Yodilit
So, In my Chemistry Class we did a Lab, in which we were to do a battery of tests to determine what an unknown substance was.
However, the tests have left me baffled.
I've got almost no idea which substance it was.
So, I'm asking for your help to determine which substance it most likely is.
Possibilities
NaNO3
NaCH3COO
H2SO4
NaOH
CaCO3
KI
All (aq)
Test Results
1 - Red Litmus Paper - Turned Blue.
2 - Blue Litmus Paper - No Change.
3 - omited
4 - Add Magnesium Oxide - Clear, few bubbles formed on top of the solution.
5 - Add Distilled Water - No Change.
6 - Add Sodium Chloride - Clear, few bubbles formed.
7 - Add Calcium Hydroxide(LimeWater) - Small amounts of white flakey-looking precipitate form in the solution.
8 - Add Zinc Nitrate - No change.
9 - Add Silver Nitrate - Small amounts of white flakey-looking precipitate form in the solution.
The solution was clear to begin with.
If you an help me with this, at all, I will love you forever.
I can't really figure out which unknown it is, at all.
thankyou very much for your time!chances are, people who help determine the substance will be rewarded with gold or art. Whichever you prefer. First of all, the litmus test in the first two sets of results tell us that the substance is basic. Easy way to remember:
Blue is
basic.
This will then narrow down the choices as to which substance the unknown may be from the given list.
NaNO3 is neither an acid or a base in the Arrhenius in the acid-base concept, but I think it has a basic pH in solution. Sodium acetate, although used as a buffer, is an acid and so it would not change the litmus to blue. H2SO4 is a strong acid and would also not change the colour of the litmus paper to blue. I know that sodium hydroxide is a base and I'm pretty sure that calcium carbonate also is. I also think potassium iodide is a base.
So, we are left with:
NaNO3
NaOH
CaCO3
KI
Therefore, the rest can be solved by using your solubility rules. For simplicity, I'll show it to you by just putting the number that corresponds to that set of results.
4. Reaction with magnesium hydroxide.
2NaNO3 + Mg(OH)2 --> 2NaOH + Mg(NO3)2
ALL PRODUCTS ARE SOLUBLE2NaOH + Mg(OH)2 --> 2NaOH + Mg(OH)2
THE MAGNESIUM HYDROXIDE PRODUCED WILL BE A PRECIPITATECaCO3 + Mg(OH)2 --> Ca(OH)2 + MgCO3
CALCIUM HYDROXIDE AND MAGNESIUM CARBONATE ARE INSOLUBLE2KI + Mg(OH)2 --> 2KOH + MgI2
ALL PRODUCTS ARE SOLUBLETherefore, because no precipitate was formed from this reaction, the unknown must be either sodium nitrate or potassium iodide.
6. Reaction with NaCl
NaNO3 + NaCl --> NaCl + NaNO3
THERE WILL BE NO PRECIPITATEKI + NaCl --> NaI + KCl
THERE WILL BE NO PRECIPITATENaOH + NaCl --> NaCl + NaOH
THERE WILL BE NO PRECIPITATECaCO3 + 2NaCl --> CaCl2 + Na2CO3
THERE WILL BE NO PRECIPITATETherefore, because no preciptate was formed by the unknown, the unknown is either sodium nitrate, potassium iodide, sodium hydroxide or calcium carbonate.
7. Reaction with Ca(OH)2
2KI + Ca(OH)2 --> 2KOH + CaI2
THERE WILL BE NO PRECIPITATE2NaNO3 + Ca(OH)2 --> 2NaOH + Ca(NO3)2
THERE WILL BE NO PRECIPITATENaOH + Ca(OH)2 --> NaOH + Ca(OH)2
CALCIUM HYDROXIDE SHOULD FORM A WHITE PRECIPITATECaCO3 + Ca(OH)2 --> Ca(OH)2 + CaCO3
CALCIUM CARBONATE AND CALCIUM HYDROXIDE SHOULD FORM WHITE PRECIPITATESAccording to your results, there was a small amount of white solid formed. This therefore leads me to believe that the unknown is either calcium carbonate or sodium hydroxide. However, in the test with magnesium hydroxide, the theoretical solubilities disprove that the unknown is either of these. So, somewhere along the way, an error has surfaced in the results. This could have been from impurities etc etc.
8. Reaction with Zn(NO3)2
2KI + Zn(NO3)2 --> 2KNO3 + ZnI2
THERE SHOULD BE NO PRECIPITATENaNO3 + Zn(NO3)2 --> Zn(NO3)2 + NaNO3
THERE SHOULD BE NO PRECIPITATE2NaOH + Zn(NO3)2 --> 2NaNO3 + Zn(OH)2
ZINC HYDROXIDE IS INSOLUBLECaCO3 + Zn(NO3)2 --> Ca(NO3)2 + ZnCO3
ZINC CARBONATE IS INSOLUBLEThe results you have support that the unknown is either potassium iodide or sodium nitrate.
9. Reaction with AgNO3
KI + AgNO3 --> KNO3 + AgI
AgI WILL FORM A WHITE PRECIPITATENaNO3 + AgNO3 --> AgNO3 + NaNO3
THERE WILL BE NO PRECIPITATENaOH + AgNO3 --> AgOH + NaNO3
SILVER HYDROXIDE WILL FORM A PRECIPITATECaCO3 + AgNO3 --> Ag2CO3 + Ca(NO3)2
SILVER CARBONATE WILL FORM A TAN COLOURED PRECIPITATEThe results show that a white precipitate was formed, which is supported by the reaction of KI with silver nitrate. Calcium carbonate forms a tan coloured precipitate when reacting with silver nitrate, so it can't be that. I'm not sure about the colour of silver hydroxide though. I think it may be white but I'm not sure.
Due to the errors that have occured in the experiment though, it is hard to determine with any certainty, what the actual unknown is. I'm thinking it may be potassium iodide, but I'm not entirely sure because the results when reacting with calcium hydroxide showed that a precipitate shouldn't have formed when there is one. However, all the other reactions are justified by the solubility rules. This inconsistancy, as I said earlier, may have just been formed by impurities in your test tube or something... I'm not entirely sure.
I hope this helps, if not, just PM me
smile
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