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Molecular, complete ionic, and net ionic equations


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 AgNO3 and NaCl. When aqueous solutions of AgNO3 and NaCl are mixed, solid AgCl and aqueous NaNO3 are formed. Using this information, we can write a balanced molecular equation for the reaction:
If we could zoom in on the contents of the reaction beaker, though, we wouldn't find actual molecules of AgNO3, NaCl, or NaNO3. Since AgNO3, NaCl, and NaNO3 are soluble ionic compounds, they dissociate into their constituent ions in water. For example, NaCl dissociates into one ion of Na+ for every ion of Cl; 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:
Notice that we didn’t change the representation of AgCl(s) since AgCl 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 NO3(aq) and Na+(aq) 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:
Doing so leaves us with the net ionic equation, which shows only the species actually involved in the chemical reaction:
This net ionic equation tells us that solid silver chloride is produced from dissolved Ag+ and Cl 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 Ag+ and Cl 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.


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, H2SO4(aq), is a strong acid that completely dissociates into H+ and SO42 ions in aqueous solution. Sodium hydroxide, NaOH(aq), is a strong base that completely dissociates into Na+ and OH ions in aqueous solution. When H2SO4(aq) and NaOH(aq) are combined, the products are water and aqueous sodium sulfate, Na2SO4(aq). This reaction is represented by the molecular equation below.
Which of the following is the balanced net ionic equation for the reaction between H2SO4(aq) and NaOH(aq)?
Choose 1 answer:

Want to join the conversation?

  • starky ultimate style avatar for user Daniel
    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?
    (31 votes)
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  • starky ultimate style avatar for user Aud Montrull
    how do you know whether or not the ion is soulable or not?
    (12 votes)
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  • blobby green style avatar for user nik.phatslap
    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.
    (6 votes)
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  • female robot amelia style avatar for user fombahj
    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?
    (8 votes)
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    • aqualine ultimate style avatar for user wchargin
      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).
      (13 votes)
  • piceratops seed style avatar for user Quinn Becker
    Why when you divide 2H+ by two do you get H+, but when you divide 2Na- by two it goes away?
    (3 votes)
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    • orange juice squid orange style avatar for user Jessica
      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
      (19 votes)
  • purple pi teal style avatar for user Siddesh Minde
    What are cation and anions
    (3 votes)
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  • blobby green style avatar for user alsalem318
    how come AgCl doesnt dissolve in water is it not a salt like NaCl? what is the difference between the two that makes one not dissolve in water?
    (2 votes)
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    • leaf red style avatar for user Richard
      While both are sodium chloride and silver chloride are ionic compounds, sodium chloride is soluble and dissolves easily in water while silver chloride is insoluble and does not dissolve much into water. The key is the energy associated with the dissolving process. An energy input is required to break bonds, and energy is subsequently released when bonds are formed. The same is true here.

      For an ionic solid to dissolve, the solvent (in this case water) has to break the ionic bonds then surround the ions and form new ion-dipole bonds in a solvation shell. For the dissolution process to be successful, we must release more energy than is required to break the bonds. If more energy is required to break the bonds than is released during bond formation, then the solid is insoluble. In sodium chloride, the ion-dipole bonds release more energy than was required to break the ionic bonds so it is soluble, but for silver chloride the ion-dipole bonds release less energy than is required to break the ionic bonds so it is insoluble.

      Hope that helps.
      (7 votes)
  • starky seedling style avatar for user Kelli Evans
    I have a question....I am really confused on how to do an ionic equation....Please Help!
    (4 votes)
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    • piceratops seed style avatar for user RogerP
      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.
      (2 votes)
  • leaf green style avatar for user Yu Aoi
    I know this may sound silly, but can we subtract or add a reactant to both sides just like in mathematics?
    (2 votes)
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    • leaf red style avatar for user Richard
      Mathematically it's completely acceptable to do so, however we have to consider the actual chemical makeup of our reaction if we do so. For ionic equations like these it's possible for us to eliminate, essentially subtract out, spectator ions from an equation. So these are ions which are present in the reaction solution, but don't really participate in the actual reaction (they don't change as a product compared to when they were a reactant). So since they're not participating in the reaction, subtract them is allowed because it doesn't affect the reaction if they're absent from the equation.

      In redox reaction it's common to add water, H+, and OH- ion to the equations when balancing them. Now in this case where we're adding chemicals to the equation it may just seem like we're adding chemicals out of convenience to make the math work out. However if it's an aqueous solution, these added chemicals are technically always present in the reaction solution and what we're actually doing is recognizing that some of them are actually part of the redox reaction. So it's not much that we're adding chemicals rather we're discovering the actual reaction occurring.

      So we can add/subtract in chemical equations, but its can't just simply make mathematical sense, it also has to make sense in a chemistry context.

      Hope that helps.
      (5 votes)
  • piceratops ultimate style avatar for user wanglx123456789
    why can the reaction in "Try it" happen at all? There is no solid in the products.
    (2 votes)
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    • leaf red style avatar for user Richard
      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.
      (3 votes)