Net Ionic Equations (NIE’s)
The purpose of writing NIE’s is to tell us what takes place in a chemical reaction. You will notice
from your experience writing NIE’s the driving forces for many chemical reactions, that is,
reactions typically form solids, liquids or gasses.
Three example of writing NIE’s that you will come across.
1) Typical NIE
2) Weak Acid NIE
3) Gas Forming NIE
For the beginner it’s helpful to write the following equations in order.
1) Molecular Equation
2) Total Ionic Equation
3) Net Ionic Equation
(Note) Special Gas Forming Reactions: Don’t forget, if you see H2CO3, H2SO3, or NH4OH on the
Product Side, to rewrite them into their gas, as well as adding water.
1) H2CO3(aq) on the product side turns into: CO2(g) + H2O(l)
2) H2SO3(aq) on the product side turns into: SO2(g) + H2O(l)
3) NH4OH(aq) on the product side turns into: NH3(g) + H2O(l)
4) H2S on the product side is always written as a gas H2S (g)
Students should know that metals plus an acid produces a gas as well.
4) M metal + HCl  H2 (g) + MCl
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Example 1: Typical NIE
AgNO3 (aq) + Li3PO4 (aq)
Step 1: Write the Molecular Equation
 Determine Products, then balance atoms
Always write or “think” of the charge (oxidation state) of each ion:
Ag+ NO3- (aq) + Li+3PO43- (aq)
So
Ag1+
NO31Li1+
PO43-
These charges are usually found on our periodic table; however, the
Octet Rule is typically used (Formal charge for 1st year science
majors) for determining charges/oxidation numbers.
Next, write ONLY the charge on the product side, don’t forget to exchange partners.
Ag1+ NO31- (aq) + Li1+3 PO43- (aq)
1+ 3-
+
1+ 1-
Next, write the ion under the charge
Ag+ NO3- (aq) + Li+3PO43- (aq)

Ag1+ PO43-
+
Li1+ NO31-
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Finish by balancing charges---chris-cross method-----using subscript, so the net charge of each
molecule has a zero charge.
Ag+ NO3- (aq) + Li+3PO43- (aq)

Ag1+ PO43-
+
Li1+ NO31-
The equation looks like this when the charges are balanced. In this example, the Ag gets a
subscript 3 to balance the PO4’s 3- charge.
AgNO3 + Li3PO4  Ag3PO4 + LiNO3
The complete, the balanced equation looks like:
3AgNO3 + Li3PO4  Ag3PO4 + 3LiNO3

Determine Solubility from a solubility table or chart (an example is found in this web
site).
3AgNO3( ? ) + Li3PO4( ? )  Ag3PO4( ? ) + 3LiNO3( ? )
AgNO3( ? ) nitrate is always soluble so the molecule gets an (aq) ending, meaning it will dissolve
and disassociate in water.
Li3PO4( ? ) all Group 1, like Lithium elements are always soluble, so the molecule gets an (aq)
ending, meaning it will dissolve and disassociate in water.
Ag3PO4( ? ) Look for the anion phosphate, phosphates are not soluble with Silver, so the
molecule gets a (s) ending, meaning it will form a solid/precipitate in water.
LiNO3( ? ) all Group 1, like Lithium elements, and nitrates are always soluble, so the molecule
gets an (aq) ending, meaning it will dissolve and disassociate in water.
The final form will look like:
3AgNO3(aq) + Li3PO4(aq)  Ag3PO4(s) + 3LiNO3(aq)
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Step 2: Write the Total Ionic Equation
If the ending is (aq) break the molecule apart in water----unless the molecule is categorized as a
weak or non-electrolyte.
3Ag+(aq) + 3NO3-(aq) + 3Li+ + PO4-3(aq)  Ag3PO4(s) + 3Li+(aq) + 3NO3-(aq)
Step 3: Write the Net Ionic Equation (NIE). The previous step is helpful, because if we look at
that equation and do not write Spectator ions. In this example since NO3- and Li+ appear on
BOTH sides (spectator ions) then do Not write them in final NIE.
3Ag+(aq) + PO4-3(aq)  Ag3PO4(s)
Notice what drove this reaction was the formation of a solid.
Example 2: Weak Acid NIE
HC2H3O2 + NaOH 
Molecular Equation. First write the charge and ions on the product side to determine the
products.
H1+C2H3O21- + Na1+OH1-
H+ OH- +
Na1+C2H3O21-
Next balance the charges to make the correct molecules. Notice the equation is also balanced.
HC2H3O2 + NaOH 
H2O +
NaC2H3O2
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Consult a solubility table for solubility rules. This is our final Molecular Equation.
HC2H3O2 (aq) + NaOH (aq) 
H2O (l) +
NaC2H3O2 (aq)
Now for the Total Ionic Equation.
Since HC2H3O2 is a weak acid, that means that it only dissociates about 5% in water, it remains
mainly as HC2H3O2 on the reactant side, though it does react.
In previous examples any (aq) means = break it apart, here though we leave the HC 2H3O2 alone.
Also, all liquids (l) remain intact.
HC2H3O2 (aq) + Na1+(aq) + OH1-(aq) 
H2O (l) + Na1+(aq) + C2H3O21- (aq)
Our Net Ionic Equation does not include any spectator ions.
HC2H3O2 (aq) + OH1-(aq) 
H2O (l) + C2H3O21- (aq)
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Example 3: Gas Forming NIE
HCl + Na2CO3 
Molecular equation
H1+Cl1-
Na1+2CO32+
+
Na1+Cl1- +
H+1CO32-
The equation where all of the ions are balanced to make molecules is:
HCl + Na2CO3  NaCl + H2CO3
Balanced, the equation is:
2HCl + Na2CO3  2NaCl + H2CO3
Note that H2CO3 appears as a product, so, it is canceled out and replaced (because it is one of
the special gas forming reactions) with CO2 (g) and H2O (l)
2HCl + Na2CO3  2NaCl + H2CO3
2HCl + Na2CO3  2NaCl + CO2(g) + H2O(l)
Look up the solubility of the rest of the molecules to complete the Molecular Equation:
2HCl (aq) + Na2CO3 (aq)
 2NaCl (aq) + CO2(g) + H2O(l)
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Next write the Total Ionic Equation:
2H1+(aq) + 2Cl1-(aq) + 2Na1+(aq) + CO32-(aq)  2Na1+(aq) + 2Cl1-(aq) + CO2(g) + H2O(l)
Next write the NIE:
2H1+(aq) + CO32-(aq)  + CO2(g) + H2O(l)
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