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NET IONIC EQUATIONS (read section 7.12)
When an ionic compound dissolves in water it is generally completely ionized; by that we mean
that the compound splits into ions that float freely in the water surrounded by water molecules.
Therefore, when ionic compounds are dissolved in water the positive ions and the negative ions no longer
have any contact with each other. When a metal or another ionic solution is added, we can show the
reaction that occurs as involving only some of the ions, while the other ions take no part. These other
ions, which remain unchanged on either side of the equation, are called spectator ions. If we then rewrite
the equation, omitting the spectator ions, we get the NET IONIC EQUATION. Follow these instructions:
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Only ionic compounds can be written as separated ions (ionized).
Ionic compounds are either salts or acids.
You can recognize salts by looking for a metal + nonmetal (the latter is sometimes a polyatomic ion).
The ammonium ion, NH4+, acts a metal ion.
Aqueous acids are also ionized ionic compounds. Look for H’s instead of metal ions.
A molecular (covalent) compound, such as water cannot be separated into ions.
Only ionic compounds dissolved in water are ionized. Therefore, insoluble ionic compounds (solids
or precipitates) cannot be separated as ions.
You should start out by writing a complete ionic equation – write everything aqueous as separated
ions.
Next you identify the spectator ions as the ions that do not change from before the reaction takes
place (left side of the arrow) to after the reaction takes place (right side of the arrow).
Finally, you rewrite the equation without the spectator ions - this is the net ionic equation.
It may happen that there are no spectator ions. In that case, the complete ionic equation and the net
ionic equation are the same.
Example 1: Lead acetate and copper sulfate solutions are mixed:
Pb(C2H3O2)2 (aq) + CuSO4 (aq)  PbSO4(s) + Cu(C2H3O2)2 (aq)
Complete ionic equation:
Pb+2(aq) + 2 C2H3O2-(aq) + Cu+2(aq) + SO4-2(aq)  PbSO4(s) + Cu+2(aq) + 2 C2H3O2-(aq)
Notice that insoluble lead (II) sulfate is not broken down into its ions.
Notice also that we still have to write two acetate ions, but since they are really are floating
separately in the solution we write them as 2 C2H3O2, and not as (C2H3O2)2.
Identify spectators: Pb+2(aq) + 2 C2H3O2-(aq) + Cu+2(aq) + SO4-2(aq)  PbSO4(s) + Cu+2(aq) + 2 C2H3O2-(aq)
Write the net ionic equation, eliminating the spectator ions: Pb+2(aq) + SO4-2(aq)

PbSO4(s)
Example 2: Potassium cyanide solution is mixed with hydrochloric acid to give hydrogen cyanide gas
and potassium chloride solution.
KCN(aq) + HCl(aq)  HCN(g) + KCl (aq)
Complete ionic equation: K+(aq) + CN-(aq) + H+(aq) + Cl-(aq)  HCN(g)
+
Identify the spectator ions: K+(aq) + CN-(aq) + H+(aq) + Cl –(aq)  HCN(g)
K+(aq)
+ Cl-(aq)
+ K+(aq) + Cl-(aq)
Write the net ionic equation, eliminating the spectator ions: CN-(aq) + H+(aq)  HCN(g)
Write the full ionic equations, underline the spectator ions, then write the net ionic equation.
1) Sulfuric acid is mixed with barium chloride
2)
Acetic acid is poured over a piece of magnesium metal
3)
Solutions of potassium phosphate and iron(II)bromide are mixed.
4)
A piece of zinc metal is put into a beaker containing a copper(II)nitrate solution.
5)
Solutions of manganese(II)sulfate and barium nitrate are mixed
6)
Calcium hydroxide solution is mixed with hydrochloric acid solution.
7)
Nitric acid is poured over calcium carbonate and it fizzes.
8)
Ammonium phosphate and potassium carbonate solutions are mixed.
9) Chlorine gas is bubbled through a solution of aluminum bromide.