Qualitative analysis is a set of procedures used to identify a particular ion or a sample. Quantitative analysis is used to identify the amount of ion presence in a solution.
- What type of reaction will use to identify the cation in part I?
Double replacement and precipitate-forming reaction will be used to identify the cation in part 1.
- The group II(alkali earth) contain Ra that is not used in experiment, why don’t we use radium?
We do not use radium in our experiment because it is rare, expensive, radioactive, and dangerous.
- Look at appendix 2 to explain why Na2No3 or Na3Po4 would not be useful reagent to help to identify cations?
Na2No3 would not be useful to identify cation because anion of NO3- is soluble with any cations. Na3PO4 would not be useful to identify cation because the anion of PO43- is not soluble and form precipitate with any cations.
Data and Observations:
Part I –Qualitative Analysis of Group 2 Elements
Table 1
Part II– Qualitative Analysis of Selected Anions
Table 2
Questions from
Part 1:
- Write net ionic equations for each combination in which a precipitate occurred.
Sr2+(aq)+ CrO42-(aq) → SrCrO4(s)
Ba2+(aq)+CrO42-(aq)→ BaCrO4(s)
Ca2+(aq)+ C2O42-(aq)→ CaC2O4(s)
Sr2+(aq)+ C2O42-(aq)→ SrC2O4(s)
Ba2+(aq)+C2O42-(aq)→ BaC2O4(s)
Sr2+(aq)+SO42-(aq)→ SrSO4(s)
Ba2+(aq)+ SO42-(aq)→ BaSO4(s)
Mg2+(aq)+ 2OH-(aq)→ Mg(OH)2(s)
Ca2+(aq)+ 2OH-(aq)→ Ca(OH)2(s)
Ba2+(aq)+SO42-(aq)→ BaSO4(s)
- State the identity of your unknown (along with its sample number). Give the reasoning you used to arrive at this conclusion.
From our observation, Unknown B should be Ca2+ because it has a same property as Ca2+ does. By the examination in table 1, we know that Ca2+ forms ppt with C2O42- and OH-. The unknown anions forms ppt with C2O42- and OH- as well, therefore we conclude that the unknown cation should be Ca2+.
Part 2:
- Write net ionic equations for each combination in which a precipitate formed or another reaction occurred.
Ba2+(aq)+ CO32-(aq)→ BaCO3(s)
Ba2+(aq)+ SO42-(aq)→ BaSO4(s)
2Ag+(aq)+ CO32-(aq)→ Ag2CO3(s)
Ag+(aq)+ Cl-(aq)→ AgCl(s)
Ag+(aq)+ I-(aq)→ AgI(s)
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Write net ionic equations for each situation in which the precipitate re-dissolved on the addition of HNO3 or NH3.
BaCO3(s)+2H+(aq)→ Ba2+(aq)+CO2(g)+H2O(l)
BaSO4(s)+2H+→Ba2+(aq)+ SO4(aq)+2H+
Ag2CO3(s)+2H+→ 2Ag+(aq)+ H2O(l)+ CO2(g)
Ag2CO3(s)+4NH3(l)→2Ag(NH3)2(aq)+CO32-(aq)
AgCl(s)+ 2NH3(aq)→ Ag(NH3)2+(aq)+ Cl-(aq
AgI(s)+2NH3→Ag(NH3)2+(aq)+I-(aq)
- State the identity of your unknown (along with its sample number). Give the reasoning you used to arrive at this conclusion.
The Unknown X should be SO42- because it has a same property as SO42- does. When SO42- is added to Ba(No3)2, and AgNo3, it forms a ppt; for the unknown anion, when it is added to Ba(No3)2, and AgNo3, it forms a ppt as well. When HNO3 is added to BaSO4, the ppt disappeared; for the unknown anion, when HNO3 is added to unknown, the ppt disappeared too, therefore we can conclude that the unknown is SO42-.
Follow-Up Questions:
- Devise a sequence of reactions to follow (using filtering or centrifuging where necessary to remove precipitates) to identify an unknown containing two or more cations of Group 2 elements.
The Group 2 elements are Mg, Ca, Sr, and Ba. To identify an unknown containing two or more cations of Group 2 elements, we first add CrO4 into the solution. Then we could identify them with their color when we separate them by filter. Next we add C2O4, if the ppt is formed, then we know Ca2+ is involved in the solution. Next we add OH into the solution, if a ppt formed, that means Mg2+ is involved in the solution.
3. Why are the reagents used to test for cations usually alkali metal salts or ammonium salts rather than salts of other metals?
The reagents used to test for cations usually alkali metal salt or ammonium salts rather than salts of other metals because the alkali metal is soluble with most anions. It won’t form a ppt with other anions. These reagents will prevent any side reaction from occurring in the solution.
4. Why are the reagents used to test for anions usually a nitrate of the cation that is reacting rather than other salts of that cation?
The reagents used to test for anions usually a nitrate of the cation that is reacting rather than other salts of that cation because the nitrate is soluble with almost every cation.
Conclusion:
In this lab, we carry out precipitation test of four cations and four anions, and use the observations to identify two unknowns.
First we mix Mg2+, Ca2+, Sr2+, Ba2+ with K2CrO4, and observed that Sr2+ and Ba2+ forms a ppt. Then when we mix (NH4)2C2O4 instead of K2CrO4, we observed that all of the cations forms a ppt except for Mg2+. Next, we did the same thing by using Na2SO4 and NaOH instead of (NH4)2C2O4 . Lastly, we examined unknown B and found that it has the same chemical properties with Ca2+. So we conclude that the unknown substance should be Ca2+. In part II, we mix CO32-, SO42-, Cl-, and I- with HNO3 to each of the test tube and notice no ppt formed. Secondly, we mix Ba(NO3)2 instead of HNO3 with the anions, and we observed that CO32- and SO42- forms a ppt. Then we add HNO3 to the ones that formed ppt, and the ppt disappeared. Next, we mix AgNO3 instead of Ba(NO3)2 with the anions, and observed ppt formed with all of the anions except for SO42-. After that, we added HNO3 and NH3 separately to the anions and we observed no change in SO42-, but the precipitates that formed in CO32-, Cl-, and I- disappeared. And there is a ppt formed when NH3 is added to SO42-, and the other ones’ precipitates turns to a lighter ppt. By using these observations, we found out the unknown ion contains the same chemical properties as SO42-.
At last, by using all observations from this lab, we concluded that if two ions are soluble to each other, there will be no ppt formed. If two ions are not soluble to each other, there will be a ppt formed.