The Eight Solution Problem

The Eight Solution Problem
This pre-lab is more involved than usual and so is worth more points than usual.
The pre-lab will be worth 5 points, and the report will be worth 10 points, for a total of 15 points
for this assignment. The Pre-Lab information is below the Procedure.
For this pre-lab, write one (legible) copy and photocopy it. Hand one copy into me at the
start of lab (it must be ready at the start of lab), and write or tape/staple/glue the other
copy into your notebook as usual. You can write directly into your lab notebook and
copy that, and give that to me, as long as it is clear!
Discussion
This experiment should teach you something about 1) using the solubility table, 2)
precipitation reactions, 3) making good observations, and 4) learning how to write molecular,
ionic and net ionic reactions. All of the combinations of substances that you will mix together
will be classified as either a precipitation reaction, which will have at least one solid product, or
“no reaction” which has no products. If both potential products are soluble, then no solid is
formed and no reaction occurs.
When writing a potential reaction, start with any two of the substances below. For
example: (not one of your reactions, just an example)
LiF (aq) + Pb(ClO 3 ) 2 (aq)
Precipitation reactions are a type of double replacement reactions.
AB(aq) + CD (aq) AD(s) + CB (aq)
The cation from the first compound combines with the anion of the second compound. Make the
products, assuming there are products (this is the molecular reaction). When writing the
products, the cation is always written first, and the anion is written second.
LiF (aq) + Pb(ClO 3 ) 2 (aq) LiClO 3 () + PbF ()
When you make the products, you need to make sure that they are neutral. In these types of
reactions, the charge on each ion does not change from reactant to product. So what are the
charges on each ion? Li comes from Group IA, and always gets a 1+ charge.
F comes from Group 7A and always gets a 1- charge.
Pb can have a 2+ or 4+ charge, so you have to look at the anion:
the chlorate ion has a 1- charge, and so the lead ion must have a 2+ charge, to make a neutral
compound. Are the proposed products in a correct ratio?
• LiClO 3 : One ion has a 1+ charge, the other a 1- charge, so they are written 1:1.
Yes, LiClO 3 is correctly written.
• PbF: Pb still has a 2+ charge (from the reactants) and
F is from the halogens and always gets a 1- charge so:
they should not be written 1:1 but 1:2. PbF 2 is the correctly written compound. So the equation
should be:

LiF (aq) + Pb(ClO 3 ) 2 (aq) LiClO 3 () + PbF 2 ()
Now the question of whether or not the reaction will actually occur. Look at the solubility table
(check the index). If one or both of the products is insoluble (becomes a solid), a precipitation
reaction will occur. If neither product is insoluble, no reaction occurs.
To start, look for the ion rather than the compound. According to the table, ionic
compounds containing the chlorate ion are all soluble, and so LiClO 3 is (aq). Looking at the
table again, look for either the Pb 2+ ion or the F - ion. All ionic compounds containing F - are
soluble, except for “fluorides of Mg 2+ , Ca 2+ , Sr 2+ , Ba 2+ , and Pb 2+ ”. This means that PbF 2 is
insoluble and a reaction should occur. Write the reaction with the states and balance it. The
order of the products does not matter.
2 LiF (aq) + Pb(ClO 3 ) 2 (aq) 2 LiClO 3 (aq) + PbF 2 (s)
If both products are soluble, then they are not made; and so you should put the products
in brackets and write “N.R.” for “no reaction”.
For example: LiI (aq) + Mg(CH 3 CO 2 ) 2 (aq)
Find the charge on each ion:
Li 1+ (see above)
F - (a halogen, so 1- charge)
Mg (in alkaline earth metals) so Mg 2+
-
CH 3 CO 2 : this is the polyatomic ion acetate, and the charge on it may be
memorized, or you can figure it out because Mg has a 2+ charge and there are two acetate ions,
so the charge on each ion must be 1-.
Compounds containing I - are soluble with the exceptions of silver ion, mercury ion and lead ion.
Since I is combined with Li, the compound is soluble. It also states in the table that all ionic
compounds containing acetate ions are soluble; therefore you should conclude that there is no
reaction. The equation can be written this way:
LiI (aq) + Mg(CH 3 CO 2 ) 2 (aq) [LiCH 3 CO 2 (aq) + MgI 2 (aq) ] N.R.
You can also put an “X” over the arrow. I just haven’t figured out how to do that on my
computer.
Procedure
For this experiment, there will be eight solutions already made. All of the substances
below have been dissolved in water, and are therefore aqueous solutions. These solutions, in no
particular order, are:
3 M NaOH 0.05 M Na 2 S 0.2 M BaCl 2
1 M NH 4 Cl 0.03 M Na 2 SO 4 0.01 M AgNO 3
0.1 M KI Distilled H 2 O

First, you will mix all combinations of two solutions, and write down what you observe,
in each case.
For instance, starting with 3 M NaOH mix 10 – 20 drops of NaOH with 10-20 drops of
NH 4 Cl and write down any color changes, precipitates formed, odors that go away, or are new.
Then mix NaOH with KI and record what you see, and so forth. Once you have completed all
combinations of two solutions containing NaOH, then start with NH 4 Cl and continue the process.
(Since you have already combined NaOH and NH 4 Cl in the previous series, you do not have to
repeat this experiment.) In all mixtures, use equal numbers of drops of the two substances, and
be consistent in your methodology.
Once you have completed all of the reactions described above, you will have available
another set of the same solutions, and this set of solutions will be labeled A, B, C, D, E, F, G and
H. Solution A does not necessarily correlate with NaOH. Your job will be to use the
information that you learned from the first series of experiments to identify which solution is
contained in each lettered bottle. For this part, you may use only the unknown solutions to figure
out the correct substance in each bottle. Also, these are individual experiments. You should not
be consulting with each other about one another’s results. This is a quiet lab.
Pre-Lab
Your pre-lab consists of this: first write the reactants for each of the potential reactions.
For instance, the first two sets of reactants could be:
NaOH (aq) + NH 4 Cl(aq)
NaOH (aq) + KI (aq)
Write these out for each possible combination. Next, use the solubility table, and figure
out which reactions should occur, and write out what the products should be and the states of
each product. If no solid is expected to form, then a reaction is not occurring. If no reaction is
expected to occur, put the would-be products in brackets and write “N.R.” for “no reaction” after
the arrow.
Remember that each one of these solutions contains an ionic compound, except for the
distilled water which is not a solution but a pure liquid. In order to correctly write out and
balance the chemical reaction, you must determine the correct charges on each ion and, when
forming a product, make sure that the products are in fact neutral! [ This is very important.]
Report
Observations of each mixture or chemical reaction, (each equation should include an
observation). Write the reactants for all 28 combinations and list the observations of each
substance by itself, and each of the 28 combinations. Include colors, textures and odors (which
sometimes appear and sometimes disappear).
Chemical Reactions – write the molecular, ionic and net ionic equation for only the actual
chemical reactions.

Discussion. Discuss any points that you think require discussion. For example, if you had
trouble figuring out the identity of one or two unknowns, what caused the uncertainty and how
did you resolve it?
Conclusion.
The conclusion is the list of letters and their corresponding compounds in solution.
As before, this report can be hand written (legibly) or typed.