Chemistry Reference
In-Depth Information
iodide, on the other hand, is a “true” product of the reaction. We could pour
the contents of our solution through a filter paper, and only the lead (II)
iodide would be left behind. An ionic reaction is a more realistic represen-
tation of what actually goes on with the substances in the reaction vessel.
Remember that the reactants were in solution, which means that they started
off as free ions. After the reaction, only the precipitate came out of solu-
tion; the rest of the substances remained in ion form. Following is an ionic
equation for the same double displacement reaction:
Pb
2+
(aq)
+ 2ClO
4
-
(aq)
+ 2Na
+
(aq)
+ 2I
-
(aq)
PbI
2(s)
+ 2Na
+
(aq)
+ 2ClO
4
-
(aq)
A net ionic reaction only shows the “true” product and the ions that
reacted to form the product. The ions that look the same on both sides of
the equation don't change or enter into a chemical reaction. Ions that don't
enter into the chemical reaction are called spectator ions. To create a net
ionic reaction, we remove the spectator ions, as shown here:
Pb
2+
(aq)
+ 2ClO
4
-
(aq)
+ 2Na
+
(aq)
+ 2I
-
(aq)
PbI
2(s)
+ 2Na
+
(aq)
+ 2ClO
4
-
(aq)
This gives us the net ionic equation: Pb
2+
(aq)
+ 2I
-
(aq)
PbI
2(s)
You may not always need to show all of these steps. That will depend
upon the question that you are answering. If we look back at our original
problem, it did not ask us to show the ionic equation, only the net ionic
equation. We could come up with the net ionic equation by crossing out
the spectator ions from the original equation. Look at what we would cross
out of our chemical reaction:
Pb(ClO
4
)
2(aq)
+ 2NaI
(aq)
PbI
2(s)
+ 2NaClO
4(aq)
Adding the proper charges to the ions that remain, we would get the
same net ionic equation that we previously derived: Pb
2+
+ 2I
-
PbI
2(s)
Now, we are ready to fill in answers C and D. Remember: The precipi-
tate is the solid that forms, in this case, the lead (II) iodide.
A.
Write a word equation.
lead (II) chlorate + sodium iodide
lead (II) iodide and sodium chlorate
B.
Write a balanced chemical reaction, with subscripts indicating state or
phase.
Pb(ClO
4
)
2(aq)
+ 2NaI
(aq)
PbI
2(s)
+ 2NaClO
4(aq)
C.
Write a net ionic equation.
Pb
2+
(aq)
+ 2I
-
(aq)
PbI
2(s)
D. Identify the precipitate.
lead (II) iodide
