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4.2 Net Ionic Equations

dylan black

dalia savy

⏱️ August 6, 2020

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Reactions in Aqueous Solution

When dealing with chemical reactions, it is important to understand that, at least for the case of AP Chemistry, the vast majority of them will be taking place in an aqueous solution, that is, dissolved in water💧. Therefore, we will not be dealing with molecules, but rather their constituent particles that are created when they dissolve.

Let's take NaCl🧂 for example. When you dissolve table salt in water, it is essentially ripped apart by the water molecules. It is important to understand that water is a polar molecule. Polar molecules have a partial positive end and a partial negative end. Therefore, when an ionic compound like NaCl dissolves, the water molecules rip it apart and trap it, per say.

In other words, ionic compounds dissociate into their ions when dissolved in water.

We can write out this dissolving of salt as: NaCl (aq) --> Na+ (aq) + Cl- (aq). Thus, if we have a reaction between two compounds in water, it will be the ions/parts of the molecules that dissolve that react together.

💡Na+ attracts to the negative oxygen atoms. Cl- attracts to the positive hydrogen atoms.

Let's look at a specific class of such reaction:

Precipitation Reactions

Sometimes, when two reactants react, one of the products is insoluble. To be insoluble means that it will NOT be ripped apart by water like NaCl would be.

👉 Solubility Rules

One example of this is the precipitation of lead (ii) iodide:

2KI (aq) +Pb(NO3)2 (aq) --> 2KNO3 (aq) + PbI2 (s).

Because lead (ii) iodide is not soluble in water (we've went over solubility rules in earlier guides [link above]), it precipitates and falls out of solution.

Here is another table of solubility rules:

Image Courtesy of Regents Chemistry Reference Table

Writing Net Ionic Equations

Example #1

When we're dealing with reactions in aqueous solution, we only care about some of the ions reacting. Others are called spectator ions, and are essentially just there for the ride. They react, but they don't actually take true part in the reaction. Let's see this in action with our previous reaction: 2KI (aq) + Pb(NO3)2 (aq) --> 2KNO3 (aq) + PbI2 (s)

If we write this reaction out with all of the ions, we get:

2K+ + 2I- + Pb2+ + 2NO3- --> 2K+ + 2NO3- + PbI2 (s)

💡Note that PbI2 does NOT dissociate, since it is insoluble. Be on the lookout for this! Only dissociate soluble compounds. Also, don't dissociate weak acids and bases🍊 (you'll go over this more in the future).

This equation, with the ions written out, is called the complete ionic equation or the total ionic equation.

We see through this that potassium and nitrate go in and come out the same way, so we can cross them out and cancel them. Potassium and Nitrate were the spectator ions👓! This leaves us with:

2I- + Pb2+ --> PbI2(s).

This is called the net ionic equation for the reaction, and tells us what really happens. Potassium and nitrate took place, but they really didn't do much for us in terms of reaction. They could have been any suitable ion and it would happen the same.

💡Remember when writing net ionics that coefficients (the numbers before each reactant) are distributed between ions.

General Steps

To memorize, here are the general steps to writing a net ionic equation:

1. Figure out which compounds are soluble and insoluble using solubility rules

2. Balance the equation

3. Complete Ionic Equation - Write soluble compounds into ions (complete the dissociation)

4. Net Ionic Equation - Cross out spectator ions and write the final result!

Example #2

Try writing the net ionic equation for the following reaction:

KOH + Fe(NO3)3 --> KNO3 + Fe(OH)3

1. Using solubility rules, you should find that: Fe(OH)3 is the solid, or the precipitate

   1. KOH (aq) + Fe(NO3)3 (aq) --> KNO3 (aq) + Fe(OH)3 (s)

2. You should notice that there is only 1 OH on left and 3 OH on the right. Same goes for NO3, but vice versa. This means we have to balance the equation.

   1. 3KOH (aq) + Fe(NO3)3 (aq) --> 3KNO3 (aq) + Fe(OH)3 (s)

   2. 📝Read: AP Chemistry - Balancing Chemical Equations

3. Complete Ionic Equation = 3K+ + 3OH- + Fe+3 + 3NO3- --> 3K+ + 3NO3- + Fe(OH)3 (s)

   1. Break them up into their ions

4. Net Ionic Equation =

   1. Spectator Ions: 3K+ + 3OH- + Fe+3 + 3NO3- --> 3K+ + 3NO3- + Fe(OH)3 (s)

   2. Answer: 3OH- (aq) + Fe+3 (aq) --> Fe(OH)3 (s)

🎥Watch: AP Chemistry - Net Ionic Equations