A standard molecular equation describes a chemical change but often fails to reflect how substances behave when dissolved in water. In aqueous solutions, many compounds dissociate into charged ions, which are the true participants in the chemical process. Ionic equations represent these reactions more accurately by focusing on the species that undergo transformation. This approach distinguishes between compounds that actively react and those that merely remain dissolved. Chemical reactions are represented in three main forms: the molecular equation, the complete ionic equation, and the net ionic equation.
Preparing the Equation: Molecular Form and Electrolytes
Writing an ionic equation begins with a correctly formulated and balanced molecular equation. This equation shows all reactants and products as electrically neutral compounds. It must include the physical state of every substance using symbols: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous (dissolved in water).
The next step is classifying each aqueous substance as a strong electrolyte, a weak electrolyte, or a non-electrolyte. This distinction determines which species will be separated into ions. Strong electrolytes (strong acids, strong bases, and most soluble salts) dissociate completely and must be written in their separated ionic form. Weak electrolytes, such as weak acids, only partially dissociate, remaining mostly as intact molecules.
Non-electrolytes, typically molecular compounds, do not produce ions when dissolved. A compound’s ability to dissociate is tied to its solubility, making general solubility rules essential for conversion. For example, all compounds containing nitrate ions (\(\text{NO}_3^-\)) or alkali metal ions (Group 1) are reliably soluble and are strong electrolytes.
Halide ions (\(\text{Cl}^-\), \(\text{Br}^-\), \(\text{I}^-\)) are generally soluble, except when paired with \(\text{Pb}^{2+}\), \(\text{Ag}^+\), or \(\text{Hg}_2^{2+}\). Most sulfate ions (\(\text{SO}_4^{2-}\)) are soluble, except those with \(\text{Ba}^{2+}\), \(\text{Sr}^{2+}\), and \(\text{Pb}^{2+}\). Conversely, compounds containing carbonate (\(\text{CO}_3^{2-}\)), phosphate (\(\text{PO}_4^{3-}\)), or hydroxide (\(\text{OH}^-\)) are generally insoluble, unless combined with alkali metals or the ammonium ion (\(\text{NH}_4^+\)).
Converting to the Complete Ionic Equation
After balancing the molecular equation and identifying all aqueous species, the next step is writing the complete ionic equation. This is done by separating only the strong electrolytes marked (aq) into their constituent ions. Compounds classified as solids (s), liquids (l), gases (g), or weak electrolytes must remain together as intact molecules.
When separating a strong electrolyte, the formula is split into its cation and anion, assigning the correct charge to each. Stoichiometry is important; any coefficient preceding a compound must be distributed to all ions produced. For example, the balanced molecular equation for lead(II) nitrate and potassium iodide is \(\text{Pb}(\text{NO}_3)_2(\text{aq}) + 2\text{KI}(\text{aq}) \rightarrow \text{PbI}_2(\text{s}) + 2\text{KNO}_3(\text{aq})\).
In this example, \(\text{Pb}(\text{NO}_3)_2\) splits into one \(\text{Pb}^{2+}\) and two \(\text{NO}_3^-\). The \(2\text{KI}\) splits into two \(\text{K}^+\) and two \(\text{I}^-\). The product \(\text{PbI}_2\) is an insoluble solid and remains intact. The soluble product \(2\text{KNO}_3\) splits into two \(\text{K}^+\) and two \(\text{NO}_3^-\). Writing all separated ions results in the complete ionic equation, which shows every species present in the solution.
Isolating the Reaction: Writing the Net Ionic Equation
The complete ionic equation must be simplified to isolate only the species directly involved in the chemical transformation. The final step is identifying and removing the spectator ions. Spectator ions are those that appear identically on both the reactant and product sides of the complete ionic equation.
Spectator ions remain dissolved in the water to maintain electrical neutrality but do not participate in forming the precipitate, liquid, or gas. In the example, the complete ionic equation is \(\text{Pb}^{2+}(\text{aq}) + 2\text{NO}_3^-(\text{aq}) + 2\text{K}^+(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow \text{PbI}_2(\text{s}) + 2\text{K}^+(\text{aq}) + 2\text{NO}_3^-(\text{aq})\). Comparing both sides shows that \(\text{K}^+\) and \(\text{NO}_3^-\) are the spectators, as they remain aqueous and unchanged.
Canceling these spectator ions leaves only the reacting species, forming the net ionic equation: \(\text{Pb}^{2+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow \text{PbI}_2(\text{s})\). A final check ensures the net ionic equation is balanced both by mass and by charge. Mass is balanced with one lead atom and two iodine atoms on each side, and the charge is balanced because the total charge on the reactant side (\(+2 + 2(-1) = 0\)) equals the charge on the product side (0).