Question

Many ionic compounds are water soluble, so, for example, we can make solutions of sodium iodide, NaI, and lead(II) nitrate, Pb(NO3)2, by simply dissolving the ionic solids in water. In the process of dissolving, the ions from the ionic solutes (sometimes referred to as salts) are split apart and hydrated (surrounded) by water molecules.a. We represent the species in the sodium iodide solution as Na+(aq) and I–(aq), and in the lead(II) nitrate solution as the species Pb2+(aq) and NO3–(aq).b. Complete the balanced overall ionic equation for sodium iodide dissolving in water. Nal(s) -----c. Complete the balanced overall ionic equation for lead(II) nitrate dissolving in water. Pb(NO3)2(s) (3 and 2 are lowered)

Answer 1

Answer :

(b) The balanced overall ionic equation will be,

$NaI(s)\rightarrow Na^+(aq)+I^-(aq)$

(c) The balanced overall ionic equation will be,

$Pb(NO_3)_2(s)\rightarrow Pb^{3+}(aq)+2NO_3^{-}(aq)$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

Part B :

When sodium iodide dissolving in water then it dissociates to give sodium ion and iodide ion.

The balanced overall ionic equation will be,

$NaI(s)\rightarrow Na^+(aq)+I^-(aq)$

Part C :

When lead(II) nitrate dissolving in water then it dissociates to give lead ion and nitrate ion.

The balanced overall ionic equation will be,

$Pb(NO_3)_2(s)\rightarrow Pb^{3+}(aq)+2NO_3^{-}(aq)$