Ionic reactions - dissolving and precipitates
Much of the chemistry around us involves the dissolution of ionic solids in water to give aqueous solutions and the precipitation of ions from aqueous solution to give precipitates (solids). So what I would like to do first is to look a little at the process of dissolving and the nature of aqueous solutions.
We will look at water later. The thing we need to know now is that the ions in water are not independent - they dissolve because they substitute interactions with water molecules for interactions with counter ions. And they stay in solution because they are insulated from each other by the water "shells" around each ion. A couple of corollaries.
- First, only so many ions will dissolve until we run out of water molecules to make the shells, then they will interact with each other and precipitate out.
- Second, the amount of a given ion which will dissolve in water depends on the competition between water and its counter ion for interaction with it. If the ion-water interactions are very strong, it will tend to be soluble, if the ion-ion interaction is very strong, then they will precipitate out at very low concentrations etc.
It is useful to remember some simple "rules" (really more like guide lines) to help in predicting reactions. For common compounds such as we see in general chemistry we can use the following rules:
Let's consider some chemical processes:
- Dissolve NaCl in water:
NaCl Na+(aq) + Cl-(aq)
- Add NaCl solution to a AgNO3 solution. First recall that each of the ions in solution is in an (aq)ueous complex, so we can write:
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) AgCl(s) + Na+(aq) + NO3-(aq)
Notice that two ions don't change. So, if they don't prticipate, why show them. Instead we write a net ionic equation:
Ag+ + Cl- AgCl(s)
Notice the ions that appeared on both sides are not shown (in mathematical terms they cancelled)
- Mix barium chloride and potassium sulfate:
Ba2+(aq) + 2 Cl-(aq) + 2 K+(aq) + SO42- BaSO4(s) + 2 K+(aq) + 2 Cl-
Again, we want to write a net ionic equation showing only the ions which reacted:
Ba2+ + SO42- BaSO4(s)
Notice that net ionic equations are very general expressions. Essentially they are saying that any time we have these species present they will react, regardless of what else happens to be there! (Sometimes folks are confused when they add ions which should react and they don't. This is usually a case where something else reacted first, so the ions of interest really weren't there!).
Since all ions in aqueous reactions are considered to be hydrated, we do not generally include (aq) as part of the formula. But remember, it is assumed!
In these reactions we see a transfer of electrons from one atom or molecule to another. First let's look at some terms.
CH4 + 2 O2 CO2 + 2 H2O
Notice that the methane is oxidized by the oxygen. We say that the carbon and hydrogen are both oxidized to give the new covalent products, water and carbon dioxide, because the electrons are not evenly shared, they are pulled toward the oxygens in each bond.
There are two common methods for balancing redox reactions: the oxidation number method and the half-reaction method. The half-reaction method works very well for ionic reactions, it is relatively easy to give partial credit, and it is the only method I will use in this class. If you know how to do the other method you are welcome to do so, but be careful to make sure you show your work or I won't be able to give partial credit!
In the half-reaction method what we do is first break an equation into two parts and then balance the parts individually. Presented stepwise:
Separate the reaction into two half-reactions.
Balance each half-reaction separately:
Add two equations together
Cancel items appearing on both sides.
Example. Balance the following equation as it occurs in acid solution:
First break the equation into two half reactions, one for Mn and one for Cl
10 e- + 16 H+ + 2 MnO8-+ 10 Cl- 2 Mn2+ + 8 H2O + 5 Cl2 + 10 e-
(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.
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© R A Paselk
Last modified 11 February 2015