Aim:
To find the unknown chemicals that are in
the labelled bottles
Equipment:
·
Test Tubes
·
Safety Glasses
·
Distilled Water
·
Eye Dropper
·
Calcium Hydroxide
·
Copper Sulfide
·
Silver Sulfate
·
Potassium Iodide
·
Silver Chloride
·
Copper Sulfate
·
Silver Nitrate
Results:
Set 1: (answers simulated as no actual
results could be found)
|
bottle
|
A
|
b
|
c
|
d
|
|
Result
|
BaCl2
|
CuSO4
|
H2SO4
|
NaCl
|
Set 3: (answers are from actual results)
|
bottle
|
M
|
n
|
o
|
p
|
q
|
r
|
|
Result
|
H2SO4
|
Na2CO3
|
BaCl2
|
HCl
|
Pb(NO3)2
|
KOH
|
Processing Data:
set 1:
Looking at all
the solutions in a test tube, we can immediately recognise one of the solutions
as being blue. Now from the data sheet, we can see that copper ions have a blue
colour in an aqueous solution, and copper sulfate is soluble in water.
Therefore this means that once copper sulfate is added to water, it will
completely dissociate to make copper ions and sulfate ions. As said before,
copper ions in an aqueous solution makes a blue colour. Therefore, bottle B has
to be copper sulfate. Now, by eliminating the copper sulfate, we are left with
3 bottles of unknown solution. If we add calcium hydroxide (limewater) to the
solutions, we will see that one of the solutions will form a white precipitate.
Looking at the data sheet, we can see that the only solution which would slightly
form white precipitates would be calcium sulfate.
H2SO4 (aq) + Ca(OH)2 (aq) --> CaSO4 (s) + H2O (l)
Now, this would be the second solution that is eliminated. Now by adding copper sulfide to the two remaining solutions, we will again see that one of the test tubes forms a white precipitate. From the data sheet, we can see that the only solution that is insoluble in water is BaS.
BaCl2 (aq) + CuS (aq) --> BaS (s) + CuCl2 (aq).
Therefore, we can say that bottle A is BaCl2. Now from the process of elimination we can see that the only remaining solution is NaCl. However, just to prove that the remaining solution is NaCl, we just need to add silver sulfate to the solution to form a white precipitate.
2NaCl (aq) + Ag2SO4 (aq) + Na2SO4 (aq) + 2AgCl (s)
H2SO4 (aq) + Ca(OH)2 (aq) --> CaSO4 (s) + H2O (l)
Now, this would be the second solution that is eliminated. Now by adding copper sulfide to the two remaining solutions, we will again see that one of the test tubes forms a white precipitate. From the data sheet, we can see that the only solution that is insoluble in water is BaS.
BaCl2 (aq) + CuS (aq) --> BaS (s) + CuCl2 (aq).
Therefore, we can say that bottle A is BaCl2. Now from the process of elimination we can see that the only remaining solution is NaCl. However, just to prove that the remaining solution is NaCl, we just need to add silver sulfate to the solution to form a white precipitate.
2NaCl (aq) + Ag2SO4 (aq) + Na2SO4 (aq) + 2AgCl (s)
Set 3:
By adding potassium iodide to the solutions
we can see that one of the precipitates has a yellow colour. This shows us that
the precipitate formed is lead iodide.
Pb(NO3)2 (aq) + 2KI (aq) --> PbI2 (s) + 2KNO3 (aq)
Now by adding silver chloride to the remaining solutions, again we can see that one of the precipitates formed has a yellow colour. From the data sheet, we can confirm that the precipitate formed is silver carbonate.
Na2CO3 (aq) + 2AgCl (aq) --> Ag2CO3 (s) + 2NaCl (aq)
By adding copper sulfate to the remaining solutions, would leave two of the solution with a green colour. However, one of those solution that is green will also have precipitates formed as well. This shows us that the solution that formed a precipitate is BaCl2 and the others are HCl and H2SO4.
BaCl2 (aq) + CuSO4 (aq) --> BaSO4 (s) + CuCl2 (aq)
2HCl (aq) + CuSO4 (aq) --> H2SO4 (aq) + CuCl2 (aq)
H2SO4 (aq) + CuSO4 (aq) --> H2SO4 (aq) + CuSO4 (aq)
Now to find out which of the two solutions is HCl and H2SO4, we just need to add silver chloride to the two solutions. One of the solutions will form a white precipitate. Therefore, according to our data sheet we can confirm that the solution that has the precipitate is HCl (2 mol) and the other solution has to be H2SO4 based on elimination.
CuCl2 (aq) + 2AgCl (aq) --> CuCl2 (aq) + 2AgCl (s)
Now the only solution that is left is KOH. To prove that the solution is KOH, we can just add silver nitrate to the solution which would form silver hydroxide precipitates
KOH (aq) + AgNO3 (aq) --> KNO3 (aq) + AgOH (s)
Pb(NO3)2 (aq) + 2KI (aq) --> PbI2 (s) + 2KNO3 (aq)
Now by adding silver chloride to the remaining solutions, again we can see that one of the precipitates formed has a yellow colour. From the data sheet, we can confirm that the precipitate formed is silver carbonate.
Na2CO3 (aq) + 2AgCl (aq) --> Ag2CO3 (s) + 2NaCl (aq)
By adding copper sulfate to the remaining solutions, would leave two of the solution with a green colour. However, one of those solution that is green will also have precipitates formed as well. This shows us that the solution that formed a precipitate is BaCl2 and the others are HCl and H2SO4.
BaCl2 (aq) + CuSO4 (aq) --> BaSO4 (s) + CuCl2 (aq)
2HCl (aq) + CuSO4 (aq) --> H2SO4 (aq) + CuCl2 (aq)
H2SO4 (aq) + CuSO4 (aq) --> H2SO4 (aq) + CuSO4 (aq)
Now to find out which of the two solutions is HCl and H2SO4, we just need to add silver chloride to the two solutions. One of the solutions will form a white precipitate. Therefore, according to our data sheet we can confirm that the solution that has the precipitate is HCl (2 mol) and the other solution has to be H2SO4 based on elimination.
CuCl2 (aq) + 2AgCl (aq) --> CuCl2 (aq) + 2AgCl (s)
Now the only solution that is left is KOH. To prove that the solution is KOH, we can just add silver nitrate to the solution which would form silver hydroxide precipitates
KOH (aq) + AgNO3 (aq) --> KNO3 (aq) + AgOH (s)
Evaluation:
From the results above, I can say that the
way we uncovered each of the unknown solutions; was done in the least number of
steps given the solutions that we were provided. If we were given objects such
as matches, would have simplified the procedures. For example, instead of
testing for precipitates, we could just test for gases. This would have been a
great use mainly in set 3. However, based on our results and how we got each
one by proving exactly what they were, I am confident that our results are
accurate. Finding what each precipitate was just purely looking at the data
sheet to see what the compound was.
Conclusion:
To conclude, the results here have been
modified because we had not written down how each solution was proven to be
what. Therefore, we also didn’t write down what we mixed each solution with.
This is why I had to work out how to get each of the solution based purely on the
data sheet.
