Answer:
[tex]2Cl^{-}(aq.)+Pb^{2+}(aq.)\rightarrow PbCl_{2}(s)[/tex]
Explanation:
KCl, [tex]Pb(NO_{3})_{2}[/tex] and [tex]KNO_{3}[/tex] are strong electrolytes. Hence they are fully ionized in aqueous solution.
[tex]PbCl_{2}[/tex] is a sparingly soluble salt. Hence it remains undissociated in aqueous solution.
So, total ionic equation:
[tex]2K^{+}(aq.)+2Cl^{-}(aq.)+Pb^{2+}(aq.)+2NO_{3}^{-}(aq.)\rightarrow PbCl_{2}(s)+2K^{+}(aq.)+2NO_{3}^{-}(aq.)[/tex]
Net ionic equation is written by omitting spectator ions from total ionic equation.
Here, [tex]K^{+}(aq.)[/tex] and [tex]NO_{3}^{-}(aq.)[/tex] ions are spectator ions as they remain present on both side of total ionic equation.
So, net ionic equation:
[tex]2Cl^{-}(aq.)+Pb^{2+}(aq.)\rightarrow PbCl_{2}(s)[/tex]
So, option (b) is correct.
