# Complete ionic and net ionic equations

How to use the molecular equation to find the complete ionic and net ionic equation.

## Introduction

As a diligent student of chemistry, you will likely see tons of reactions that occur in water, or perhaps you are here because you are already drowning in them! When ions are involved in a reaction, the chemical equation can be written with various levels of detail. Depending on which part of the reaction you are interested in, you might write a molecular, complete ionic, or net ionic equation.
These terms are particularly useful for double-replacement (a.k.a. exchange), precipitation, gas-forming, and neutralization reactions, though you can technically use them for any reaction that uses water as the solvent.

## Definitions of molecular, complete ionic, and net ionic equations

A molecular equation is sometimes simply called a balanced equation. In a molecular equation, all ionic compounds and acids are represented as neutral compounds using the molecular formula. The state of each substance is indicated in parentheses after the formula.
The state of the substance can be solid ($s$), liquid ($l$), gas ($g$), or dissolved in water ($aq$). Word problems will often tell you the state of the reactants and sometimes the products as well. In cases where you have to figure out the state of the substances yourself, you will often require knowledge of solubility rules.
Let’s consider the reaction that occurs when aqueous solutions of $\text {NaCl}$ and $\text {AgNO}_3$ are mixed. We write the reactants on the left side of the reaction arrow: $\text {NaCl}(aq)$ and $\text {AgNO}_3(aq)$. The reaction produces $\text {NaNO}_3(aq)$ and $\text {AgCl}(s)$. With this information, we can write the molecular equation for this reaction and make sure our equation is balanced!
This is a double-replacement reaction, sometimes called an exchange reaction. It is also a precipitation reaction since we are forming a solid product from ions in solution.

$\text {NaCl}(aq) + \text{AgNO}_3(aq) \rightarrow \text{NaNO}_3(aq) + \text{AgCl}(s)$
If we imagine zooming in on the contents of our reaction beaker, however, we won’t find molecules of $\text {NaCl}$, $\text {AgNO}_3$, or $\text {NaNO}_3$. Since $\text{NaCl}$, $\text{AgNO}_3$, and $\text{NaNO}_3$ are soluble ionic compounds, they dissociate into their constituent ions in water. For example, any molecule of $\text{NaCl}$ will be dissociated into one ion of $\text{Na}^+$ and one ion of $\text{Cl}^-$; these ions are stabilized by ion-dipole interactions with the surrounding water molecules.
In case you haven't learned the solubility rules yet (or need a quick refresher):
• All sodium, potassium, ammonium, and nitrate salts are soluble in water.
• Most chloride salts are soluble except for silver halides.
If you aren't sure how to break apart the soluble ionic compounds into the cation(s) and anion(s), you can check out the article on common polyatomic ions.
Picture of sodium chloride crystal next to picture of chloride and sodium ions dissociation in water. Each chloride ion is interacting with multiple water molecules through the positive dipole of the water, and each sodium ion is interacting with water molecules through the negative dipole of the water, from the oxygen atoms.
Figure 1. Picture of sodium chloride crystal next to chloride and sodium ions resulting from dissociation in water. Each ion is surrounded by polar water molecules. Image from "Salts", by OpenStax Anatomy and Physiology, CC-BY-NC-SA 4.0.
We can write the soluble ionic compounds as dissociated ions to get the complete ionic equation, sometimes simply called the ionic equation:
$\purpleC{\text{Na}^+(aq)} + \text{Cl}^-(aq) + \text{Ag}^+(aq) + \greenD{\text{NO}_3^-(aq)} \rightarrow \purpleC{\text{Na}^+(aq)} + \greenD{\text{NO}_3^- (aq)} + \text{AgCl}(s)$
Notice that we didn’t change the representation of our product $\text{AgCl}(s)$ since $\text{AgCl}$ is insoluble in water. As a general rule, soluble ionic compounds, strong acids, and strong bases should be written as dissociated ions in the complete ionic equation. Insoluble salts and weak acids are written using the neutral molecular formula since they only dissociate to a very small extent in water.
If we take a closer look at our complete ionic equation, we see that $\purpleC{\text{Na}^+(aq)}$ and $\greenD{\text{NO}_3^- (aq)}$ are present on both sides of the reaction arrow. Ionic species such as $\purpleC{\text{Na}^+}$ and $\greenD{\text{NO}_3^-}$ that are not changed in the reaction are called spectator ions. We can cancel out spectator ions from our complete ionic equation since they appear on both sides of the equation, similar to how we cancel out equal elements on either side of a math equation:
$\cancel{\purpleC{{\text{Na}^+(aq)}}} + \text{Cl}^-(aq) + \text{Ag}^+(aq) + \cancel{\greenD{\text{NO}_3^-(aq)} }\rightarrow \cancel{\purpleC{\text{Na}^+(aq)}} + \cancel{\greenD{\text{NO}_3^- (aq)}} + \text{AgCl}(s)$
This gives us our net ionic equation, which includes only the species involved in the chemical reaction:
The net ionic equation tells us that solid silver chloride may be produced from dissolved $\text{Cl}^-$ and $\text{Ag}^+$ ions, regardless of the source of these ions. In comparison, the complete ionic equation tells us about all the ions in solution during your reaction, and the molecular equation tells us about the ionic compounds that were used as sources of and $\text{Ag}^+$.
Problem-solving tip: By convention, the net ionic equation is written with the stoichiometric coefficients as the smallest possible integer values. That means that sometimes we might have to divide all of the stoichiometric coefficients by a common divisor as the last step to get the final form of the net ionic equation.

## Summary

The net ionic equation shows only the chemical species that are involved in a reaction, while the complete ionic equation also includes spectator ions. We can find the net ionic equation using the following steps:
1. Write the balanced molecular equation, including the state of each substance.
2. Use your knowledge of solubility rules, strong acids, and strong bases to rewrite the equation as the complete ionic equation showing which compounds are dissociated into ions.
Strong acids and strong bases will completely dissociate into ions in solution. Any acid or base that is not a strong acid or strong base is called a weak acid (or base).
1. Cancel out the spectator ions that appear on both sides of the equation to get the net ionic equation.
Make sure your net ionic equation is balanced and that the sum of the positive and negative charges is the same on both sides of your equation. If this is not the case, go back to your previous steps to make sure the complete ionic equation and molecular equations have balanced atoms and charges.

Kotz, J. C., Treichel, P. M., Townsend, J. R., and Treichel, D. A. (2015). Net Ionic Equations. In Chemistry and Chemical Reactivity, Instructor's Edition (9th ed., pp. 112-114). Stamford, CT: Cengage Learning.

## Try it!

Sulfuric acid, $\text{H}_2 \text{SO}_4(aq)$, is a strong acid that completely dissociates into $\text H^+$ and $\text {SO}_4^{2-}$ in aqueous solution. Sodium hydroxide, $\text{NaOH}(aq)$, is a strong base that completely dissociates into $\text{Na}^+$ and $\text{OH}^-$ in aqueous solution. When $\text{H}_2 \text{SO}_4(aq)$ and $\text{NaOH}(aq)$ are combined, the products are water and soluble sodium sulfate, $\text{Na}_2 \text{SO}_4(aq)$.
$\text{H}_2 \text{SO}_4(aq)+2\text{NaOH}(aq) \rightarrow 2\text H_2 \text O(l) + \text{Na}_2 \text{SO}_4(aq)$
What is the net ionic equation for the chemical reaction between $\text{H}_2 \text{SO}_4(aq)$ and $\text{NaOH}(aq)$?
$\text{Na}_2 \text{SO}_4$ is soluble in water, and it dissociates into $2\text{Na}^+(aq)$ and $\text{SO}_4^{2-} (aq)$.
$\text{SO}_4^{2-}$ and $\text{Na}^+$ are spectator ions because they remain in solution during the reaction.
$2\text{H}^+ (aq)+\text{SO}_4^{2-}(aq)+2\text{Na}^+ (aq)+2\text{OH}^-(aq) \rightarrow 2\text H_2 \text O + 2\text{Na}^+(aq) + \text{SO}_4^{2-} (aq)$
Once the spectator ions are eliminated from the complete ionic equation and we make sure all the stoichiometric coefficients are the lowest integer value—we can divide them all by $2$—we get the net ionic equation:
$\text{H}^+ (aq)+\text{OH}^-(aq) \rightarrow \text H_2 \text O(l)$