Lighthouse Scientific
Education Presents
a lecture in the
Chemical Reaction series.
The topic; Net Ionic
Equations Part 1.
Part 2 is in the
Solutions series and uses
the solubility rules.
Not all students will study
this topic to that depth.
Know what is expected of you.
Material in this lecture relies
on an understanding of the
previous lectures:
Ionic Bonding and
Balancing Chemical Equations.
The lecture begins
with some definitions
including chemical equation,
total ionic equation,
net ionic equation
and spectator ions.
Then it is on to the steps for
producing net ionic equations
with 3 example problems that
highlight some of the issues
that come up with
net ionic equations.
Definitions: First up is
the chemical equation and
this definition is in the
context of net ionic equations.
Such chemical equations
may be given with balanced
coefficients or have balancing
as a step prior to the
process of getting
net ionic equations.
At this point all ionic
compounds are shown
as neutral compounds.
Below is an acid/base reaction
written as a chemical equation.
Note the 'aq' subscript.
As a refresher, aq is the phase
description that stands for
aqueous or dissolved in water.
Another more useful term
associated with the aq
subscript is dissociates.
That indicates that the cations
and anions are separated
from each other and found
in solution surrounded
by water molecules.
In this reaction both
KOH and HNO3 are dissolved
in water. KOH is an ionic
compound that dissolves
or dissociates into its
constituent ions.
HNO3 is an ionic compound
that is dissolves or dissociates
into its constituent ions.
Another way of saying this
is that KOH and HNO3 are
soluble which follows from
the definition of solubility.
It is the ability of a substance
to dissolve in a solvent.
For net ionic equations
that solvent will be water.
So the term soluble means
able to dissolve in water
and insoluble means unable
to dissolve in water.
Solubility is a topic for
the Solution lecture in the
Solutions series of lectures.
The definition of total
ionic equation is the
balanced chemical equation
in which ionic compounds
are shown in their
predominant aqueous form.
The KOH in the chemical
equation is shown as
the potassium cation and the
hydroxide anion since this
how KOH is found in water.
It is a strong base.
The aq subscript is necessary
here since ions do not exist
for any length of time
outside of their solvent.
Importantly, all of
the compounds with the
aq subscript are written as
ions in the total ionic equation.
The nitric acid and the salt
potassium nitrite are also
shown in their
predominate aqueous form.
It is the aq subscript that
is the indicator to break the
ionic compounds into
their constituent ions.
In the second lecture on net
ionic equations a set of rules
is offered that can also
be used in determining
soluble ionic compounds.
The last definition, and the
topic of this lecture, is the
net ionic equation. It is
the balanced chemical equation
that only shows species that
participate in the reaction.
It is the total ionic
equations that omits
or ignores spectator ions.
And just what are
spectator ions?
They are ions that remain
unchanged in a reaction.
They do not participate
in the reaction.
They are there but do not react.
Referring back to the
total ionic equation we see
that the potassium ion is on
the reactant side
and the product side.
It is unchanged
in the reaction.
The same can be said
of the nitrate ion.
It too is unchanged.
These ions are spectators,
or onlookers, to the reaction.
Not including them in the net,
or remaining, ionic equation
leaves just the reactants that
participate in the reaction and
the products that they form.
The net ionic equation just
shows the actual reaction.
A quick note
on the products found
in a net ionic equation.
They need to be stable
molecules or compounds.
They can be a gas or
a liquid or a solid.
What they cannot be is
a soluble ionic compound.
KNO3 is not a product in the
net ionic equation because
it does not form
a stable compound.
The potassium and the nitrate
most certainly need to be
in the reaction because
a hydrogen ion or a
hydroxide ions cannot
be added by itself.
Ions are added as
neutral compounds.
So why spend the time
and energy getting a
net ionic equation?
By comparing the
net ionic equation to either
the chemical equation or
the total ionic equation
we can see its attraction.
The net ionic equation just
shows the relevant parts of
the reaction which makes
it easier to follow.
Simplicity is its goal.
Getting a net ionic equation
from a chemical equation
really just follows from
the definitions listed here.
Net ionic equations are usually
just limited to reactions that
take place in water and
have soluble ionic compounds.
Be sure that is the
case before proceeding.
The first step in getting
a net ionic equation is to
write out the balanced
chemical equation.
Write out the
reactants and products.
Be sure to include the
subscripts of state of matter.
Especially the
aqueous subscript.
Then rewrite the chemical
equation using ions for
compounds listed as aqueous.
Essentially, write out
the total ionic equation.
It should be pointed out here
that this step is modified
in the more advanced
Net Ionic Equation
Part II lecture.
In that lecture the student
is responsible for determining
which compounds are soluble.
They do not rely on
the aq subscript.
To dramatically simplify
the process a table of
solubility rules is provided.
If the student's course work
regarding net ionic equations
involve using this rules
then follow this lecture up
with the part II lecture.
If not then continue on.
And that involves noting,
in the total ionic equation,
which ions participate
in the reaction and which
are spectator ions.
That allows for step 3.
Rewrite the total
ionic equation
without spectator ions.
Finally, double check that
the equation is still balanced.
Also double check the total
charge is balanced too.
A balanced charge is one in
which the total charges on
the reactants is equal to the
total charges on the products.
The charge contribution of
an ion is found by multiplying
its charge by its coefficient.
Onto some practice problems.
The process begins by writing
out the chemical equation
with its states and
phase subscripts.
Problem 1 has aqueous
silver nitrate reacting with
aqueous potassium chloride to
produce solid silver nitrate
and aqueous potassium nitrate.
A question to always
ask when dealing with
chemical equations is
whether or not the
equation is balanced.
It cannot be assumed that
every equation handed to a
student is balanced. In fact,
if there are no coefficients
attached to the equation
than that is a red flag for a
potentially
unbalanced equation.
Inspection of this equation
shows each element, ion and
polyatomic ion coming up once
as a reactant and
once as a product.
The equation is balanced.
That allows us to rewrite the
chemical equation using ions
for compounds listed as
aqueous. Write a total ionic
equation. The first reactant
in the chemical equation is
AgNO3 and it is dissociates
in water as indicated by the
aq subscript.
The total ionic equation will
therefore write the
compound as dissolved ions.
Ag+ and NO3-.
Aqueous subscripts
are included.
The KCl is also in
the aqueous phase so it
is written as dissolved
K+ and Cl-.
Aqueous subscripts are included.
That is the reactant side.
For the products,
AgCl is a solid
it is placed in the total
ionic equation unchanged.
That is not the case for
the other product KNO3.
It is in the aqueous
phase and written as the
dissociated K+ ion and
the negative nitrate ion.
Aqueous subscripts
are included.
The total ionic
equation is complete.
While we have it here
we should note which ions
participate in the reaction and
which ones are spectator ions.
A good place to start is to
look at the formed product.
The solid AgCl tells us that
the silver ion in the reactant
is a participating ion
as is the chloride ion.
These ions react to form AgCl.
What about the spectator ions?
Are there ions that remain
unchanged in this reaction?
Yes, the nitrate is found
as a dissolved ion on the
reactant and products sides.
So is the potassium ion.
We are now ready for the
third type of equation
the net ionic equation.
We will take the
total ionic equation and
rewrite it without the
spectator ions.
In other words remove the
nitrate ion from both sides
and remove the potassium ion
from both sides leaving the
reaction aqueous silver ion
plus aqueous chloride ion react
to form solid silver chloride.
The reaction has been
reduced to its essentials.
Is this equation balanced?
There is 1 silver and 1 chloride
on each side so yes.
If the answer is no
than return to the
chemical equation and
verify that it is balanced.
What about the charge.
Is the total charge on the
reactant side equal to the
total charge on
the product side?
Silver chloride is
a neutral compound
so its net charge is 0.
On the reactant side
there is a single positive
charge and a single
negative charge which
combine for a net charge of 0.
Yes, charges are balanced
and the equation checks out.
The problem is complete.
The second practice
problem begins the same way.
Write out the balanced
chemical equation.
Be sure to include the
subscript state of matter.
Especially the
aqueous subscript.
That chemical equation
is aqueous barium chloride
reacting with aqueous
sodium sulfate to produce
solid barium sulfate and
aqueous sodium chloride.
Is the equation balanced?
The lack of coefficients
means we should check.
Since this is not a lecture
on balancing chemical equations
we will cut to the chase
and add a 2 in front of
the sodium chloride. Now
that is a balanced equation.
Proceed to step 2 and rewrite
the chemical equation using
ions for compounds
listed as aqueous.
Write the total ionic equation.
Starting with the first
reactant: aqueous
barium chloride; BaCl2.
It is an ionic compound that
dissociates in water and
is written as an aqueous
Ba+2 ion and
2 aqueous Cl- ions.
Problem 2 brings in a
new element in that there
are compounds that have
multiple anions or cations.
There is only 1 barium chloride
compound in this reaction
but it has 2 chloride ions and
the total ionic equation needs
to reflect that ratio.
It does so by taking that
subscript 2 and putting
it as a 2 coefficient.
This same feature
can be seen with
the aqueous sodium sulfate.
As a dissolved ionic
compound it is found in the
total ionic equation as
2 dissolved sodium ions
and 1 sulfate ion.
Again the 2 coefficient in
front of the sodium comes
from the 2 subscript in
the sodium sulfate.
Now to the products.
Barium sulfate is a solid
and remains unchanged in the
total ionic equation.
The other product
aqueous sodium chloride is
a dissolved ionic compound
and written in the total
ionic equation as 2 solvated
sodium ions and 2
solvated chloride ions.
The second new element in
this problem is the presence
of a non-'1' coefficient
in front of a compound.
The total ionic equation
is a balanced equation.
That 2 in front of the NaCl
has to be include in the
total ionic equation.
There are 2 sodium ions
in 2 sodium chlorides.
There are 2 chloride ions
in 2 sodium chlorides. So,
when including a dissolved ion
in the ionic equation it must
be given a coefficient that
represents both the
coefficient and subscript
of that ion when written
in the compound form.
To further clarify this
statement consider a situation
in which the aqueous
sodium sulfate had
a coefficient of 3 in a
balance chemical equation.
Not this equation but
a hypothetical reaction.
In writing this compound as an
ion in a total ionic equation
what coefficient would
the sodium ion receive?
There are 3 sodium
sulfates and each
one has 2 sodium ions.
The way to get the value
is to take the coefficient of
the compound, 3, and multiply
it by the subscript on the sodium.
The correct coefficient
for the sodium ion is 6.
Using the same equation
for the sulfate, coefficient
of 3 times the implied
subscript of 1
gives 3 sulfates.
Be sure not to
confuse the subscript 4
on the oxygen of
the sulfate as the
subscript in this math.
There is only 1 sulfate
polyatomic ion in the
sodium sulfate compound.
Back to the problem and it
is time to identify which ions
participate in the reaction
and which ones are spectators.
Taking the product solid
barium sulfate as a guide,
the reactant barium ion and the
reactant sulfate ion are both
require for this reaction.
Ions found unchanged on both
sides of the reaction include
the chloride ion
and the sodium ion.
The third equation, the net
ionic equation is a rewrite
of the total ionic equation
without the spectator ions.
The chloride ions are removed
as are the sodium ions.
What remains is the
net ionic equation.
Aqueous barium ion reacts
with aqueous sulfate ion to
form solid barium sulfate.
Sweet and simple.
Is this reaction balanced?
Well, there is 1 barium and
1 sulfate on the reactant side
and 1 barium and 1 sulfate
on the product side.
It is balanced.
What about charge?
The product barium sulfate
is a neutral compound
so its net charge is 0.
On the reactant side the
barium has a plus 2 charge
which is matched by the
minus 2 charge on the sulfate.
Both sides of the reaction have
a net charge of 0.
The charges are balanced.
Our final practice problem
has us write out the
balanced equation with
phase and state subscripts.
Solid magnesium reacts with
aqueous hydrochloric acid to
produce aqueous magnesium
chloride and hydrogen gas.
Is the equation balanced? Well
there are 2 hydrogens on the
product side but 1 on the
reactant side. So, no it isn't.
Adding a 2 to the HCl takes
care of business and it is on to
step 2 which has us rewrite
the chemical equation using
ions for compounds
listed as aqueous.
Write the total ionic equation.
The first reactant is
the solid magnesium.
This goes unchanged in
the total ionic equation.
Not so for the aqueous HCl.
It is found dissociated ions
in solution and is
added to the equation as
2 H+ and 2 Cl- ions.
Remember to include
the coefficient in front
of the compound.
On the product side there
is the aqueous MgCl2.
In the total ionic equation
these aqueous ions are written
as Mg with a plus 2 charge
and 2 Cl- ions.
Be sure to incorporate
the 2 subscript as a coefficient.
The last compound is the
hydrogen gas.
It is not an ionic compound
and is left unchanged. This
brings us to a generalization
Solids and gasses will be
part of the net ionic equation.
And this brings us
to the point where we
identify ions that participate
in the chemical reaction and
those that do not. Unlike
problems 1 and 2 this one
doesn't have a solid as a
product. It does, however,
have a stable, formed
compound; hydrogen gas.
This means that
the hydrogen ion
on the reactant side
participates in the reaction.
While the products do
not contain a solid the
reactants do. Magnesium
solid is transformed into
magnesium ion. Going
from an elemental form
to an ionic form is
a chemical reaction.
The magnesium ion is a
participant in this reaction.
And that leads to
another generalization;
look in the products
for participating ions.
Not all participating ions
will be in the reactants.
As for spectators, there
is a chloride ion on the
reactant and on product side.
It is unchanged and
therefore denoted as spectator.
With these two equations we
are ready to generate the net
ionic equation by rewriting
the total ionic equation
without the spectator ions.
The chloride ion is removed
giving a net ionic equation of
solid magnesium reacts
with aqueous hydrogen ion
producing magnesium
ion and hydrogen gas.
This is redox reaction in
which elections are transferred
from the magnesium metal
to the hydrogen ion.
Is it balanced? Well, there
is one magnesium on both
sides of the reaction.
That there has been a loss
of electrons does not mean the
atom in the reactants is not
the ion in the products.
There are 2 hydrogens on
the reactant side and
2 on the product side.
So, yes, it's balanced.
What about charge? Is the
net charge on the reactants
the same as the net
charge on the products?
On the reactant side
there are 2 plus 1 ions for
a total charge of plus 2.
On the product side there
is 1 ion with a plus 2 charge.
Both sides have a total charge
of plus 2 and are balanced.
And that leads to our
final generalization;
Net ionic equations do
not need to be neutral.
They are not full equations
and omit the charges on the
spectator ions. Therefore,
they can have reactants and
products with
charge as long as the
total charge is balanced.
Recapping the lecture; To
get a net ionic equation begin
by writing out the
balanced chemical equation.
Chemical equations are
balanced equations in which
the ionic compounds are
shown as whole compounds.
Include state of
matter and phase.
Be sure to include
the aq if possible.
Solubility was defined as
the ability of a substance to
dissolve in a solvent.
This lecture dealt strictly
with water as the solvent.
Then rewrite the chemical
equation using ions from the
compounds listed as aqueous.
Write the total ionic equation
That is defined as a balanced
chemical equation in which
ionic compounds are shown in
their predominant aqueous form.
If the compound dissociates
in water write it's ions as
aqueous ions. Be sure
to keep tract of the
coefficients and subscripts
of the compounds so that
dissolved ions come
in the same ratio.
With the total ionic equation
note which ions participate
in reaction and which
are spectator ions.
Spectator ions are the ions
that remain unchanged in
a chemical reaction. These
are dissociated ions that
do not participate in
the chemical reaction.
They are unchanged
in the reaction.
Next rewrite total
ionic equation
WITHOUT SPECTATOR IONS.
The net ionic equation is
the balanced chemical equation
that only shows species that
participate in the reaction.
Finally, double check to see if
the equation is still balanced.
Also check to see that the
reactants and products have
the same net charge.
And that concludes the
lecture on net ionic equations.
If your courses work has
solubility rules attached to
the net ionic equation
go to the follow up
lecture 'Solubility Rules
and Net Ionic Equation II'.
It is in the Water
and Solutions series
[MUSIC]
