9.9: Complete Ionic and Net Ionic Equations - More Examples

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Learning Objectives

Convert a formula (molecular) equation into a full ionic or net ionic equation.

When converting a formula (molecular) equation into an ionic equation, remember the following guidelines:

Always start with a balanced formula (molecular) equation.
Only soluble ionic compounds dissociate into ions. We will deal with acids--the only significant exception to this guideline--next term.
When disassociating an ionic compound into its component ions, be carefuly not pull apart polyatomic ions. For example, if a compound contains nitrate ions, don't convert NO₃^-(aq) to N^3-(aq) and 3O²^-(aq). Neither nitride ions nor oxide ions are stable in an aqueous environment. Also, examination of the charges should convince you that nitrate is not composed of oxide and nitride ions.
Be careful with ionic compounds that have multiple monatomic ions in a formula unit. For example, CaCl₂ dissociates into one calcium ion and two chloride ions: Ca²⁺(aq) + 2Cl^-(aq). Don't be tempted to write Cl₂^-(aq). The formula Cl₂^- would indicate a polyatomic ion formed from two chlorine atoms that has an overall charge of -1; it does not indicate two chloride ions.

It might be helpful to look at a few more examples.

Example \(\PageIndex 1\)

In the following section, we will examine the reaction that occurs when a solid piece of elemental magnesium in placed in an aqueous solution of copper(II) chloride:

\[ \ce{ CuCl_2(aq) + Mg(s) \rightarrow Cu(s) + MgCl_2(aq)} \]

Write the full ionic and net ionic equations for this reaction.

Solution

The elemental metals (magnesium on the reactant side, copper on the product side) are neutral solids. They therefore appear unaltered in the full ionic equation. Only the aqueous ionic compounds (the two chloride salts) are written as ions:

\[ \ce{ Cu^2+(aq) + 2Cl^- (aq) + Mg(s) \rightarrow Cu(s) + Mg^2+(aq) + 2Cl^- (aq)} \]

The chloride ions are spectator ions. To get the net ionic equation, we cancel them from both sides of the equation:

\[ \ce{ Cu^2+(aq) + Mg(s) \rightarrow Cu(s) + Mg^2+(aq)} \]

Note that when variable-charge metals such as copper appear as part of a compound, we have to determine the charge on the cation by looking at the number of anions and their charge. This is the same process we followed when naming a compound with a variable-charge metal in chapter 4.

Example \(\PageIndex 2\)

When an excess of an aqueous hydroxide salt is added to a solution containing ammonium ions, ammonia gas is formed:

\[ \ce{ 2NH_4Cl(aq) + Ba(OH)_2(aq) \rightarrow 2NH_3(g) + BaCl_2(aq) + 2H_2O(l)} \]

Write the full ionic and net ionic equations for this reaction.

Solution

Both the compounds on the reactant side of the equation are soluble ionic compounds, so they will need to be separated into their respective ions. In this case, both compounds contain a polyatomic ion. Remember, these polyatomic ions maintain their integrity in solution; do not separate them into ions. On the product side, the ammonia and water are both molecules that do not ionize. Only the barium chloride is separated into ions:

\[ \ce{ 2NH_4^+ (aq) + 2Cl^- (aq) + Ba^2+ (aq) + 2OH^- (aq) \rightarrow 2NH_3(g) + Ba^2+ (aq) + 2Cl^- (aq) + 2H_2O(l)} \]

Both the barium ions and the chloride ions are spectator ions. The net ionic equation results from cancelling them from the full ionic equation:

\[ \ce{ 2NH_4^+ (aq) + 2OH^- (aq) \rightarrow 2NH_3(g) + 2H_2O(l)} \]

In the case of this net ionic equation, the stoicheometric coefficients can be reduced by dividing through by two:

\[ \ce{ NH_4^+ (aq) + OH^- (aq) \rightarrow NH_3(g) + H_2O(l)} \]

The mechanism of the reaction becomes more clear by inspecting the net ionic equation: the ammonia molecule is created from the ammonium ion when the hydroxide ion strips a hydrogen away from it. By gaining a hydrogen (and a unit of charge) the hydroxide ion transforms into a water molecule.