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. 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).

A) Complete the balanced overall ionic equation for sodium iodide dissolving in water. NaI(s)...

b) Complete the balanced overall ionic equation for lead(II) nitrate dissolving in water. Pb(NO3)2(s)

c)What will happen if we combine the solution of sodium iodide and the solution of lead(II) nitrate? The solutions will mix, so all of the ions will be evenly distributed throughout the entire solution. But sometimes mixing solutions containing dissolved ions leads to a precipitation reaction: the formation of a new insoluble compound. Based on the information presented in that table, complete the balanced overall ionic equation for the mixing of these two solutions. 2Na^+(aq)+2I^-(aq)+pB^2+(aq)+2NO^-3(aq)

Salts are electrolytes so they dissociate in water to form ions. The overall ionic equation includes all the ions and the species that do not dissociate in water.

A) Complete the balanced overall ionic equation for sodium iodide dissolving in water.

NaI(s) → Na⁺(aq) + I⁻(aq)

B) Complete the balanced overall ionic equation for lead(II) nitrate dissolving in water.

Pb(NO₃)₂ → Pb²⁺(aq) + 2 NO₃⁻(aq)

C) What will happen if we combine the solution of sodium iodide and the solution of lead(II) nitrate?

This is a double displacement reaction, in which insoluble PbI₂ is formed.

2 Na⁺(aq) + 2 I⁻(aq) + Pb²⁺(aq) + 2 NO₃⁻(aq) → PbI₂(s) + 2 Na⁺(aq) + 2 NO₃⁻(aq)