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## AP®︎/College Chemistry

### Unit 4: Lesson 1

Net ionic equations

# Molecular, complete ionic, and net ionic equations

AP.Chem:
TRA‑1 (EU)
,
TRA‑1.B (LO)
,
TRA‑1.B.1 (EK)
,
TRA‑1.B.2 (EK)
,
TRA‑1.B.3 (EK)

## Introduction

As a diligent student of chemistry, you will likely encounter tons of reactions that occur in aqueous solution (perhaps you are already drowning in them!). When ions are involved in a reaction, the equation for the reaction 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.

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

A molecular equation is sometimes simply called a balanced equation. In a molecular equation, any ionic compounds or acids are represented as neutral compounds using their chemical formulas. The state of each substance is indicated in parentheses after the formula.
Let's consider the reaction that occurs between start text, A, g, N, O, end text, start subscript, 3, end subscript and start text, N, a, C, l, end text. When aqueous solutions of start text, A, g, N, O, end text, start subscript, 3, end subscript and start text, N, a, C, l, end text are mixed, solid start text, A, g, C, l, end text and aqueous start text, N, a, N, O, end text, start subscript, 3, end subscript are formed. Using this information, we can write a balanced molecular equation for the reaction:
start text, A, g, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis, plus, start text, N, a, C, l, end text, left parenthesis, a, q, right parenthesis, right arrow, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, plus, start text, N, a, N, O, end text, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis
If we could zoom in on the contents of the reaction beaker, though, we wouldn't find actual molecules of start text, A, g, N, O, end text, start subscript, 3, end subscript, start text, N, a, C, l, end text, or start text, N, a, N, O, end text, start subscript, 3, end subscript. Since start text, A, g, N, O, end text, start subscript, 3, end subscript, start text, N, a, C, l, end text, and start text, N, a, N, O, end text, start subscript, 3, end subscript are soluble ionic compounds, they dissociate into their constituent ions in water. For example, start text, N, a, C, l, end text dissociates into one ion of start text, N, a, end text, start superscript, plus, end superscript for every ion of start text, C, l, end text, start superscript, minus, end superscript; these ions are stabilized by ion-dipole interactions with the surrounding water molecules.
Image of crystalline sodium chloride next to image of chloride and sodium ions dissociated 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.
Sodium chloride dissociates into sodium and chloride ions in water, and these ions become solvated by the highly polar water molecules. Image credit: "Salts: Figure 1" by OpenStax Anatomy and Physiology, CC BY 4.0.
From the molecular formula, we can rewrite the soluble ionic compounds as dissociated ions to get the complete ionic equation:
start text, A, g, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start color #11accd, start text, N, O, end text, start subscript, 3, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, end color #11accd, plus, start color #ca337c, start text, N, a, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, end color #ca337c, plus, start text, C, l, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, right arrow, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, plus, start color #ca337c, start text, N, a, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, end color #ca337c, plus, start color #11accd, start text, N, O, end text, start subscript, 3, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, end color #11accd
Notice that we didn’t change the representation of start text, A, g, C, l, end text, left parenthesis, s, right parenthesis since start text, A, g, C, l, end text is insoluble in water. As a general rule, soluble ionic compounds, strong acids, and strong bases should be separated into their constituent ions in a complete ionic equation, while insoluble salts and weak acids should remain as one unit.
If we take a closer look at our complete ionic equation, we see that start color #11accd, start text, N, O, end text, start subscript, 3, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, end color #11accd and start color #ca337c, start text, N, a, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, end color #ca337c are present on both sides of the reaction arrow. Ionic species that remain unchanged like this during a reaction are called spectator ions. Since spectator ions appear on both sides of the equation, we can cancel them out, similar to how we can cancel out equal terms on either side of a math equation:
start text, A, g, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start cancel, start color #11accd, start text, N, O, end text, start subscript, 3, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, end color #11accd, end cancel, plus, start cancel, start color #ca337c, start text, N, a, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, end color #ca337c, end cancel, plus, start text, C, l, end text, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, right arrow, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis, plus, start cancel, start color #ca337c, start text, N, a, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, end color #ca337c, end cancel, plus, start cancel, start color #11accd, start text, N, O, end text, start subscript, 3, end subscript, start superscript, minus, end superscript, left parenthesis, a, q, right parenthesis, end color #11accd, end cancel
Doing so leaves us with the net ionic equation, which shows only the species actually involved in the chemical reaction:
start text, A, g, end text, start superscript, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, start text, C, l, end text, start superscript, −, end superscript, left parenthesis, a, q, right parenthesis, right arrow, start text, A, g, C, l, end text, left parenthesis, s, right parenthesis
This net ionic equation tells us that solid silver chloride is produced from dissolved start text, A, g, end text, start superscript, plus, end superscript and start text, C, l, end text, start superscript, minus, end superscript ions, regardless of the source of these ions. In comparison, the complete ionic equation tells us about all of the ions present in solution during the reaction, and the molecular equation tells us about the ionic compounds that were used as the sources of start text, A, g, end text, start superscript, plus, end superscript and start text, C, l, end text, start superscript, minus, end superscript for the reaction.
Problem-solving tip: By convention, a net ionic equation is written with the smallest possible integer values for the stoichiometric coefficients. 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

A net ionic equation shows only the chemical species that are involved in a reaction, while a complete ionic equation also includes the spectator ions. We can find the net ionic equation for a given reaction using the following steps:
1. Write the balanced molecular equation for the reaction, including the state of each substance.
2. Using your knowledge of solubility rules, strong acids, and strong bases, rewrite the molecular equation as a complete ionic equation that shows which compounds are dissociated into ions.
3. Identify and cancel out the spectator ions (the ions that appear on both sides of the equation).

## Try it!

Sulfuric acid, start text, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, left parenthesis, a, q, right parenthesis, is a strong acid that completely dissociates into start text, H, end text, start superscript, plus, end superscript and start text, S, O, end text, start subscript, 4, end subscript, start superscript, 2, minus, end superscript ions in aqueous solution. Sodium hydroxide, start text, N, a, O, H, end text, left parenthesis, a, q, right parenthesis, is a strong base that completely dissociates into start text, N, a, end text, start superscript, plus, end superscript and start text, O, H, end text, start superscript, minus, end superscript ions in aqueous solution. When start text, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, left parenthesis, a, q, right parenthesis and start text, N, a, O, H, end text, left parenthesis, a, q, right parenthesis are combined, the products are water and aqueous sodium sulfate, start text, N, a, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, left parenthesis, a, q, right parenthesis. This reaction is represented by the molecular equation below.
$\ce{H2SO4}(aq) + \ce{2NaOH}(aq) \rightarrow \ce{2H2O}(l) + \ce{Na2SO4}(aq)$
Which of the following is the balanced net ionic equation for the reaction between start text, H, end text, start subscript, 2, end subscript, start text, S, O, end text, start subscript, 4, end subscript, left parenthesis, a, q, right parenthesis and start text, N, a, O, H, end text, left parenthesis, a, q, right parenthesis?

## Want to join the conversation?

• Just to be clear, in the problem H and OH are not spectator ions because they form a compound with a covalent bond as a product, rather than one with an ionic bond?
• Yup! To be more specific, they form a covalent molecule as opposed to a soluble ionic compound (if they made an insoluble ionic compound, they would not get cancelled out as spectator ions either).
• how do you know whether or not the ion is soulable or not?
• You need to know the dissociation constant but it is not uncommon for ionic salts to dissolve in water
• In getting the net iconic equation from the above equation, why did we have to get rid of the stoichiometric coefficient in front of each chemical species in the net ionic reaction in order for the answer to be correct?
• I'm assuming that you're talking about the last reaction—H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O (ℓ).

When we break this up into its ions, we get H2 (aq) + SO4(2−) (aq) + 2 Na+ (aq) + 2 OH− (aq) → 2 Na+ = SO4(2−) (aq) + 2 H2O(ℓ). So we can "cancel" the 2 Na+ (aq) and the SO4(2−) (aq) on each side. Okay.

Then the resulting reaction is H2 (aq) + 2 OH− (aq) → 2 H2O (ℓ). But when diatomic hydrogen dissolves in water to form H2 (aq), the protons are separated, so we really just have 2 H+ (aq) + 2 OH− (aq) → 2 H2O (ℓ). Now the factor of two is redundant, so we factor it out to get the result listed: H+ (aq) + OH− (aq) → H2O (ℓ).
• How can we tell if something is a strong base or acid? It seems kind of important to this section, but hasn't really been spoken about until now.
• Memorize the six common strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, and HClO₄.
At this stage, if your acid isn't one of these, it is almost certainly a weak acid.
Most metal hydroxides are strong bases.
If your base isn't one of these, it is probably a weak base.
• Why when you divide 2H+ by two do you get H+, but when you divide 2Na- by two it goes away?
• You're not dividing the 2Na- to make it go away. It goes away because it's a spectator ion (it's unchanged during the reaction so it is present on both sides of the equation and you can cross them out).

You should end up with:
2H+(aq) + 2OH-(aq) --> 2H20

You can then divide all of these by 2 (like simplifying a math problem) to get:
H+(aq) + OH-(aq) --> H20
• What are cation and anions
• Cations are atoms that have lost one or more electrons and therefore have a positive charge.
Anions are atoms that have gained one or more electrons and therefore have a negative charge.
• I have a question....I am really confused on how to do an ionic equation....Please Help!
• Without specific details of where you are struggling, it's difficult to advise. But often, if you don't understand one tutor's presentation then it's worth seeking out other teachers who might explain the topic differently.

If you put ionic equations into YouTube there are lots of videos that may help you.
• why can the reaction in "Try it" happen at all? There is no solid in the products.
• A solid precipitate isn't the only thing you look for in net ionic equations, you also look for neutral covalent compounds like water forming. In acid/base reaction it's common for the H+, OH-, and H2O to be the only species left in a net ionic equation after all the other spectator ions have been eliminated. Hope this helps.
• How can you tell which ions will react with which to produce a compound that won't dissociates, which turns into a solid, hence, it won't be canceled out? Like the example above, how do you know that AgCl is a solid and not NaNO3?