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.

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 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

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?

it depends on how much is the product soluble in the solvent in which your reaction occurs. for example in water, AgCl is not very soluble so it will precipitate. NaNo3 is very soluble in water and it will dissociate into Na+ and NO3-.

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.

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.

I know this may sound silly, but can we subtract or add a reactant to both sides just like in mathematics?

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.

Main content

Chemistry library

Course: Chemistry library > Unit 5

Lesson 5: Types of chemical reactions

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)

Google Classroom

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.

[Huh?]

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

[What kind of reaction is this?]

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.

[I don't get it!]

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:

Write the balanced molecular equation for the reaction, including the state of each substance.
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.
[I haven't learned about strong acids and bases yet!]
Identify and cancel out the spectator ions (the ions that appear on both sides of the equation).
[Want to double check your work?]

[Attributions and references]

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?

(Choice A)
$\ce{H2SO4}(aq) + \ce{2OH-}(aq) \rightarrow \ce{2H2O}(l) + \ce{SO4^2-}(aq)$
(Choice B)
$\ce{SO4^2-}(aq) +\ce{2Na+}(aq) \rightarrow \ce{Na2SO4}(aq)$
(Choice C)
$\ce{HSO4-}(aq) \rightarrow \ce{H+}(aq) + \ce{SO4^2-}(aq)$
(Choice D)
$\ce{H+}(aq) + \ce{OH-}(aq) \rightarrow \ce{H2O}(l)$

[Hint 1]

[Hint 2]

[Hint 3]

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Daniel
Posted 8 years ago. Direct link to Daniel's post “Just to be clear, in the ...”
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?
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(27 votes)
Answer
- yuki
  Posted 7 years ago. Direct link to yuki's post “Yup! To be more specific,...”
  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).
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  (42 votes)
Audrey Harmon-Montrull
Posted 7 years ago. Direct link to Audrey Harmon-Montrull's post “how do you know whether o...”
how do you know whether or not the ion is soulable or not?
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(11 votes)
Answer
- Matt B
  Posted 7 years ago. Direct link to Matt B's post “You need to know the diss...”
  You need to know the dissociation constant but it is not uncommon for ionic salts to dissolve in water
  Comment on Matt B's post “You need to know the diss...”
  (12 votes)
fombahj
Posted 7 years ago. Direct link to fombahj's post “In getting the net iconic...”
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?
Button navigates to signup pageComment on fombahj's post “In getting the net iconic...”
(7 votes)
Answer
- wchargin
  Posted 7 years ago. Direct link to wchargin's post “I'm assuming that you're ...”
  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).
  Button navigates to signup page
  (13 votes)
nik.phatslap
Posted 7 years ago. Direct link to nik.phatslap's post “How can we tell if someth...”
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.
Button navigates to signup pageComment on nik.phatslap's post “How can we tell if someth...”
(5 votes)
Answer
- Ernest Zinck
  Posted 7 years ago. Direct link to Ernest Zinck's post “Memorize the six common s...”
  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.
  Button navigates to signup page
  (17 votes)
Quinn Becker
Posted 8 years ago. Direct link to Quinn Becker's post “Why when you divide 2H+ b...”
Why when you divide 2H+ by two do you get H+, but when you divide 2Na- by two it goes away?
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(3 votes)
Answer
- Jessica
  Posted 8 years ago. Direct link to Jessica's post “You're not dividing the 2...”
  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
  Comment on Jessica's post “You're not dividing the 2...”
  (19 votes)
Siddesh Minde
Posted 7 years ago. Direct link to Siddesh Minde's post “What are cation and anion...”
What are cation and anions
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(2 votes)
Answer
- Ernest Zinck
  Posted 7 years ago. Direct link to Ernest Zinck's post “Cations are atoms that ha...”
  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.
  Comment on Ernest Zinck's post “Cations are atoms that ha...”
  (6 votes)
Kelli Evans
Posted 5 years ago. Direct link to Kelli Evans's post “I have a question....I am...”
I have a question....I am really confused on how to do an ionic equation....Please Help!
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(4 votes)
Answer
- RogerP
  Posted 5 years ago. Direct link to RogerP's post “Without specific details ...”
  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.
  Button navigates to signup page
  (2 votes)
wanglx123456789
Posted 2 years ago. Direct link to wanglx123456789's post “why can the reaction in "...”
why can the reaction in "Try it" happen at all? There is no solid in the products.
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(2 votes)
Answer
- Richard
  Posted 2 years ago. Direct link to Richard's post “A solid precipitate isn't...”
  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.
  Comment on Richard's post “A solid precipitate isn't...”
  (3 votes)
minhthuhoang2000
Posted 6 years ago. Direct link to minhthuhoang2000's post “How can you tell which io...”
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?
Button navigates to signup pageComment on minhthuhoang2000's post “How can you tell which io...”
(3 votes)
Answer
- skofljica
  Posted 6 years ago. Direct link to skofljica's post “it depends on how much is...”
  it depends on how much is the product soluble in the solvent in which your reaction occurs. for example in water, AgCl is not very soluble so it will precipitate. NaNo3 is very soluble in water and it will dissociate into Na+ and NO3-.
  Button navigates to signup page
  (2 votes)
Yu Aoi
Posted a year ago. Direct link to Yu Aoi's post “I know this may sound sil...”
I know this may sound silly, but can we subtract or add a reactant to both sides just like in mathematics?
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(2 votes)
Answer
- Richard
  Posted a year ago. Direct link to Richard's post “Mathematically it's compl...”
  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.
  Button navigates to signup page
  (3 votes)