Question

Write balanced complete ionic and net ionic equations for each reaction. a. \(\operatorname{HCl}(a q)+\operatorname{LiOH}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\operatorname{LiCl}(a q)\) b. \(\mathrm{MgS}(a q)+\mathrm{CuCl}_{2}(a q) \longrightarrow \mathrm{CuS}(s)+\mathrm{MgCl}_{2}(a q)\) c. \(\mathrm{NaOH}(a q)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaNO}_{3}(a q)\) d. \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NiCl}_{2}(a q) \longrightarrow \mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{NaCl}(a q)\)

Short answer

a. Net ionic: \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l). b. Net ionic: \text{Cu}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{CuS}(s). c. Net ionic: \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l). d. Net ionic: 2\text{Ni}^{2+}(aq) + 3\text{PO}_4^{3-}(aq) \rightarrow \text{Ni}_3(\text{PO}_4)_2(s).

Step-by-step solution

Complete Ionic Equation for Reaction a

The first step in writing the complete ionic equation is to separate the strong electrolytes into their constituent ions. Hydrochloric acid (\text{HCl}) and lithium hydroxide (\text{LiOH}) both dissociate completely in water. The complete ionic equation is: \[ \text{H}^+(aq) + \text{Cl}^-(aq) + \text{Li}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) + \text{Li}^+(aq) + \text{Cl}^-(aq) \]

Net Ionic Equation for Reaction a

The net ionic equation includes only the ions and molecules directly involved in the reaction. Spectator ions, like \text{Li}^+ and \text{Cl}^-, which do not participate in the reaction, are omitted. The net ionic equation is: \[ \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \]

Complete Ionic Equation for Reaction b

Separate soluble ionic compounds into their ions. \text{MgS} and \text{CuCl}_2 dissociate completely. However, since \text{CuS} is insoluble, it remains as a solid precipitate and does not dissociate. The complete ionic equation is: \[ \text{Mg}^{2+}(aq) + \text{S}^{2-}(aq) + \text{Cu}^{2+}(aq) + 2\text{Cl}^-(aq) \rightarrow \text{CuS}(s) + \text{Mg}^{2+}(aq) + 2\text{Cl}^-(aq) \]

Net Ionic Equation for Reaction b

The net ionic equation only shows the precipitate formation, excluding the spectator ions \text{Cl}^- and \text{Mg}^{2+}. The net ionic equation is: \[ \text{Cu}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{CuS}(s) \]

Complete Ionic Equation for Reaction c

Write the ions that dissociate in solution. Sodium hydroxide (\text{NaOH}) and nitric acid (\text{HNO}_3) are strong electrolytes that dissociate completely. The complete ionic equation is: \[ \text{Na}^+(aq) + \text{OH}^-(aq) + \text{H}^+(aq) + \text{NO}_3^-(aq) \rightarrow \text{H}_2\text{O}(l) + \text{Na}^+(aq) + \text{NO}_3^-(aq) \]

Net Ionic Equation for Reaction c

Omit the spectator ions (\text{Na}^+ and \text{NO}_3^-). The net ionic equation is: \[ \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \]

Complete Ionic Equation for Reaction d

Disassociate strong electrolytes into their ions, except for insoluble compounds. \text{Na}_3\text{PO}_4 and \text{NiCl}_2 dissociate, but \text{Ni}_3(\text{PO}_4)_2 is insoluble. The complete ionic equation is: \[ 3\text{Na}^+(aq) + \text{PO}_4^{3-}(aq) + 2\text{Ni}^{2+}(aq) + 4\text{Cl}^-(aq) \rightarrow \text{Ni}_3(\text{PO}_4)_2(s) + 6\text{Na}^+(aq) + 4\text{Cl}^-(aq) \]

Net Ionic Equation for Reaction d

Remove the spectator ions (\text{Na}^+ and \text{Cl}^-) to obtain the net ionic equation, which shows the formation of the insoluble compound. The net ionic equation is: \[ 2\text{Ni}^{2+}(aq) + 3\text{PO}_4^{3-}(aq) \rightarrow \text{Ni}_3(\text{PO}_4)_2(s) \]

Key concepts

Complete Ionic Equations

Complete ionic equations are an essential tool in chemistry for understanding the nuances of chemical reactions. These equations provide a more detailed depiction of the process by breaking down soluble ionic compounds into their constituent ions when they dissolve in water. Consider a reaction between hydrochloric acid (\text{HCl}) and lithium hydroxide (\text{LiOH}). When written as a complete ionic equation, each compound is displayed as separate ions like this:

\[\begin{equation}\text{H}^+(aq) + \text{Cl}^-(aq) + \text{Li}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) + \text{Li}^+(aq) + \text{Cl}^-(aq)\end{equation}\]
This allows us to see which ions are actively participating in the reaction, such as the hydrogen ions and hydroxide ions, and which ones remain unchanged. Emphasizing the disassociation of strong electrolytes into ions, complete ionic equations lay the groundwork for identifying the agents of chemical change.

Spectator Ions

Spectator ions are like the audience in a play; they watch the action unfold without participating themselves. These ions remain unchanged and do not take part in the actual chemical reaction. They are present in the reaction mixture but do not affect the outcome, and hence, they are omitted when writing the net ionic equation.

For example, in the reaction between \text{HCl} and \text{LiOH}, lithium (\text{Li}^+) and chloride (\text{Cl}^-) ions are spectator ions. Their presence is accounted for when writing the complete ionic equation, but they are excluded in the net ionic equation:\[\begin{equation}\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)\end{equation}\]
Understanding which ions are spectators is crucial in simplifying equations and focusing on the substances that actually contribute to the reaction's products.

Precipitate Formation

When two aqueous solutions react and produce an insoluble substance, this is called precipitate formation. A precipitate is a solid that emerges from a liquid solution as a result of a chemical reaction.

Distinguishing precipitates is vital in net ionic equations because they highlight an essential change during the reaction. For instance, when \text{MgS} reacts with \text{CuCl}_2, the product \text{CuS} is insoluble and forms a precipitate:\[\begin{equation}\text{Cu}^{2+}(aq) + \text{S}^{2-}(aq) \rightarrow \text{CuS}(s)\end{equation}\]
Here, \text{CuS} precipitates out of the solution, visibly demonstrating the chemical transformation. Through precipitation reactions, we can observe and measure chemical phenomena that might otherwise go unnoticed in completely aqueous environments.