Preparing dissolutions and
developing specific
reactions
Students: Ana
Experiment´s date:
Zubiarrain and Maite
08/11/11
Cortés
Report´s date: from
Teacher: Nuria Aguado
13/11/11 to
18/11/11
Index
Introduction page ..........................................................................
Materials ...........................................................................................
Calculations .....................................................................................
Methods .............................................................................................
Results...............................................................................................
Conclusion ........................................................................................
Bibliography .....................................................................................
2
Introduction page
In this experiment we have worked with dissoluions by preparing specific
reactions. For the preparation of this dissolutions we had to know how to calculate and
use concentrations expressed in different ways.These are the ways that we learnt:
 Molarity:
is the most common unit of solution concentration. It is defined
as the number of moles of solute per one liter of solution. It´s important to
remember that you have to use liter and not milliliter or kiloliter. One liter of
solution contains solute and solvent. This is the formula: M=n/L
 Mass percent: it is defined based on the grams of solute per 100 grams of
solution. This is the formula: % (m/m)=mass solute/mass solution x 100
 Volume percent: it is defined as milliliters of solute per 100mL of
solution. This is the formula: %(v/v)= volume solute/volume solution x 100
 Mass-volume percent: it is also very common. this solution is indicated
by w/v % and is defined as the grams of solute per 100 milliliters of solution.
this is the formula: %(m/v)=mass solute/volume solution x 100
 Fraction: it indicates how many parts of the solutions are the solute, for
example if you have 1:4 it means that from all the solution only 1 part from
every 4 parts are the solute. If our solution is of 8g the solute would be 2g.
Using this ways of expressing concentration we prepared some reactions. A
chemical reaction is a process that leads to the transformation of one set of chemical
substances to another. There are two types of reactions: spontaneous, which happened
without any input of energy, and non-spontaneous, which need an input of energy,
usually heat, light or electricity. Chemical reactions encompass changes that involve the
motion of electrons in the forming and breaking of chemical bonds. The substance or
substances that you have before happening the chemical reaction are called reagents and
the element or elements that you obtain after the chemical reaction has happened are
called products.
In this experiment most of the reactions were double replacement, except this one:
Na2CO3 + 2HCL à2NaCl + H2o + CO2 ↑ Double replacement reactions are a
molecular process which involve the exchange of bonds between the two reacting
chemical species, which results in the creation of products with similar or identical
bonding affiliations. We can represent this by this general reaction scheme: AX + BY
→ BX + AY
Materials
Test tube x7
Test tube rack
Beaker X2
3
Droplet
Watch glass
Washing flask
Spatula
Graduated cylinder
Soap
Brush
Stirring rod
Analytes
NaCl
Na2SO4
Na2CO3
CuSO4
AgNO3
Pb(NO3)2
FeCl3
Reagents
AgNO3
BaCl2
HCl
NH4OH
KI
KSCN
Calculations
 A1 100ml NaCl – 0.2M
o 𝑛𝑁𝑎𝐶𝑙 = 0.2𝑀 · 0.1𝑙 = 0.02𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
o 0.02𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 ·
57.5𝑔 𝑁𝑎𝐶𝑙
1𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
= 1.15𝑔 𝑁𝑎𝐶𝑙
 A2 100ml Na2SO4 – 0.1M
o 𝑛𝑁𝑎2𝑆𝑂4 = 0.1𝑀 · 0.1𝑙 = 0.01𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
o 0.01𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4 ·
142𝑔 𝑁𝑎2 𝑆𝑂4
1𝑚𝑜𝑙 𝑁𝑎2 𝑆𝑂4
= 1.42𝑔 𝑁𝑎2 𝑆𝑂4
 A3 100ml Na2CO3 – 0.2M
4
o 𝑛𝑁𝑎2𝐶𝑂3 = 0.2𝑀 · 0.1𝑙 = 0.02𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
106𝑔 𝑁𝑎2 𝐶𝑂3
o 0.02𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3 ·
1𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
= 2.12𝑔 𝑁𝑎2 𝐶𝑂3
 A4100ml CuSO4 – 0.1M
o 𝑛𝐶𝑢𝑆𝑂4 = 0.1𝑀 · 0.1𝑙 = 0.01𝑚𝑜𝑙 𝐶𝑢𝑆𝑂4
o 0.01𝑚𝑜𝑙 𝐶𝑢𝑆𝑂4 ·
160𝑔 𝐶𝑢𝑆𝑂4
1𝑚𝑜𝑙 𝐶𝑢𝑆𝑂4
= 1.6𝑔 𝐶𝑢𝑆𝑂4
 A550ml AgNO3 – 0.1M
o 𝑛𝐴𝑔𝑁𝑂3 = 0.1𝑀 · 0.05𝑙 = 0.005𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
o 0.005𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3 ·
169.9𝑔 𝐴𝑔𝑁𝑂3
1𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
= 0.85𝑔 𝐴𝑔𝑁𝑂3
 A6 50ml Pb(NO3)2 – 0.1M
o 𝑛𝑠 = 0.1𝑀 · 0.05𝑙 = 0.005𝑚𝑜𝑙 𝑃𝑏(𝑁𝑂3 )2
o 0.005𝑚𝑜𝑙 𝑃𝑏(𝑁𝑂3 )2 ·
331.2𝑔 𝑃𝑏(𝑁𝑂3 )2
1𝑚𝑜𝑙 𝑃𝑏(𝑁𝑂3 )2
= 1.66𝑔 𝑃𝑏(𝑁𝑂3 )2
 A7 100ml FeCl3 – 0.1M
o 𝑛𝑠 = 0.1𝑀 · 0.1𝑙 = 0.01𝑚𝑜𝑙 𝐹𝑒𝐶𝑙3
o 0.01𝑚𝑜𝑙 𝐹𝑒𝐶𝑙3 ·
162.3𝑔 𝐹𝑒𝐶𝑙3
1𝑚𝑜𝑙 𝐹𝑒𝐶𝑙3
= 1.623𝑔 𝐹𝑒𝐶𝑙3
 R150ml AgNO3 – 1% (m/m)
o 50𝑔 𝑑𝑖𝑠𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ·
1𝑔 𝐴𝑔𝑁𝑂3
100𝑔 𝑑𝑖𝑠𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 0.5𝑔 𝐴𝑔𝑁𝑂3
 R2 50ml BaCl2 – 5% (m/m)
o 50𝑔 𝑑𝑖𝑠𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ·
5𝑔 𝐵𝑎𝐶𝑙2
100𝑔 𝑑𝑖𝑠𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 2.5𝑔 𝐴𝑔𝑁𝑂3
 R3 60ml HCl – 1:5
1
o 60𝑚𝑙 𝐻𝐶𝑙 · = 12𝑚𝑙 𝐻𝐶𝑙
5
 R4 100ml NH4OH – 1:4
1
o 100𝑚𝑙 𝑁𝐻4 𝑂𝐻 · = 25𝑚𝑙 𝑁𝐻4 𝑂𝐻
4
 R5 50ml KI – 0.5M
o 𝑛𝐾𝐼 = 0.5𝑀 · 0.05𝑙 = 0.025𝑚𝑜𝑙 𝐾𝐼
o 0.025𝑚𝑜𝑙 𝐾𝐼 ·
166𝑔 𝐾𝐼
1𝑚𝑜𝑙 𝐾𝐼
= 4.15𝑔 𝐾𝐼
 R6 50ml KSCN – 0.1M
o 𝑛𝐾𝑆𝐶𝑁 = 0.1𝑀 · 0.05𝑙 = 0.005𝑚𝑜𝑙 𝐾𝑆𝐶𝑁
o 0.005𝑚𝑜𝑙 𝐾𝑆𝐶𝑁 ·
97𝑔 𝐾𝑆𝐶𝑁
1𝑚𝑜𝑙 𝐾𝑆𝐶𝑁
= 0.485𝑔 𝐾𝑆𝐶𝑁
5
Methods
First of all we took all the material that we needed and washed it in order not to
pollute any of the dissolutions. After washing everything we marked the seven test
tubes with the special marker from A1 to A7, to know what dissolutions we were going
to have in each of them. Each group had to prepare two dissolutions, an analyte and a
reagent, except one pair that had to prepare a single dissolution. Those dissolutions were
then used by all of us. The next thing we did was to prepare our dissolutions, the A2 and
the R2. First, we calculated how many grams of Na2SO4 were needed to prepare 0.1M
dissolution of 100mL of water, which were 1.42g. We had a little problem with the
balance though, and after putting a bit of the compound on the watch glass we saw that
the balance was not working, and since we could not start the process again we worked
with the amount that Nuria told us (which was calculated at guess). We placed the
Na2SO4 in a beaker and then poured 100mL of water in it. We had to do the same thing
with the reagent, 50mL dissolution of BaCl2 (5%). After calculating that we needed
2.5g of solute, we weighed them and put them in another beaker where we poured
50mL of water. We mixed each dissolution with a stirring rod (which we cleaned
between each use). When we finished that, we took some drops of the A2 and put them
in its corresponding test tube; we cleaned the droplet and took some other drops of the
matching reagent and placed them in the same test tube. We did the same with all the
analytes and reagents and we annotated the observations. We had to be very careful and
wash the droplet after each use, because if not, the dissolutions would get polluted.
These are the analytes and their matching reagents:
A1R1
A2R2
A3R3
A4RA
A5R5
A6R5
A7R6
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Results
Analyte
Reagent
observations
A1
AgNO3
white, similar to the milk
A2
BaCl2
white with precipitations
A3
HCl
bubbles
A4
NH4OH
navy blue
A5
KI
whitish yellow
A6
KI
bright yellow
A7
KSCN
blood red
Reaction
NaCl + AgNO3  NaNO3 + AgCl
Na2SO4 + BaCl2  2NaCl + BaSO4
Na2CO3 + 2HCL 2NaCl + H2o + CO2 ↑
CuSO4+ 2 NH4OH  Cu(OH)2 + (NH4)2SO4
AgNO3 + KI  AgI + KNO3
Pb(NO3)2 + KI PbI2 + 2KNO3
FeCl3 + 3KSCN  Fe(SCN)3 + 3KCl
1-. In this reaction AgCl and NaNO3 are formed, and the colour of the product is
white. The reason why it has that colour is that both compounds are a white solid, but
whereas NaNO3 is very soluble in water, AgCl is not, which produces precipitates.
2-. The products of this double replacement reaction are NaCl and BaSO4, and the
colour of the final dissolutions is white with some precipitates. This is caused because
the BaSO4 is a white crystalline solid that is insoluble in water.
3-. The only remarkable thing about this reaction is that some bubbles appeared
when the two dissolutions were mixed together. It is due to the CO2 that appears in the
product, because it is a gas and when it "scapes" from the liquid bubbles appear.
7
4-. The product of this reaction is navy blue. Here you have the reason for it to have that
colour: when an aqueous CuSO4 dissolution (which is blue) is mixed with a small
quantity of (NH4)2SO4, there is a reaction between the available OH- ions and the
hydrated copper ions and insoluble Cu(OH)2 is produced. That compound is a pale
blue, gelatinous solid. When more (NH4)2SO4 is added to the dissolution, the insoluble
Cu(OH)2 dissolves forming a soluble complex ion, Cu(NH3)4]2+. This ion has a deep
blue colour, the colour we saw in our reaction.
5-.This reaction produces a whitish yellow product. The compounds that form the
product are AgI and KNO3. The first one is a yellow, inorganic, photosensitive
compound; and the second one is a white salt.
6-. This dissolution is bright yellow, and that is due to the fact that PbI2, one of
the products, is a bright yellow solid at room temperature (it becomes brick red by
heating). The other product is KNO3, and as mentioned above, it is a white salt.
7-. In this reaction Fe(SCN)3 and KCl are formed, being the colour of the product
blood red. That colour appears due to the presence of ferric thiocyanate, Fe(SCN)3. This
compound used to be known as rhodanide (from a Greek word for rose) because of the
red colour of its complexes with iron.
Conclusion
By carrying this experiment out, we have learnt how we could detect some
compounds in dissolutions we do not know. For example, let's say that we have a KSCN
and NaCl dissolution but we do not know that. If we wanted to see if it has KSCN we
could pour some FeCl3, and since it would turn red as seen in reaction 7, we would
know that it does have KSCN. Another example could be that our teacher gives as a
dissolution and tells us that we have to prove if it has Na2SO4. To do that we could do
the same as in reaction 2, prepare a BaCl2 dissolution and put some drops of it in the
dissolution we have been handed. If it did not turn white we would know that it does not
have NaSO4, but if it did turn white, we would know for sure that it has Na2SO4.
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Bibliography
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http://www2.caes.hku.hk/schoolscience/files/2010/09/SSHKS_WrongMatchinDiet_Ch
ungYikSham.pdf
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http://answers.yahoo.com/question/index?qid=20070913201304AASVxKy
http://www.dmgrad.com/?p=2849
http://uk.answers.yahoo.com/question/index?qid=20110412070253AAqcFlu
http://answers.yahoo.com/question/index?qid=20070130094432AASSN7m
http://mx.answers.yahoo.com/question/index?qid=20090902090302AAs50lO
http://www.ebooksread.com/authors-eng/augustus-price-west/experimental-organicchemistry-hci/page-12-experimental-organic-chemistry-hci.shtml
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http://en.wikipedia.org/wiki/Thiocyanate
http://en.wikipedia.org/wiki/NaNO3
http://en.wikipedia.org/wiki/AgCl
http://en.wikipedia.org/wiki/Barium_sulfate
http://en.wikipedia.org/wiki/Cu(OH)2
http://en.wikipedia.org/wiki/AgI
http://en.wikipedia.org/wiki/PbI2
http://en.wikipedia.org/wiki/Thiocyanate
http://en.wikipedia.org/wiki/Chemical_reaction
http://en.wikipedia.org/wiki/Double_replacement_reaction
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