# Net Ionic Equations

When you go to a baseball game, there are plenty of spectators cheering (or booing) from the stands. They aren't a part of the game, but it wouldn't be the same without them. Reactions are the same way. Some reactions have spectator ions, which, while a part of the total reaction, don't actually participate directly in it. When we want just the main reaction, we write a net ionic equation.

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If someone asked you to describe a baseball game you went to, they probably only care about the actual players/game, not the spectators. This is the same concept, just for chemistry!

In this article, we will be learning about net ionic equations and walking through step-by-step how to write them.

• We will learn about why we use net ionic equations, and what they represent.
• Next, we will walk through how to write a net ionic equation.
• Lastly, we will learn how to write these equations for liquid and gas products.

## Net Ionic Equation Definition

A net ionic equation is a chemical equation that only shows the ions, elements, and compounds that are directly participating in a reaction.

So when we say that a spectator ion "doesn't directly participate", what do we mean? Well, for these types of reactions, we are looking at aqueous solutions. In an aqueous solution, the compounds have been dissolved in water, so they are present as their ions.

Fig. 1 - When NaCl is dissolved in water, the ions are attracted to the water molecules. Na+ is attracted to oxygen, which is slightly negative. Cl- is attracted to hydrogen, which is slightly positive.

As you can see above, when a solid is dissolved in water, its ions are attracted to the water molecules. The positive ions are attracted to the partially negative oxygen, and the negative ions are attracted to the partially positive hydrogen (in this example, these are shown by sodium and chlorine respectively). The bonds between the ionic solid are broken, and new bonds are formed with water. This attraction is why these solids are soluble, and why they are ions in solution.

When two aqueous compounds combine, the products are either also going to be aqueous or they will form a precipitate (a solid not soluble in water). The "reaction" we are talking about is the formation of the solid (or any non-aqueous species, as we will see later). The other ions are just hanging out in solution the whole time, not really doing much.

So how do we know which compounds are soluble and insoluble? For ionic solids, we follow the solubility rules. These rules tell us whether ions/ion pairs are soluble. For example, compounds with Group I metals (Li+, Na+, etc.) are typically soluble. Also, compounds with halides (Cl-, Br-, etc.) are typically soluble, this is why we know that NaCl is soluble in water (other than observing it, of course!).

## Net Ionic Equation Example

Now that we've covered what a net ionic equation is, what does it look like? Here is an example of a chemical reaction and its net ionic equation:

$$NaCl+AgNO_3\rightarrow NaNO_3+AgCl$$

$$Ag^+_{(aq)}+Cl^-_{(aq)}\rightarrow AgCl_(s)$$

The image below illustrates what is happening in this reaction.

Fig. 2 - In a reaction of AgNO3 and NaCl, the products are NaNO3 (aqueous) and AgCl (solid).

We have two reactants: NaCl and AgNO3. Since they are both aqueous, their ions are in solution. When they react, they form NaNO3 (aqueous) and AgCl (solid). AgCl is insoluble in water, so it forms a solid and is therefore not in its ionic form. Na+ and NO3- are our spectator ions, since they stay as ions the whole time. The net ionic equation only shows us the reaction that forms the solid.

Most halide salts (Cl-, Br-, etc.) are soluble in water. However, there are a few exceptions, namely, AgCl. When Ag+ and Cl- interact in water, a white powder will form, which is the precipitate.

## Balanced Net Ionic Equation

Now let's walk through how we got that net ionic equation from the original chemical equation

Write the net ionic equation for the reaction of NaCl and AgNO3

The first thing we need to do is write out the balanced reaction

$$NaCl_{(aq)}+AgNO_{3\,(aq)}\rightarrow NaNO_{3\,(aq)}+AgCl_{(s)}$$

Next, we need to break up every aqueous solution into its ions. All solids stay untouched.

$$Na^++Cl^-+Ag^++NO_3^-\rightarrow Na^++NO_3^-+AgCl$$

Now we cancel out any ions that are on both sides of the equation, so what we are left with is the balanced net ionic equation.

$$\cancel{Na^+}+Cl^-+Ag^++\cancel{NO_3^-} \rightarrow \cancel{Na^+}+\cancel{NO_3^-}+AgCl$$

$$Ag^+_{(aq)}+Cl^-_{(aq)}\rightarrow AgCl_{(s)}$$

You'll note that we swapped the order of Ag and Cl in the final step. This was done to simplify adding them together, but it isn't necessary. . Based on the example above, you'll see that there are only a few steps to writing a net ionic equation. To summarize, these are:
1. Write the balanced reaction equation.
2. Write out every aqueous solution as their ions.
3. Cancel out ions that are present on both sides.
4. Simplify the equation.

## HCl NaOH Net Ionic Equation

Net ionic equations aren't just for solids. If a liquid is formed when two aqueous solutions are reacted together, we can still utilize a net ionic equation. Liquids, like solids, aren't in their ionic form in water. Also, liquids (for example, HCl) can be aqueous. Like with solids, it all depends on whether they are soluble in water. Let's write the HCl + NaOH net ionic equation as an example:

First, let's write out the molecular equation.

$$HCl_{(aq)}+NaOH_{(aq)}\rightarrow H_2O_{(l)}+NaCl_{(aq)}$$

Now we can split our aqueous compounds into ions.

$$H^++Cl^-+Na^++OH^-\rightarrow H_2O+Na^++Cl^-$$

Next, we'll cancel out like ions to have the net equation.

$$H^++\cancel{Cl^-}+\cancel{Na^+}+OH^-\rightarrow H_2O+\cancel{Na^+}+\cancel{Cl^-}$$

$$H^+_{(aq)}+OH^-_{(aq)}\rightarrow H_2O_{(l)}$$

It makes sense that water is considered a liquid here, since it can't dissolve itself. However, there are plenty of liquids that aren't soluble in water, like octanol (C8H18O). Net ionic equations always focus on products made from aqueous solutions, no matter the state they are in (besides aqueous).

## Net Ionic Equations for Gases

Lastly, we can use net ionic equations for gas products as well. This is, of course, as long as the gas isn't soluble in water. Let's look at an example:

Write the net ionic equation for the reaction of chlorine gas (Cl2) and NaBr.

We follow the same steps as before.

$$Cl_{2\,(g)}+2NaBr_{(aq)}\rightarrow Br_{2\,(g)}+2NaCl_{(aq)}$$

$$Cl_2+2Na^++2Br^-\rightarrow Br_2+2Na^++2Cl^-$$

$$Cl_2+\cancel{2Na^+}+2Br^-\rightarrow Br_2+\cancel{2Na^+}+2Cl^-$$

$$Cl_{2\,(g)}+2Br^-_{(aq)}\rightarrow Br_{2\,(g)}+2Cl^-_{(aq)}$$

One thing to note is that both reactants don't have to be aqueous, just one of them. The main thing is that there are spectator ions that can be canceled out. Also, always double-check that your equation is balanced.

## Practice Problems on Net Ionic Equations

Now that we've covered each of the three types, let's work on a few extra problems.

Based on the equation below, write the net ionic equation:

$$2Fe(NO_3)_{3\,(aq)}+3Na_2CO_{3\,(aq)}\rightarrow Fe_2(CO_3)_{3\,(s)}+6NaNO_{3\,(aq)}$$

For equations like this, make sure you are multiplying the subscripts (little number) by the coefficients (number in front). Otherwise, the equation won't be balanced.

$$2Fe^{3+}+6NO_3^-+6Na^++3CO_3^{2-}\rightarrow Fe_2(CO_3)_3+6Na^++6NO_3^-$$

$$2Fe^{3+}_{(aq)}+3CO_{3\,(aq)}^{2-}\rightarrow Fe_2(CO_3)_{3\,(s)}$$

Now for another problem:

Based on the equation below, write the net ionic equation:

$$H_3PO_{4\,(aq)}+KOH_{(aq)}\rightarrow H_2O_{(l)}+K_3PO_{4\,(aq)}$$

Now, for this equation, it isn't actually balanced. Always make sure your initial equation is balanced before trying to write the net ionic equation.

$$H_3PO_{4\,(aq)}+3KOH_{(aq)}\rightarrow 3H_2O_{(l)}+K_3PO_{4\,(aq)}$$

$$3H^++PO_4^{3-}+3K^++3OH^-\rightarrow 3H_2O+3K^++PO_4^{3-}$$

$$3H^++3OH^-\rightarrow 3H_2O$$

$$H^+_{(aq)}+OH^-_{(aq)}\rightarrow H_2O_{(l)}$$

We also want the equation to be as simple as possible. Since every species was divisible by 3, we simplified it as you see above. Now for one last example!

Based on the equation below, write the net ionic equation:

$$2HCl_{(aq)}+Na_2S_{(aq)}\rightarrow H_2S_{(g)}+2NaCl_{(aq)}$$

$$2H^++2Cl^-+2Na^++S^{2-}\rightarrow H_2S+2Na^++2Cl-$$

$$2H^+_{(aq)}+S^{2-}_{(aq)}\rightarrow H_2S_{(g)}$$

Hopefully, after all of that practice, the concept has gotten a little easier. As long as you remember your steps, writing net ionic equations is easy!

## Net Ionic Equations - Key takeaways

• A net ionic equation is a chemical equation that only shows the ions, elements, and compounds that are directly participating in a reaction.
• We write these equations so that we are only focusing on the direct reaction and not the spectator ions.
• Spectator ions are ions that, while present in the solution, are not a part of the reaction.
• To write a net ionic equation, you write out all aqueous compounds as their ions, then cancel out "like" ions to get the net equation.
• These equations can be used for solid, liquid, and gas products, as long as they aren't soluble in water.

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How do you write net ionic equations?

First, you write out the balanced chemical equation. Next, you break up all aqueous solutions into their ions. Lastly, you cancel the ions that are present on both sides to get the balanced net ionic equation.

what is a net ionic equation?

net ionic equation is a chemical equation that only shows the ions, elements, and compounds that are directly participating in a reaction.

Why do we use net ionic equations?

We use them so that we can look at the direct chemical reaction without spectator ions.

Which is the net ionic equation for the reaction: NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)

a) Na+ + Cl- + Ag+ + NO3- ->  Na+ + NO3- + AgCl

b) Ag+ + Cl-  ->  AgCl

c)Na+ + NO3- -> NaNO3

The answer is b. The first option is the unbalanced ionic equation, not the net ionic equation. Option c isn't creating the right product. Net ionic equations focus on the insoluble product made, NaNO3 is soluble.

What are spectator ions?

Spectator ions are the ions that, while present in the solution, don't directly participate in the reaction.

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