<< Hide Menu
Dylan Black
Dalia Savy
Dylan Black
Dalia Savy
When dealing with chemical reactions, it is important to understand that, at least in 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 se.
In other words, ionic compounds dissociate into their ions when dissolved in water. To tie this back to unit three and intermolecular forces, when you dissolve NaCl, you will see ion-dipole interactions. The positive end of the water molecule (hydrogen) is attracted to the negative chloride anion, while the negative end of the water molecule (oxygen) is attracted to the positive sodium cation.
Let's look at a specific class of such reaction:
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. In other words, an insoluble product is one that precipitates out as a solid and does not dissolve in water.
One example of a precipitation reaction where one product is insoluble 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 (according to the general solubility rules linked above), it precipitates and falls out of the solution.
Here is another table of solubility rules, but you do not need to memorize every single one of these. College Board has stated that only the following solubility rule will be assessed on the AP exam: "All sodium, potassium, ammonium, and nitrate salts are soluble in water."
Let's see these concepts 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⁺ (aq) + 2I⁻ (aq) + Pb²⁺ (aq) + 2NO₃⁻ (aq) → 2K⁺ (aq) + 2NO₃⁻ (aq) + PbI₂ (s). This was obtained by dissociating, or separating, each soluble compound into its constituent parts.
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 learn more about this more in the future).
This equation, with the ions written out, is called the complete ionic equation or the total ionic equation. It shows every ion in the solution that is present during the reaction, including spectator ions.
Now let's take the spectator ions out of the reaction. When looking at the total ionic equation, we see 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👓!
Once we eliminate the spectator ions, we are left with 2I⁻ (aq) + Pb²⁺ (aq) → PbI₂(s). This is what we call the net ionic equation for the reaction, and it tells us what really happens. Potassium and nitrate were in the flask, but they really didn't do much for us in terms of the formation of the insoluble compound. They could have been any suitable ion and the reaction would have had the same outcome.
Now that you have an idea of what a net ionic equation is, it is time to learn how to write one out! First, you must know the basics of representing a chemical reaction. Make sure you review those in the last study guide before moving on.
To work through and learn this process, here are the general steps to writing a net ionic equation:
Try writing the net ionic equation for the following reaction: KOH + Fe(NO₃)₃ → KNO₃ + Fe(OH)₃
© 2024 Fiveable Inc. All rights reserved.