Net ionic equation Calculator

How to calculate a net ionic equation?

You can follow the following simple steps while calculating a net ionic equation.

1. Write the complete ionic equation from the chemical reaction equation.
2. Identify the spectator ions from the complete ionic equation.
3. Cancel out the spectator ions on both sides of the equation.
4. Obtain the net ionic equation.

Let’s understand the process of writing net ionic equations with the help of examples.

Step 1: Write the complete ionic equation from the chemical reaction equation.

For example, a chemical reaction takes place between copper (II) sulfate and sodium chloride in aqueous forms.

The chemical formula for copper (II) sulfate is CuSO₄ while the chemical formula for sodium chloride is NaCl.

The chemical reaction between CuSO₄ and NaCl can be represented by the balanced chemical equation shown below.

On the reactant side, CuSO₄ dissociates into Cu²⁺ and SO₄^2- ions in an aqueous solution. Similarly, 2 NaCl dissociates into 2 Na⁺ and 2 Cl^– ions in the aqueous solution.

On the product side, Na₂SO₄ dissociates into 2 Na⁺ and SO₄^2- while CuCl₂ is a precipitate (it is in the solid form) so it will not dissociate into its respective ions and stay as it is.

So, the chemical equation given above can be written as a complete ionic equation, as shown below.

Step 2: Identify the spectator ions from the complete ionic equation.

In the equation above, you can see that 2 Na⁺ ions and an SO₄^2- ion stays unchanged on both the reactant and the product sides.

As these ions stay uninvolved during the chemical reaction therefore they are marked as spectator ions.

Step 3: Cancel out the spectator ions on both sides of the equation

Make sure to check that the net ionic equation is a balanced equation.

Step 4: Obtain the net ionic equation

So, the balanced net ionic equation for this reaction is:

Using all the steps given above, practice with us some more examples for a better understanding of calculating net ionic equations.

Examples of Net ionic equations

Example 1

A complete neutralization reaction takes place between sodium hydroxide (NaOH) and hydrochloric acid (HCl). A balanced chemical equation for the reaction is given below.

 NaOH dissociates into Na⁺ and OH^– ions while HCl dissociates into H⁺ and Cl^– on the reactant side. While on the product side, NaCl is also water soluble so it breaks down into Na⁺ and Cl^– ions in the presence of H₂O.  So the complete ionic equation for this reaction is:

This net ionic equation shows that an acid-base neutralization reaction actually takes place between hydrogen (H⁺) and hydroxide (OH^–) ions only.

Example 2

A neutralization reaction takes place between magnesium hydroxide Mg(OH)2 and sulfuric acid (H2SO4). A balanced chemical equation for the reaction is given below.

Mg(OH)₂ dissociates into Mg²⁺ and 2 OH^– ions while H₂SO₄ dissociates into 2 H⁺ and SO₄^2- on the reactant side. While on the product side, MgSO₄ formed is a water-soluble ionic compound so it dissociates into Mg²⁺ and SO₄^2- ions under aqueous conditions. Thus, the complete ionic equation for this reaction is:

Example 3

Lead (II) nitrate Pb(NO3)2 reacts with potassium iodide (KI) to yield potassium nitrate and a bright yellow precipitate of lead iodide (PbI2).

Pb(NO₃)₂ dissociates into Pb²⁺ and 2 NO₃^– ions while 2 KI dissociates into 2 K⁺ and 2 I^– ions on the reactant side. 2 KNO₃ dissociates into 2 K⁺ and 2 NO₃^– ions while PbI₂ stays intact on the product side as it is a precipitate. So the complete ionic equation for this reaction is:

Example 4

Sodium chloride (NaCl) reacts with silver nitrate (AgNO3) to yield a white silver chloride (AgCl) precipitate while sodium nitrate (NaNO3) is formed as the by-product.

NaCl dissociates into Na⁺ and Cl^– ions while AgNO₃ dissociates into Ag⁺ and NO₃^– ions on the reactant side. NaNO₃ dissociates into Na⁺ and NO₃^– ions while AgCl stays intact on the product side. So the complete ionic equation for this reaction is:

Notice that we have placed the Ag⁺ ion first and then the Cl^– ion is written in the net ionic equation. This is because the convention is that the positively charged cation is written prior to the negatively charged ion on the reactant side while writing net ionic equations. However, this rule is not compulsory.

You should also keep in mind that in addition to the same moles of an element on both reactant and product sides, the positive and negative charges should also be balanced, as you can ensure from any of the net ionic equations we have discussed in this article.

Now, let’s see another example.

Example 5

Copper (II) chloride (CuCl2) reacts with sodium phosphate (Na3PO4) to produce a blue solid precipitate known as copper (II) phosphate and sodium chloride.

3 CuCl₂ dissociates into 3 Cu²⁺ and 6 Cl^– ions while 2 Na₃PO₄ dissociates into 6 Na⁺ and 2 PO₄^3- ions on the reactant side. 6 NaCl dissociates into 6 Na⁺ and 6 Cl^– ions while Cu₃(PO₄)₂ stays unchanged on the product side. So the complete ionic equation for this reaction is:

Positive and negative charges on both sides of the ionic equation are balanced.

You may note that the positive and negative charges are also balanced in this net ionic equation i.e., +6 and -6 respectively on both sides of the equation.

Example 6

Lead nitrate Pb(NO3)2 reacts with lithium bromide (LiBr). As a result, lithium nitrate (LiNO3) and lead bromide (PbBr2) are produced, as shown in the chemical equation given below.

Now in the above equation, the physical state symbols of all the reactants and products are not given. In such a situation, you need to identify which of the reactants and products are water soluble while others are not.

Applying the solubility rules, we will identify that lead bromide appears as a white solid as r.t.p. It has very low solubility in water i.e., 0.455 g per 100 g H₂O_. Therefore, in the above chemical reaction, PbBr₂ is the precipitate obtained. All the other compounds are water soluble so they appear as aqueous solutions.

Pb(NO₃)₂ dissociates into Pb²⁺ and 2 NO₃^– ions while 2 LiBr dissociates into 2 Li⁺ and 2 Br^– ions on the reactant side. 2 LiNO₃ dissociates into 2 Li⁺ and 2 NO₃^– ions while PbBr₂ stays undissociated on the product side. So the complete ionic equation for this reaction is:

Example 7

A chemical reaction between sodium iodide (NaI) and calcium nitrate Ca(NO3)2 produces an aqueous solution of sodium nitrate (NaNO3) and calcium iodide (CaI2).

Now in this reaction, no precipitates are formed. So, all the ionic compounds will dissociate into their ions both on the reactant as well as the product side.

2 NaI dissociates into 2 Na⁺ and 2 I^– ions while Ca(NO₃)₂ dissociates into Ca²⁺ and 2 NO₃^– ions on the reactant side. While on the product side, 2 NaNO₃ dissociates into 2 Na⁺ and 2 NO₃^– ions while CaI₂ breaks down to produce Ca²⁺ and 2 I^– ions.

All the ions here prove to be spectator ions, so all the ions are canceled out. Thus, there is no such reaction. So no net ionic equation.

FAQ

What does the net ionic equation represent?

A net ionic equation represents only the ions actually taking part in the chemical reaction, eliminating spectator ions.

What is a net ionic equation calculator?

The net ionic equation calculator is an online tool that shows the “chemical equation”, “complete ionic equation”, “Cancellation of spectator ions”, and a “Net ionic equation”.

What is the difference between the Balanced and net ionic equation?

The balanced equation shows all the details about species that are in the system. It gives the actual number of molecules of each reactant and product. Whereas the net ionic equation shows only species that take part in the reaction, it doesn’t show the spectator’s ions. You can easily balance the chemical equation by using this tool given below –