Complete Ionic Equations - Online Tutor, Practice Problems & Exam Prep

Understanding ionic equations is crucial for grasping the nuances of chemical reactions in aqueous solutions. In a molecular equation, all reactants and products are listed, including aqueous compounds, solids, liquids, and gases. However, only aqueous compounds dissociate into ions. Using solubility rules helps identify which compounds are aqueous. A complete ionic equation, derived from the molecular equation, breaks down aqueous compounds into their respective ions, applying coefficients to determine the correct number of ions present. The net ionic equation further refines this by eliminating spectator ions—ions that appear unchanged on both the reactant and product sides of the reaction—highlighting the ions that actively participate in the chemical transformation. This progression from molecular to complete to net ionic equations reveals the underlying ionic activity driving the observable chemical changes.

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 two moles of lithium phosphate aqueous producing 6 moles of lithium bromide as an aqueous compound plus one mole of calcium phosphate solid.

All right, so remember we can only break up aqueous compounds, so only these three 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 three is going to get distributed to the number of calciums and the number of bromines. So it gets distributed to California and branch when this breaks up into its ions, we know that we're going to have calcium ion aqueous plus bromide ion aqueous, then we distribute the coefficient.

So that's going to give us three calcium ions and then this is 3 * 2 six bromide ions. Plus also remember that when we have an ion, it's an aqueous phase within a solution. This two 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 * 3 + 2 phosphate ions.

Remember phosphate is a polyatomic ion, produces 6 gets distributed to lithium and bromide. So 6 lithium plus one ions +6 bromide ions. This is a solid so it stays together. Does not break up into ions so plus one, so one. Not 6/1 calcium phosphate solid O. 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 ion's 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 comes from the complete ionic equation. So remember, we started out with our molecular equation. From there we're able to determine our complete ionic equation and then from our complete ionic equation we now are looking at our net ionic equation.

O. 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 reactance, provide both the molecular equation and the complete ionic equation O. Here we have ammonium sulfate reacting with calcium chloride. So for step zero, we're just going to follow steps one to four that we've learned in the past to 1st 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 and the sulfate ion. Don't worry about the fact that we have two 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 and chloride ion. Now remember we're going to swap ionic partners because opposite charges attract. This plus one charge is attracted to this -1 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 plus then we're going to have here c

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

Copper (II) Bromide and Lithium Hydroxide

• A. Cu⁺ (aq) + OH^– (aq) → CuOH (s)
• B. Cu²⁺ (aq) + 2 OH^– (aq) → Cu(OH)₂ (s)
• C. Cu²⁺ (aq) + 2 H₂O (aq) → Cu(OH)₂ (s) + 2 H⁺ (aq)
• D. Cu²⁺ (aq) + OH^– (aq) → Cu(OH)₂ (s)

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?

A complete ionic equation is a chemical equation that shows all of the species as they are actually present in solution. This means that compounds that dissociate and separate into their constituent ions in a solution are written as those individual ions. For example, if sodium chloride (NaCl) is dissolved in water, it separates into Na⁺ and Cl^- ions, and the complete ionic equation would reflect this dissociation.

It's important to note that the complete ionic equation does not include the spectator ions, which are ions that do not participate in the actual chemical reaction. To write a complete ionic equation, you start with a balanced molecular equation, then rewrite it showing all soluble strong electrolytes as ions. This provides a more detailed picture of the chemical reaction by illustrating which ions are involved in the reaction and which ones are merely present in the solution.

How do you write a complete ionic equation?

Writing a complete ionic equation involves representing all strong electrolytes (substances that dissolve in water to form a solution that conducts electricity) as ions. Here's how to do it step by step:

1. Start with a balanced molecular equation. Write down the chemical reaction with all the reactants and products in their molecular form, ensuring that the number of atoms of each element is the same on both sides of the equation.
2. Identify the strong electrolytes. Strong acids, strong bases, and most salts are considered strong electrolytes. They will dissociate completely into ions in solution.
3. Separate the strong electrolytes into ions. Write out all the ions that each strong electrolyte would form in solution. For example, NaCl would be written as Na⁺ and Cl^-.
4. Indicate the states of matter. Beside each compound or ion, write (s) for solid, (l) for liquid, (g) for gas, or (aq) for aqueous (dissolved in water).
5. Cancel out the spectator ions. Spectator ions are ions that appear unchanged on both the reactant and the product side of the equation. They do not participate in the actual chemical reaction and can be omitted to write the net ionic equation.

Here's an example

What information does the complete ionic equation give?

A complete ionic equation provides detailed information about the species present in a chemical reaction that takes place in an aqueous solution. Unlike a standard chemical equation that shows the reactants and products as whole compounds, a complete ionic equation breaks down soluble ionic compounds into their individual ions. It indicates which substances dissociate into ions in water and shows the actual forms of the reactants and products as they exist in solution.

For example, in the reaction between sodium chloride (NaCl) and silver nitrate (AgNO₃), the complete ionic equation would show the sodium (Na⁺), chloride (Cl^-), silver (Ag⁺), and nitrate (NO₃^-) ions separately. This level of detail helps you understand which ions are spectator ions (ions that do not participate in the reaction) and which ions combine to form the precipitate or the products of the reaction. It's particularly useful for predicting the outcomes of precipitation reactions and for understanding the movement of individual ions in electrochemical cells.