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An Introduction to Qualitative Analysis

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Lab 16E -An Introduction to Qualitative Analysis Purpose: In this lab, we would tests the ions of Mg2+, Ca2+, Sr2+, and Ba2+, and enable each to be identified separately. And by using these observations, we could also identify an unknown. We also would tests the ions of SO42-, CO32-, Cl-, and I-, that also enable each to be identified separately, and to be use to identify the unknown. Safety: Chemicals include toxic. Remember to wash hands after the experiment. Procedure Summary; Part I - Qualitative Analysis of Group 2 Elements We first mix 0.02M K2CrO4 with each Mg(NO3)2, Ca(NO3)2, Sr(NO3)2 and Ba(NO3)2 together. Secondly, we mix 0.1M (NH4)2C2O4 instead of 0.02M K2CrO4 together with the same reactants used before. Thirdly, we mix 0.1M Na2SO4 as the added reagent. Then, we mix 0.1M NaOH as the added reagent before. Precipitates should forms for some, and be recorded in Table 1 with observations. At last, we obtain an unknown solution to identify the cation by mixing it with the 4 reagents we used before. Part II- Qualitative Analysis of Selected Anions We first mix 1M HNO3 with each Na2CO3, Na2SO4, NaCl and NaI together. Repeat these steps by placing 0.1M Ba(NO3)2 instead of 1M HNO3. Then mix1M HNO3 for the onces that formed precipitates. Repeat the first step by placing 0.1M AgNO3 instead of 1M HNO3. ...read more.


immediately - 6M NH3 added to precipitates from AgNO3 light ppt (yellow) immediately light ppt (yellow) immediately light ppt (yellow) immediately light ppt (yellow) immediately light ppt (yellow) immediately Questions from Part 1: 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) 2. 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: 1. 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) 2. 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) 3. State the identity of your unknown (along with its sample number). ...read more.


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. 1 2003/3/10 ...read more.

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