Complete Ionic Equations: Study with Video Lessons, Practice Problems & Examples

Complete ionic equations represent aqueous compounds as their constituent ions, while solids, liquids, and gases remain intact. To derive a net ionic equation, spectator ions—those present in both reactants and products—are removed from the complete ionic equation. This process begins with a molecular equation, leading to the complete ionic equation, and finally to the net ionic equation. Understanding these relationships is crucial for analyzing chemical reactions, particularly in acid-base reactions and precipitation reactions, where identifying active ions is essential for predicting outcomes.

Complete Ionic Equations

Complete ionic equations show aqueous compounds as associated ions. Now remember, in a molecular equation we have our aqueous compounds, but we also possess solids, liquids, and gases. When it comes to solid, liquids, and gases, they themselves never break up into ions. It's only the aqueous compounds. And remember, we use the solubility rules to determine if something is aqueous or not. Now when it comes to the complete ionic equation, we're going to say it comes from the molecular equation, and because of that, you're going to have to remember to distribute the coefficient of each compound to determine the correct number of ions. So as we go into going from the molecular equation to our complete ionic equations, we'll see what that entails.

Complete Ionic Equations Example 1

In this example question, it says, convert the following molecular equation into a complete ionic equation. So here we have 3 moles of calcium bromide aqueous reacting with 2 moles of lithium phosphate aqueous, producing 6 moles of lithium bromide as an aqueous compound plus 1 mole of calcium phosphate solid. Alright. So remember, we can only break up aqueous compounds. So only these 3 compounds will break up. If you are solid, a liquid, or gas, you will not break up in the complete ionic equation. Also, remember that the coefficient gets distributed to each one of these compounds. Now what exactly does that mean? Well, here that means that this 3 is going to get distributed to the number of calciums and the number of bromines. So it gets distributed to Ca and Br. When this breaks up into its ions, we know that we're going to have Ca²⁺(aq) plus Br^-(aq). Then we distribute the coefficient, so that's going to give us 3 calcium ions. And then this is 3 times 2, 6 bromide ions. Plus, also remember that when we have an ion it's in aqueous phase when within a solution. This 2 is going to get distributed to here and to here, so we know that's going to give us 6 lithium ions because it is 2 times 3 plus 2 phosphate ions. Remember phosphate is a polyatomic ion, produces 6 gets distributed to lithium and bromide, so 6 lithium plus 1 ions, plus 6 bromide ions. This is a solid, so it stays together. It does not break up into ions. So plus, 1, so 1, not 6, 1 calcium phosphate solid. So this here represents our complete ionic equation. Remember to break up only aqueous compounds and remember to distribute the ions to each one of well, distribute the number of coefficients to each one of those ions that you form.

Complete Ionic Equations

Now a net ionic equation shows the ions participating in the chemical reaction by removing what we call the spectator ions. These spectator ions represent compounds that are both reactants and products. So they're found on both sides of an equation, but they are not part of a net ionic equation. Now when it comes to the net ionic equation, we're going to say it comes from the complete ionic equation. So remember, we started out with our molecular equation. From there, we were able to determine our complete ionic equation. And then from our complete ionic equation, we now are looking at our net ionic equation. So, this is the path you have to take to get to our net ionic equation. Now that we know how they're connected, let's move on to the next video and take a look at an example question.

Complete Ionic Equations Example 2

Based on the given reactants, provide both the molecular equation and the complete ionic equation. So here we have ammonium sulfate reacting with calcium chloride. So for step 0, we're just going to follow steps 1 to 4 that we've learned in the past to first give the molecular equation. Now remember, when it comes to these steps, we're first going to break up each of these compounds into their ionic forms. So ammonium sulfate breaks up into the ammonium ion (NH₄⁺) and the sulfate ion (SO₄^2-). Don't worry about the fact that we have 2 ammoniums here. We'll discuss that later on as we're going into the complete ionic equation. And then calcium chloride breaks up into calcium ion (Ca²⁺) and chloride ion (Cl^-). Now remember, we're gonna swap ionic partners because opposite charges attract. This ⁺ charge is attracted to this ^- charge. So when they connect together, remember when the numbers are the same, they just simply cancel out. So here that will give me NH₄Cl, then we're gonna have here calcium ion and sulfate ion. The numbers and the charges again are the same, so they're going to cancel out to give me calcium sulfate (CaSO₄). At this point, we have to remember our solubility rules to see if we created a solid, liquid, or gas. Otherwise, no reaction has occurred. So remember based on our solubility rules, anything connected to the ammonium ion will be soluble. So this is aqueous. And then we have calcium sulfate. Based on the rules that we learned about calcium sulfate, remember it's gonna form a precipitate if sulfate ion is connected to CBS. Remember, CBS stands for calcium, barium, or strontium. Since sulfate here is connected to calcium, it's gonna form a precipitate. Since we made a solid, a reaction has occurred.

Next, we just have to balance out this molecular equation. We have 2 ammoniums here and 2 chlorines here, so I have to put a 2 right here. The sulfate, there's just 1 sulfate and 1 sulfate, 1 calcium and 1 calcium. So here we have coefficients of 1 for the other compounds. Now we're going to go to step 5, where we break up only the aqueous compounds into their respective ions. So everyone breaks up into ions except for calcium sulfate, which is a solid. So here remember the coefficient gets distributed to each of these ions. We're going to have here 1 ammonium ion aqueous. But remember, there's a little 2 here so there's actually 2 Ammonium ions plus we're going to have Sulfate ion aqueous plus 1 calcium ion aqueous plus, we have a 2 here, 2 chloride ions aqueous gives us the 2 gets distributed to each one of these ions here, so we're gonna have 2 ammonium ions aqueous plus 2 chloride ions aqueous, plus 1 calcium sulfate solid.

Here we're gonna cancel out the spectator ions from the complete ionic equation in order to isolate the net ionic equation. At this point, what we have is the complete ionic equation. For the net ionic, let's remove the spectator ions. Remember that spectator ions are the compounds that exist as both reactants and products at the same time. So here, ammonium ion here matches up with ammonium ion here. Those will not be part of the net ionic equation. Also, we have here 2 chlorides and 2 chlorides. They also will not be part of the net ionic equation. That means all that's left at the end when we bring things down will be SO42- aqueous, plus Ca ion, aqueous, gives CaSO4 solid. So this here represents my net ionic equation. So just remember, a lot of what we learned in terms of obtaining the net ionic equation has to first deal with determining the molecular equation. From there, we have to look at the complete ionic equation, and by removing the spectator ions, that's how we're able to isolate our net ionic equation at the very end.

Provide the net ionic equation that occurs when the following aqueous compounds are mixed together: 

Copper (II) Bromide and Lithium Hydroxide

Here in this practice question, it states: Provide the net ionic equation that occurs when the following aqueous compounds are mixed together. So, they tell us here that copper(II) bromide and lithium hydroxide are aqueous compounds. Here, copper(II) means we have Cu2+. Bromide is in group 7A, so it has a -1 charge. We have lithium, which is in group 1A, so it has a +1 charge. Hydroxide is a polyatomic ion with a -1 charge. So, for our reactants, what we initially have is CuBr2 (aqueous) plus 2 LiOH (aqueous). Now, we need to determine what our products will be. So, we switch partners: the -1 from bromide and the +1 from lithium, attracted to one another, combine to form lithium bromide which, based on our solubility rules, is aqueous. Then we have the copper(II) ion and the hydroxide ion combining. The charges are different, so they crisscross, giving us Cu(OH)2. Hydroxide typically forms a precipitate except with certain ions, none of which are present here. Therefore, Cu(OH)2 is a solid. This represents our molecular equation. Next, we convert our molecular equation into a complete ionic equation. Only the aqueous compounds are broken up: Equation: Cu2+ (aqueous) + 2 Br- (aqueous) + 2 Li+ (aqueous) + 2 OH- (aqueous) → 2 Li+ (aqueous) + 2 Br- (aqueous) + Cu(OH)2 (solid) For the net ionic equation, remove the spectator ions (those that appear in both reactants and products in the same form). What's left is: Cu2+ + 2 OH- → Cu(OH)2 This represents the net ionic equation based on the two aqueous compounds given to us.

Which of the following reagents could be used to separate the two anions from a solution containing magnesium nitrate and cesium hydroxide?

• A. NH₄CN
• B. NaCl
• C. KNO₃
• D. ZnBr₂
• E. CsBrO₃

Which of the following reagents could be used to separate the two cations from a solution containing Lead (IV) acetate and cesium permanganate?

• A. Sr(NO₃)₂
• B. TiC₂H₃O₂
• C. K₂S
• D. NaClO₄
• E. KNO₃

Here’s what students ask on this topic:

What is a complete ionic equation and how is it different from a molecular equation?

A complete ionic equation represents all of the ions present in an aqueous solution, showing aqueous compounds as their constituent ions. In contrast, a molecular equation shows the compounds in their complete, undissociated form. For example, the molecular equation for the reaction between sodium chloride and silver nitrate is:

${\mathrm{NaCl}}_{(aq)}+{\mathrm{AgNO}}_{(aq)}\to {\mathrm{NaNO}}_{(aq)}+{\mathrm{AgCl}}_{(s)}$

The complete ionic equation for the same reaction is:

${\mathrm{Na}}_{(aq)}+{\mathrm{Cl}}_{(aq)}+{\mathrm{Ag}}_{(aq)}+{\mathrm{NO}}_{(aq)}\to {\mathrm{Na}}_{(aq)}+{\mathrm{NO}}_{(aq)}+{\mathrm{AgCl}}_{(s)}$

How do you write a complete ionic equation from a molecular equation?

To write a complete ionic equation from a molecular equation, follow these steps:

1. Write the balanced molecular equation.
2. Identify the aqueous compounds and dissociate them into their constituent ions.
3. Keep solids, liquids, and gases in their molecular form.
4. Distribute the coefficients to each ion to ensure the correct number of ions is represented.

For example, for the reaction between barium chloride and sodium sulfate:

Molecular equation:

${\mathrm{BaCl}}_{(aq)}+{\mathrm{Na}}_{(aq)}\to {\mathrm{BaSO}}_{(s)}+{\mathrm{NaCl}}_{(aq)}$

Complete ionic equation:

${\mathrm{Ba}}_{(aq)}+{\mathrm{Cl}}_{(aq)}+{\mathrm{Na}}_{(aq)}+{\mathrm{SO}}_{(aq)}\to {\mathrm{BaSO}}_{(s)}+{\mathrm{Na}}_{(aq)}+{\mathrm{Cl}}_{(aq)}$

What are spectator ions and how do you identify them in a complete ionic equation?

Spectator ions are ions that appear on both the reactant and product sides of a complete ionic equation but do not participate in the actual chemical reaction. To identify them, follow these steps:

1. Write the complete ionic equation.
2. Compare the ions on both sides of the equation.
3. Identify the ions that remain unchanged in both the reactants and products.

For example, in the reaction between sodium chloride and silver nitrate:

Complete ionic equation:

${\mathrm{Na}}_{(aq)}+{\mathrm{Cl}}_{(aq)}+{\mathrm{Ag}}_{(aq)}+{\mathrm{NO}}_{(aq)}\to {\mathrm{Na}}_{(aq)}+{\mathrm{NO}}_{(aq)}+{\mathrm{AgCl}}_{(s)}$

Here, Na⁺ and NO₃⁻ are spectator ions because they appear unchanged on both sides of the equation.

How do you derive a net ionic equation from a complete ionic equation?

To derive a net ionic equation from a complete ionic equation, follow these steps:

1. Write the complete ionic equation.
2. Identify and remove the spectator ions, which are ions that appear unchanged on both sides of the equation.
3. Write the remaining ions that participate in the reaction.

For example, for the reaction between sodium chloride and silver nitrate:

Complete ionic equation:

${\mathrm{Na}}_{(aq)}+{\mathrm{Cl}}_{(aq)}+{\mathrm{Ag}}_{(aq)}+{\mathrm{NO}}_{(aq)}\to {\mathrm{Na}}_{(aq)}+{\mathrm{NO}}_{(aq)}+{\mathrm{AgCl}}_{(s)}$

Net ionic equation:

${\mathrm{Ag}}_{(aq)}+{\mathrm{Cl}}_{(aq)}\to {\mathrm{AgCl}}_{(s)}$

Why are complete ionic equations important in chemistry?

Complete ionic equations are important in chemistry because they provide a detailed view of the ions present in a reaction, helping to understand the behavior of substances in aqueous solutions. They are particularly useful in identifying the actual species involved in chemical reactions, such as in acid-base reactions and precipitation reactions. By showing the dissociation of aqueous compounds into ions, complete ionic equations help predict the formation of products, identify spectator ions, and derive net ionic equations. This detailed understanding is crucial for analyzing reaction mechanisms, predicting reaction outcomes, and solving problems in various fields of chemistry.