*For the hearing impaired or just if you prefer to read along, we have provided a transcript of the video below as an alternative to close-captioning*
In this segment we’ll discuss a special class of ionic compounds’ reactions: double exchange reactions or metathesis reactions. We’ll focus on double exchange reactions in which a precipitate (insoluble solid) is formed. Reactions in which two of the ions present in solution combine in an insoluble product (precipitate) are called metathesis reactions with formation of precipitate (other types of metathesis reactions involve gas evolution or formation of a non-dissociated compound)
A general pattern for metathesis reactions with precipitate formation is
AB(aq)+CD(aq)→AD(s) + CB(aq),
In which a solution of compound AB reacts with a solution of another compound CD to result in at least one insoluble product (AD). We see that the cation A+ of the first compound binds to the anion D– of the second compound.
One can tell which of the products is insoluble based on tabulated data, or other solubility guidelines.
A typical metathesis reaction is that of Potassium Iodide with Lead Two Nitrate. When the lead two nitrate solution is added dropwise to the potassium iodide solution, a beautiful yellow precipitate forms. The chemical equation for this process is the one in which lead two nitrate reacts with potassium iodide to result in lead two iodide precipitate, and a solution of potassium nitrate
We notice that the equation is balanced.
This equation is known as the MOLECULAR EQUATION
This metathesis reaction can be better understood and analyzed using the KembloX system
For starters, we use the ion chart to build the models for the two reactants using KembloX blocks. The appropriate chemical identity has been assigned to all the blocks involved (I used an erasable marker.
Through dissolution, as discussed in another segment in this series, the ions of soluble substances separate completely. Thus, the two solutions contain the lead two ions and the nitrate ions, and the potassium ions and the iodide ions, respectively.
When the two solutions are mixed, the metathesis reaction occurs and the yellow lead two iodide precipitate is formed.
The equation for this process reflects the ions involved, and the insoluble product.
This equation is the FULL IONIC EQUATION
By comparing the situation before and after the reaction, it must be noted that some ions remain in the same form before and after reaction
In our example, those are the potassium ions and the nitrate ions, circled red.
The full ionic equation, written with those ions in red, is
One lead two ion and two nitrate ions and two potassium ions and two iodide ions result in solid lead two iodide and two potassium ions and two nitrate ions
Pb2++ 2NO3– + 2K+ + 2I– → PbI2 + 2NO3– + 2K+
Those ions in red are called SPECTATOR IONS because they do not take part in the precipitate formation.
We may disregard the spectator ions, keeping only the ions that participate in the formation of the precipitate
In this case the equation becomes much simpler. It is called NET IONIC equation, and it reflects the fact that precipitation would occur if the original compounds contained other counterions. For example, if we started the process with sodium iodide, instead of potassium iodide, or with lead two acetate, instead of lead two nitrate.
We are now familiar with:
- Metathesis reactions, in which two ionic compounds exchange ions and one of the products is insoluble. One compound’s anion binds to the second compound’s cation, and vice versa
AB(aq)+CD(aq)→AD(s) + CB(aq)
We also became familiar with:
Molecular equations that show the ionic compounds involved in the metathesis reaction
Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
Full ionic equations that show all the ions involved in the metathesis reaction
Pb2++ 2NO3– + 2K+ + 2I– → PbI2 + 2NO3– + 2K+
Finally, we also became familiar with the concepts of:
Spectator ions – ions NOT involved in the formation of the precipitate
Pb2++ 2NO3– +
2K+ + 2I– → PbI2
+ 2NO3– + 2K+
Red characters – spectator ions
And with the concept of
Net ionic equations that show only the ions involved in the formation of the precipitate
Pb2+(aq)+ 2I–(aq) → PbI2(s)
Before we go, here are Important Observations:
Number one: When dealing with METATHESIS REACTIONS, chemical equations MUST be BALANCED.
Number two: The precipitates form irrespective of the identity of the spectator ions (in our example, the same precipitate would have formed if we used sodium iodide instead of potassium iodide, or if we used lead two acetate instead of lead two nitrate