1. The labels fell off two test tubes onto a student\'s laboratory bench. One te
ID: 789329 • Letter: 1
Question
1. The labels fell off two test tubes onto a student's laboratory bench. One test tube contained sodium oxalate and the other contained barium chloride. Beside these two tubes, the student had labeled solutions of calcium sulfate and aluminum carbonate.
(a) Could the student identify the unlabeled solutions on the basis of their color? Briefly explain.
(b) Could the student identify the solutions in the unlabled tubes by adding calcium sulfate solution to a sample of each? If so, write the complete ionic and net ionic equations describing his observations. If not, explain why not.
(c) Could the student use the aluminum carbonate solution to identify the solutions in the unlabeled tubes? If so, write the complete ionic and net ionic equations describing his observations when he added the solution to each of his unlabeled solutions. If not, explain why not.
Explanation / Answer
1. Could the student identify the unlabeled solutions on the basis of their color?
answer is no, though because none of them have a color, nor can make a color with what you have been given
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2. Could the student identify the solutions in the unlabeled tubes by adding calcium sulfate solution to a sample of each?
no... because both produce white precititates
molecular: BaCl2 (aq) & CaSO4 (aq) --> BaSO (s) & CaCl2 (aq)
ionic: Ba+2(aq) & 2Cl-1(aq) & Ca+2(aq) & (SO4)-2 (aq) --> BaSO (s) & Ca+2(aq) & 2Cl-1 (aq)
net ionic: Ba+2 (aq) & (SO4)-2 (aq) --> BaSO4 (s) a wite ppt
molecular: Na2C2O4 (aq) & CaSO4 (aq) -- CaC2O4 (s) & Na2SO4 (aq)
ionic: 2Na+(aq) & (C2O4)-2(aq) & Ca+2(aq) &(SO4)-2(aq) -- CaC2O4 (s) & 2Na+ & (SO4)-2(aq)
net ionic:Ca+2 (aq) & (C2O4)-2 (aq) --> CaC2O4 (s) a white ppt
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3. Could the student use the aluminum carbonate solution to identify the solutions in the unlabeled tubes?
yes ,
BaCO3 will precipitate its Ksp = 5.1 X 10^-9
aluminum will not precipitate oxalate, it is soluble in water
molecular: 3BaCl2 (aq) & Al2(CO3)3 (aq) --> 3BaCO3 (s) & 2AlCl3 (aq)
ionic: 3Ba+2(aq) & 6Cl- (aq) & 2Al+3(aq) & 3(CO3)-2(aq) --> 3BaCO3(s) & 2Al+3(aq) &6Cl-1(aq)
net ionic: Ba+2(aq) & (CO3)-2 (aq) --> BaCO3(s)
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