1. What exactly are “molecular
equations”, “ionic equations” and “net ionic equations”?
2. Why is the first equation
called a “molecular equation”?
3. Why do we need ionic
equations and net ionic equations?
4. How do we write molecular
equations?
5. Starting with a balanced
molecular equation, how do we get to the ionic equation?
6. Drill on converting the correct formulas into
their ionic forms in the ionic equations:
7.
Once you have the ionic equation, how do you get to the net ionic
equation?
8.
How to do a quick check on the accuracy of the net ionic equation?
1. What exactly are “molecular
equations”, “ionic equations” and “net ionic equations”?
Using the simple example of the reaction when aqueous solutions of sodium chloride and silver nitrate are mixed. The chemical equation you would usually write is as follows: (Note: Arrows may not show up properly on your monitor. If you see something odd (like ¾ R), it’s supposed to be a reaction arrow.)
Molecular equation: NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)
(or Overall equation)
Ionic equation: Na+ (aq)
+ Cl- (aq) + Ag+ (aq) +
NO3- (aq) ¾® AgCl (s) + Na+ (aq)
+ NO3- (aq)
(or total ionic equation)
Net ionic equation: Cl-(aq) + Ag+ (aq) ¾® AgCl (s)
which is usually re-written as Ag+ (aq) + Cl-(aq) ¾® AgCl (s)
2. Why is the first equation
called a “molecular equation”?
NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)
It is actually rather odd to call this equation “molecular” since there are no molecules in the reaction (other than the solvent, water). All of the reactants and products are ionic compounds. None of them are molecules.
How do
you know whether the reactants and products are ionic or molecular? If you have not learned this yet, or perhaps
forgotten this, you should certainly learn it immediately. To tell whether a compound is molecular or
ionic, the general rule is that an ionic compound is made
of metal cations and nonmetal
anions. A molecular compound
is made of nonmetals only.
You would only need to check the first element listed in the
formula. If it is a metal, it is most
likely to be ionic. Of course, this is
only a general rule, and there are exceptions.
One of the important exceptions is the ammonium ion (NH4+),
which is a cation, but does not contain a metal.
Ionic
compounds: Metal cations + Nonmetal anions
Molecular
substances: All nonmetals
Important
exception: Compounds with NH4+
are ionic even though they have no metals.
Why then is “NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)” called a “molecular equation” when there are no molecules? The reason is because the formulas are shown
with no charges even though the compounds contain ions with charges. For example, “NaCl (aq)” represents Na+
and Cl- ions, each surrounded by water
molecules. Technically speaking we
should really write Na+Cl- (aq) and Ag+NO3-
(aq). But, by convention, we don’t. These compounds are written as if they
were molecules, with no charges showing. These equations are therefore called
“molecular equations”. They are useful
in that they show the correct stoichiometry and physical states without the
cumbersome charges. The formulas are, of
course, written so that the positive charge from the cations equal to the
negative charge from the anions. In
other words, the formulas of ionic compounds must have a net charge of
zero. At this point, you should have
already learned how to write these formulas: Na+ and SO42-
make Na2SO4 and not NaSO4-
and not NaSO4. (If you
need practice writing formulas correctly, click
here.)
3. Why do we need ionic
equations and net ionic equations?
Ionic equations are written to help you get to the net ionic equations, which are equations that show the essence of the reaction. When you become experienced at this, you would not have to write ionic equations in order to determine what the net ionic equations are. But, for now, you should learn to do this stepwise.
Going back to our original molecular equation:
NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)
The essence of this
reaction is the Ag+ cations combining with the Cl- ions to form the insoluble AgCl precipitate.
Ag+ (aq) + Cl- (aq) ¾® AgCl (s)
This is called a net ionic equation.
The Na+ and NO3- ions are of no concern. They are called spectator ions. The cation (Na+) can be replaced
by other cations such as K+, NH4+, Mg2+,
etc., and the anion (NO3-) can be replaced by other anions such as ClO3-, C2H3O2-, and the essence of the reaction would still be the same: Ag+ (aq) + Cl- (aq) ¾® AgCl (s).
4. How do we write molecular
equations?
To get a full discussion of this, you must refer to your textbook. I can only give you a summary:
In writing molecular equations:
1. Predict the formula of the products solely according to the charges of the ions.
· For metal cations that have variable charges, you can deduce their charges by looking at the charge of the anion.
For example, in FeCl2, since you know the chloride ion has a charge of 1– each, the iron must have a charge of 2+.
·
In some cases it may not be possible to predict
what the charge is on the cation. In that
case you may refer to the activity series to see which charge to use.
For example, in the reaction (Fe + HCl ¾® ?), you cannot deduce the
charge of Fe after the reaction. By
examining the Activity Series, you see that Fe is likely to form 2+, so you
would expect the product to be FeCl2.
·
· Most of the charges of the anions you should already know (such as sulfate having a charge of 2–, bromide having a charge of 1-, etc.)
· Occasionally you will run into anions that you had not encountered before. In such cases you can deduce the charge of the anion by looking at the charge of the caiton.
For example, in Na2S2O3, since you know each sodium ion is 1+, the anion must have a charge of 2–. Incidentally this is the thiosulfate ion, S2O32–.
Do not try to balance the equation yet!!
2. Decide whether the reaction will take place.
For precipitation reactions,
use solubility rules to decide whether a ppt (precipitate) will form.
Generally, all neutralization reactions will occur.
For redox (single
replacement-type), use the activity series to decide.
3. Write in physical states of each product according to solubility rules, and physical states of pure substances. If you have not learned the formulas and physical states of pure substances (element not bonded to another element) you should study Unit II of the Nomenclature Tutorial. You will find it invaluable! Click here to get to the Nomenclature Tutorial.
4. Balance the equation by adding only coefficients, not by changing subscripts.
Example: Predict the products to this reaction:
Na2CO3 (aq) + Cu (s) ¾®
Since copper has a variable charge, you should look up in the activity series, the charge for the copper ion (which turns out to be Cu2+). Thus the product would be CuCO3 (s) + Na (s). (You should know that the carbonate has a charge of 2–) and Na should not be written as Na2.
Na2CO3 (aq) + Cu (s) ¾® CuCO3 (s) + Na (s)
To balance the number of Na, write “2” in front of Na.
Na2CO3 (aq) + Cu (s)
¾® CuCO3
(s) + 2 Na (s)
Will the reaction occur? Examine the
Activity Series. Cu is lower than Na, so
Cu cannot replace Na.
No, the reaction will not occur. Be careful you do not say that the reaction will occur “because a precipitate has formed!” This is not a precipitation-type reaction, so we are not concerned as to whether a solid is formed.
5. Starting with a balanced
molecular equation, how do we get to the ionic equation?
The ionic equation, or total ionic
equation, is the intermediary step between molecular equation
and net ionic equation.
You take the molecular equation and convert all the
strong electrolytes
into their ionic form.
What
are strong electrolytes? They are
defined as compounds that completely dissociate into ions when placed in water,
and thus conduct electricity well. They
consist of mainly two types of compounds.
Learn this general rule well!!
1) Ionic
compounds that are soluble and dissolved
in water.
2) Strong
acids dissolved in water.
Note
that both are dissolved in water, and they are easily recognized in a molecular equation
in that they MUST have the physical state (aq). In addition to having (aq) designated, you must check that they are ionic and not
molecular, OR they must be a strong acid.
What
is a strong acid? How do you know
whether it is a strong acid?
The
answer is simple. First of all you can
easily recognize what is an acid.
Their formulas usually start with H.
For
example: HF, HCl, HNO2, H2SO4, H3PO4
How
do you know whether they are strong
acids? The answer is also very simple: You memorize
the names and formulas of the 6 common strong acids. All other acids can be assumed to be weak.
The
6 common strong acids are: HNO3, H2SO4, HClO4,
HCl,
Take
the time to memorize this now! You will
find it MOST helpful to remember this list throughout the semester and the next
(in Gen Chem II and organic chemistry).
HERE IS THE IMPORTANT RULE FOR DECIDING WHETHER TO SEPARATE A FORMULA
INTO IONS.
The scientific term for separating a formula into ions is
"dissociate" which means to "split" into ions:
In
writing total ionic equations
(your textbook may refer to them as just ionic equations):
Convert
into separate ions, only the
following:
1.
Ionic compounds that
are dissolved in water (indicated by (aq)).
2.
Strong acids
dissolved in water (also indicated as (aq)).
All
others should be left intact!
6. Drill on converting the correct formulas into
their ionic forms in the ionic equations:
In the following
exercise write down what you think the answers should be, then scroll down to
check your answers.)
e.g. K3PO4 (aq) becomes ________________
3Na2CO3 (aq) becomes ________________
CH3OH (aq) becomes ________________
Ba(CH3CO2)2 (aq) becomes ________________
2(NH4)2SO4 (aq) becomes ________________
NaCl (s) becomes ________________
H2O (l)
becomes ________________
H2S (g) becomes ________________
HNO2 (aq) becomes ________________
C12H22O11 (aq) (This is table sugar which you well know can dissolve in water.)
becomes ________________
Zn (s) becomes _________________
H2 (g) becomes _________________
Answers to above exercise:
K3PO4 (aq) becomes 3K+ (aq) + PO43- (aq)
3Na2CO3 (aq) becomes 6Na+ (aq) + 3CO32- (aq)
CH3OH (aq) becomes CH3OH (aq) (unchanged because it is not ionic)
Note that this is not an acid.
Ba(CH3CO2)2 (aq) becomes Ba2+ (aq) + 2 CH3CO2- (aq)
2(NH4)2SO4 (aq) becomes 4NH4+ (aq) + 2SO42- (aq)
NaCl (s) becomes NaCl (s) (unchanged because it is not dissolved in water)
H2O (l) becomes H2O (l) (unchanged because H2O is not on our list of strong acids)
H2S (g) becomes H2S (g) (unchanged because it is not dissolved in water, also it is not on our list of strong acid)
HNO2 (aq) becomes HNO2 (aq) (unchanged because it is not on our list of strong acids)
C12H22O11 (aq) (sugar dissolved in water)
becomes C12H22O11 (aq) (unchanged because it is not ionic)
Zn (s) becomes Zn (s) (unchanged because Zn is not an ionic compound)
It does NOT become Zn2+!
H2 (g) becomes H2 (g) (unchanged because is not ionic)
It does NOT become H+!
For example, for the molecular equation
Sr (s) + 2H2O (l) ¾® Sr(OH)2 (aq) + H2 (g)
the ionic equation is
Sr (s) + 2H2O (l) ¾® Sr2+ (aq) + 2OH– (aq)
+ H2 (g).
It is not...
Sr2+ (s) + 2H2O (l) ¾® Sr2+ (aq) + (OH)22– (aq) + 2H+ (g).
Sr (s) is not ionic! It is made of neutral atoms.
Neither is H2, which is a molecule.
It is not made up of ions.
Note also
that we do not write (OH)22– because you should recognize
(OH) as the hydroxide ion
When you write (OH)22– you are implying that the two OH’s are bonded together: (HOOH)2–.
Instead we are talking about two hydroxide ions:
2OH– which stands for 2 individual ions: OH– and OH–.
Take the
example of HNO2 (aq).
First you should recognize it as an acid. Second you should recognize it as being a
WEAK acid (because it is not on our list of strong acids).
It should therefore not be dissociated and should be written intact as HNO2 (aq).
7. Once you have the ionic
equation, how do you get to the net ionic equation?
The hardest part is writing the ionic equation. Once you have the correct ionic equation, the net ionic equation is easy to arrive at. All you have to do is cancel out the "spectator ions", which are ions that appear on both sides of the equation. What remains is the net ionic equation. Be sure to check that the coefficients are reduced to the lowest ratio.
Example: NaCl (s) + Ag+ (aq) + NO3- (aq) ¾® AgCl (s) + Na+ (aq) + NO3- (aq)
becomes NaCl (s) + Ag+ (aq) ¾® AgCl (s) + Na+ (aq)
Note
that this is different from the reaction described at the top of the page. Here, the reactant NaCl is provided as the solid
sodium chloride rather than the aqueous solution.
Example:
molecular equation: H2SO4 (aq) + 2KOH (aq) ¾® 2H2O (l) + K2SO4 (aq)
ionic equation: 2H+ (aq) + SO42- (aq) + 2K+ (aq) + 2OH- (aq) ¾® 2H2O (l) + 2K+ (aq) + SO42- (aq)
net ionic equation: 2 H+(aq) + 2OH- (aq) ¾® 2H2O (l)
which needs to be reduced to H+ (aq)
+ OH-(aq) ¾® H2O (l)
Spectator ions are SO42- and K+.
8. How to do a quick check on the accuracy of
the net ionic equation?
The charges in the net ionic equation are conserved:
The net charge on the left side of the equation must equal
to the net charge on the right side.
This gives
you a quick check on whether you made any careless mistakes.
For example,
NaCl (s) + Ag+ (aq) ¾® AgCl (s) + Na+ (aq)
the left side has a net charge of +1, and so does the right side.
In H+ (aq) + OH-(aq) ¾® H2O (l), the net charge on the left side is zero, and so is the net charge on the right.
9. What do you do with weak acids?
There is
no new rule that applies to weak
acids. Weak acids do not fit the
criterion for being dissociated. As
pointed out in an earlier example, HNO2 (aq) is such a
case. It does dissolve in water, but as
a weak acid, it exists primarily as the molecule HNO2 and does not
split up into ions.
For example,
HNO2
(aq) + NaOH (aq) ¾® H2O (l) + NaNO2
(aq)
becomes HNO2
(aq) + Na+ (aq) + OH-
(aq) ¾®
H2O (l) + Na+ (aq) + NO2-
(aq)
Net ion eqn: HNO2 (aq) + OH-(aq) ¾® H2O (l) + NO2- (aq)
The spectator ions are Na+.
Incidentally, did you ever wonder why H2O is not considered an acid
(with H written in front like the other acids)?
Well, it actually is an acid, just a very, very weak acid. You will work with H2O as a weak
acid in Gen Chem II.
What do you with weak bases just as NH3 (aq)? The answer is, no different from any formulas
that do not fit the criteria for dissociation.