Reaction between lead nitrate and sodium iodide
An aqueous solution of lead nitrate $Pb^{2+}$ $+$ $2NO_3^-$ is added to an aqueous solution of sodium iodide $Na^{+}$ $+$ $I^-$. Lead iodide is insoluble, so it precipitates: $Pb^{2+}$ $+$ $2I^-$ $\longrightarrow$ $Pb^{2+}(I^-)_2(s)$ Adding the "spectator" ions (which take not part in the reaction), we have: $Pb^{2+}$ $+$ $2NO_3^-$ $+$ $2Na^+$ $+$ $2I^-$ $\longrightarrow$ $Pb^{2+}(I^-)_2(s)$ $+$ $2Na^+$ $+$ $2NO_3^-$ A precipitate of lead iodide appears (a yellow solid) in a (colourless) solution of sodium nitrate.
When two ions are added which form an insoluble ionic substance, precipitation will occur.