An Introduction to Qualitative Analysis Essay

Procedure office staff I Qualitative summary of Group 2 ElementsMix 0.02M K2CrO4 with each Mg(NO3)2, Ca(NO3)2, Sr(NO3)2 and Ba(NO3)2 in concert. Secondly, scuffle 0.1M (NH4)2C2O4 alternatively of 0.02M K2CrO4 together with the same reactants used forwards. Thirdly, ripple 0.1M Na2SO4 with those reactants. Then, mix 0.1M NaOH with the same reactants used before again. Some precipitates should mannequins ,record the observations in Table 1. At last, identify those 2 cabalistic solution.Part II- Qualitative Analysis of Selected AnionsFirst 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 mix 1M HNO3 for the reactants that formed precipitates. ingeminate the first step by placing 0.1M AgNO3 instead of 1M HNO3. Then add 6M NH3 to those mixtures that contains precipitates in, and 1M HNO3 the mixtures contains precipitates . utilise these observation, identify an un completen union. Recording all the observations in Table 2.3. state of matter the identity of your unknown (along with its sample number). Give the reasoning you used to perplex at this conclusion.The Unknown Z should be SO42- because it has a same proportion 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 disappe ard for the unknown anion, when HNO3 is added to unknown, the ppt disappe ared too, consequently we canconclude that the unknown is SO42-.Follow-Up Questions1. Devise a successiveness of reactions to follow (using filtering or centrifuging where necessary to remove precipitates) to identify an unknown containing cardinal or more(prenominal) 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, first add CrO4 into the solution. Then we could identify them with their colour. Next we add C2O4, if the ppt is formed, and then we know Ca2+ is involved in the solution. Next we add OH into the solution, if a ppt formed, that way of life Mg2+ is involved in the solution.2.Devise a sequence of reaction to follow (using filtering or centrifuging where necessary to remove precipitates)to identify an unknown consisting of two or more of two or more of the anions tested in Part 2.The anions are CO32-,SO42-,Cl-,I-.To identify an unknown consisting of two or more anions in Part 2.First add HNO3 into solution. Then we could identify them with the observations. Then we add Ba(NO3)2,if ppt formed, then we know that CO32-is involved in the solution. Next we add HNO3 into the solution, if ppt formed, that means SO42- is involved I the solution.3. Why are the reagents used to test for cations commonly alkali coat salts or ammonium salts rather than salts of different 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 wont 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 either cation.5.For fast and accurate identification of substances, major research or examination laboratories now use very sophisticated (and expensive )equipment. Find come in the name of one of the instruments now used for analysis, and briefly describe its regularity of operation.Use glass pane. When we do the lab, we make a table on the paper, then put the glass pane on the paper sheet. aft(prenominal) that, we only drop one or two drops of each chemic. On the glass pane . Its easier to observe the color of ppt because the glass is transparent and its also a economic way.ConclusionIn this lab, we carry step forward 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 situation 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 ins tead 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 by the piece 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-.And 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.