}$                D) $3.5\times {{10}^{-8}}$

The equilibrium is $Pb{{l}_{2}}\Leftrightarrow P{{b}^{2}}+2{{I}^{-}}$ On the basis of this equation, the concentration of $P{{b}^{2+}}$ions will be half of the concentration of ${{\text{I}}^{-}}$ions. Thus, $[{{I}^{-}}]=2.6\times {{10}^{-3}}M$and$[P{{b}^{2+}}]=1.3\times {{10}^{-3}}M.$  ${{K}_{sp}}=[P{{b}^{2+}}]{{[{{I}^{-}}]}^{2}}=[1.3\times {{10}^{-3}}M]{{[2.6\times {{10}^{-3}}M]}^{2}}$ ${{K}_{sp}}=8.8\times {{10}^{-9}}.$ Hence, the correct option is [b].