Writing Chemical Equations
Chemical equations for solution reactions can be written in
th different
three
diff
t forms;
f
molecular
l l equations,
ti
complete
l t ionic
i i
equations, and net ionic equations. In class, so far, we have
been writing molecular equations.
This tutorial will show you the differences between these
types of equations, and give you some practice in writing
equations in the different forms.
The three types
yp of equations
q
are:
I. The molecular equation that gives the overall reaction
stoichiometry but not necessarily the actual forms of the
stoichiometry,
reactants and products in solution.
II. The complete ionic equation that represents all strong
electrolyte products and reactants in their ionic form.
III. The net ionic equation that gives only those reactant
p
that undergo
g change
g in the
and pproduct components
reaction, and does not mention other components,
called spectator ions, that do not undergo change.
The Molecular Equation
Let’s look at the reaction where aqueous silver nitrate and
aqueous sodium chloride react together to form a
silver chloride precipitate and sodium nitrate that remains in
solution.
First let’s connect names with molecular formulas:
aqueous silver nitrate = AgNO3(aq)
aqueous sodium chloride = NaCl(aq)
silver chloride precipitate = solid silver chloride = AgCl(s)
sodium nitrate in solution = aqueous sodium nitrate
= NaNO3(aq)
The Molecular Equation
If we combine these molecular formulas into a reaction
we get the molecular equation:
AgNO3(aq) + NaCl(aq)
NaCl(aq)=AgCl(s)
AgCl(s) + NaNO3(aq)
Fortunatelyy for yyou this simple
p example
p is alreadyy balanced
as written, so you don’t have to put any extra effort into
balancing the equation.
The Complete Ionic Equation
The molecular equation contained three different ionic
compounds
p
in aqueous
q
solution. All three of these ionic
compounds are strong electrolytes and would ionize,
so in reality the molecules AgNO3(aq), NaCl(aq),
and NaNO3(aq) don’t
don t exist!
The better way to represent the true chemistry in this
reaction
ti to
t break
b k all
ll strong
t
electrolytes
l t l t into
i t their
th i componentt
ions. Thus,
AgNO3(aq) = Ag+ (aq) + NO3- (aq)
NaCl(aq)= Na+ (aq) + Cl-(aq)
NaNO3(aq) = Na+ (aq) + NO3- (aq)
The Complete Ionic Equation
When we break all strongg electrolyes
y into their component
p
ions we get the complete ionic equation. For this reaction
the compete ionic equation is:
Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl-(aq)=
AgCl(s) + Na+ (aq) + NO3- (aq)
The Complete Ionic Equation
Notice the only thing you can separate into ions are aqueous
ionic compounds, emphasis on aqueous, emphasis on ionic.
! Never try to separate covalent compounds into ions.
! Never try to separate a solid(s) , liquid (l) or a gas(g)
into ions.
In this equation AgCl(s) is ionic, but it doesn’t get changed
into ions because it is a solid.
solid
Don’t forget this. It will trip you up if you don’t remember,
andd that
h will
ill mean points
i off.
ff
The Net Ionic Equation
The complete ionic equation,
Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl-(aq)=
AgCl(s) + Na+ (aq) + NO3- (aq)
contains two ions that appear on both sides of the chemical
equation, Na+(aq) and NO3-(aq). Things that appear
unchanged on both sides of a chemical reaction are called
spectators. Essentially the showed up and watched the
reaction,, but theyy didn’t actuallyy take ppart in it.
The Net Ionic Equation
In the Net Ionic reaction we remove the spectator ions.
Thi helps
This
h l us to emphasize
h i the
h important
i
molecules
l l andd ions
i
that are undergoing the reaction, and generally makes the
equation shorter and easier to write down.
The net ionic equation for this reaction is:
Ag+ (aq) + Cl-(aq)=AgCl(s)
Example problem
Write the balanced molecular, complete
p
ionic and net ionic
equations for the reaction of barium nitrate and potassium
chromate in solution, where they react to form a barium
chromate precipitate and aqueous potassium nitrate
nitrate.
Write
W
it the
th equations
ti
for
f the
th reaction
ti off barium
b i
nitrate
it t andd
potassium chromate to form solid barium chromate and
aqueous potassium nitrate.
First, what are the molecules?
Barium nitrate = Ba(NO3)2
Potassium chromate = K2CrO4
Barium chromate = BaCrO4
Potassium nitrate = KNO3
Write
W
it the
th equations
ti
for
f the
th reaction
ti off barium
b i
nitrate
it t andd
potassium chromate to form solid barium chromate and
aqueous potassium nitrate.
Combining the molecules and putting in the physical forms
to get an unbalanced molecular equation:
Ba(NO3)2 (aq) + K2CrO4 (aq)=BaCrO4(s) +KNO3 (aq)
Ba and CrO4are balanced, K and NO3 will balance if we
make it 2 KNO3
Balanced molecular equation:
Ba(NO3)2 (aq) + K2CrO4 (aq) = BaCrO4(s) +2 KNO3 (aq)
Write
W
it the
th equations
ti
for
f the
th reaction
ti off barium
b i
nitrate
it t andd
potassium chromate to form solid barium chromate and
aqueous potassium nitrate.
Breaking all aqueous ionic compounds into their component ions
to get the complete ionic equation we have:
Ba(NO3)2 (aq) = Ba2+(aq)+ 2 NO3-(aq)
K2CrO4 (aq) = 2 K+(aq) + CrO42- (aq)
2 KNO3(aq) = 2 K+(aq) + 2NO3- (aq)
and our complete ionic equation becomes:
Ba2+ (aq) + 2 NO3- (aq) + 2 K+(aq) + CrO42- (aq) =
BaCrO4(s) +2 K+(aq) + 2NO3- (aq)
Write
W
it the
th equations
ti
for
f the
th reaction
ti off barium
b i
nitrate
it t andd
potassium chromate to form solid barium chromate and
aqueous potassium nitrate.
In making the complete ionic equation:
2 (aq)
Ba2+ (aq)
B
( ) + 2 NO3- (aq)
( ) + 2 K+(aq)
( ) + CrO
C O42( )=
BaCrO4(s) +2 K+(aq) + 2NO3- (aq)
Remember that you can’t break the solid into its component
ions.
In addition, notice that this equation contained the covalent
ions NO3- and CrO4- . Covalent ions cannot be broken into
smaller
ll ionic
i i pieces,
i
so you have
h
to
t remember
b what
h t your
covalent ions are, and never break them down into smaller pieces.
Write
W
it the
th equations
ti
for
f the
th reaction
ti off barium
b i
nitrate
it t andd
potassium chromate to form solid barium chromate and
aqueous potassium nitrate.
Finallyy we remove the spectator
p
ions that occur on both sides
of the complete ionic equation to get the net ionic equation:
Ba2+ (aq) + CrO42- (aq) = BaCrO4(s)
One last problem….
Write the molecular, complete ionic and net ionic equations
for the reaction of barium hydroxide and nitric acid.
I stepped it up a notch here. I told you the reactants, but I
didn’t ggive yyou’re the pproducts,, that is up
p to yyou.
Actually this is not as bad as it sounds. The ‘hydroxide’ part
of the name ‘barium
barium hydroxide
hydroxide’ tells you that this compound
is a base. The name ‘nitric acid’ tells you the other reactant
is an acid. So you have an acid base reaction, that means you
should be looking for H+ and OH- to be reacting to make H2O
somewhere in this reaction.
Write the molecular, complete ionic and net ionic equations
for the reaction of barium hydroxide and nitric acid.
Associating molecular formula’s with names
R t t
Reactants:
Barium hydroxide = Ba(OH)2
Nitric Acid = HNO3
Products:
?
H2O (product of an acid-base reaction)
Write the molecular, complete ionic and net ionic equations
for the reaction of barium hydroxide and nitric acid.
Let’s write the reactants and the only known product to see
if we can figure
g
out the missingg pproduct
Ba(OH)2 + HNO3 = ? + H2O
In acid base reactions like this, the OH of one molecule reacts
with the H of the other molecule to make water. Let’s
combine the other parts of each reactant (Ba and NO3) to make
the other product.
BaNO3
Write the molecular, complete ionic and net ionic equations
for the reaction of barium hydroxide and nitric acid.
But BaNO3 doesn’t look right.
Ba should be a +2 ion, and NO3 is a -1 ion, so the charges
don’t balance. The proper product should be:
Ba(NO3)2
The unbalanced molecular equation then is :
B (OH)2 + HNO3 = Ba(NO
Ba(OH)
B (NO3)2 + H2O
Write the molecular, complete ionic and net ionic equations
for the reaction of barium hydroxide and nitric acid.
Balancing we get:
Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
And adding physical forms to get our molecular equation
we have:
h
(
)2 ((aq)
q) + 2 HNO3 ((aq)
q) = Ba(NO
( 3)2 ((aq)
q) + 2 H2O(l)
()
Ba(OH)
Write the molecular, complete ionic and net ionic equations
for the reaction of barium hydroxide and nitric acid.
Breaking the aqueous ionic compounds into their ions we get:
Ba(OH)2 (aq) = Ba2+(aq) + 2OH-(aq)
2 HNO3 (aq)
( ) = 2H+(aq)
( ) + 2 NO3-(aq)
( )
Ba(NO3)2 (aq) = Ba2+(aq) + 2 NO3-(aq)
(And I hope you didn’t try to break the H2O down.
It is NOT ionic and it is NOT aqueous)
The complete ionic equation then is:
Ba2+(aq) + 2OH-(aq) + 2H+(aq) + 2 NO3-(aq) =
Ba2+(aq) + 2 NO3-(aq) + 2H2O(l)
Write the molecular, complete ionic and net ionic equations
for the reaction of barium hydroxide and nitric acid.
Removing the spectators we have the net ionic equation:
2OH-(aq)
( ) + 2H+(aq)
( ) = 2H2O(l)
Taking all terms to the lowest common denominator we have
the final net ionic equation:
OH-(aq) + H+(aq) = H2O(l)
(Which is a true acid-base
acid base reaction)
Practice Problems
1. Write the molecular, complete ionic and net ionic
equations that describe the solution reaction of sodium
sulfate with lead nitrate. The solid product of this reaction
is lead sulfate.
2. Write the molecular, complete ionic and net ionic
equations that describe the solution reaction of sulfuric acid
with the aluminum hydroxide.
(Try them first,
first before you see my answers)
Write
W
it the
th equations
ti
that
th t describe
d
ib the
th solution
l ti reaction
ti off
sodium sulfate with lead nitrate. The solid product of this
reaction is lead sulfate.
Molecules involved:
sodium sulfate = Na2SO4
lead nitrate = Pb(NO3)2
lead sulfate = PbSO4
Some other unnamed product?
1st try at molecular equation:
Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + ?
Combining
C
bi i the
h other
h halves
h l
off the
h reactants
to make the other aqueous product
Na2SO4((aq)
q) + Pb(NO
( 3)2 ((aq)
q) = PbSO4 ((s)) + NaNO3((aq)
q)
Balancing:
Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + 2 NaNO3(aq)
Write
W
it the
th equations
ti
that
th t describe
d
ib the
th solution
l ti reaction
ti off
sodium sulfate with lead nitrate. The solid product of this
reaction is lead sulfate.
Starting guess:
Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + ?
Combining the halves of the reactants that weren’t used in
Making PbSO4 you predict that missing product is NaNO3.
Next guess
Na2SO4((aq)
q) + Pb(NO
( 3)2 ((aq)
q) = PbSO4 ((s)) + NaNO3((aq)
q)
Balancing for final molecular equation:
Na2SO4(aq) + Pb(NO3)2 (aq) = PbSO4 (s) + 2 NaNO3(aq)
Write
W
it the
th equations
ti
that
th t describe
d
ib the
th solution
l ti reaction
ti off
sodium sulfate with lead nitrate. The solid product of this
reaction is lead sulfate.
Breaking up the aqueous ionic compounds:
Na2SO4(aq) = 2Na+(aq) + SO42-(aq)
Pb(NO3)2 (aq) = Pb2+(aq) + 2 NO3-(aq)
2 NaNO3(aq) = 2 Na+(aq) + 2 NO3-(aq)
Complete ionic equation:
2Na+(aq) + SO42-(aq) + Pb2+(aq) + 2 NO3-(aq) =
PbSO4 ((s)) + 2 Na+((aq)
q) + 2 NO3-((aq)
q)
Net ionic equation:
SO422-(aq) + Pb22+(aq) = PbSO4 (s)
Write the equations that describe the solution reaction
of sulfuric acid with the aluminum hydroxide.
Reactants:
Sulfuric acid = H2SO4
Aluminum hydroxide = Al(OH)3
Products:
?
Acid and base, so H2O
Combing the other halves of the molecules
Al3+ + SO42But the charges don
don’tt work for a 1:1 complex
so Al2(SO4)3 is better
Write the equations that describe the solution reaction
of sulfuric acid with the aluminum hydroxide.
Unbalanced molecular equation:
H2SO4 + Al(OH)3 = H2O + Al2(SO4)3
Balancing:
3 H2SO4 + 2 Al(OH)3 = 6 H2O + Al2(SO4)3
Addingg pphysical
y
forms to gget our final molecular equation:
q
3 H2SO4 (aq) + 2 Al(OH)3 (aq) = 6 H2O(l) + Al2(SO4)3(aq)
Write the equations that describe the solution reaction
of sulfuric acid with the aluminum hydroxide.
Separating aqueous ions:
3 H2SO4 (aq) = 6H+(aq) + 3 SO422-(aq)
2 Al(OH)3 (aq) = 2Al3+(aq) + 6 OH-(aq)
Al2(SO4)3(aq) = 2Al3+(aq) + 3 SO42-(aq)
Putting into complete ionic equation:
6H+((aq)
q) + 3 SO42-((aq)
q) + 2Al3+((aq)
q) + 6 OH-((aq)
q) =
6H2O(aq) + 2Al3+(aq) + 3 SO42-(aq)
Write the equations that describe the solution reaction
of sulfuric acid with the aluminum hydroxide.
Removing spectators for net ionic:
6H+(aq)+ 6 OH-(aq) = 6H2O(aq)
Taking to lowest common denominator to get our final
net ionic equation:
1 H+(aq)+ 1 OH-(aq) = 1 H2O(aq)