Exp #10 - perform some precipitation because HgOH has a...

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Experiment #10 Qualitative Analysis Group 1 Cations Conclusion: The cations in the unknown #2 solution are the Pb and Ag ions. DATA ANALYSIS: Solution s Known Unknown #2 Parts precipitate Ions exist precipitate Ions exist Part 2 Yes Pb ions Yes Pb ions Part 3 Yes Hg ions No None Part 4 Yes Ag ions Yes Ag ions In the part 3, there is no precipitate in the unknown solution, so it only has the Pb and Ag ions in it. Discussion:
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In this experiment, in the part one all of three cations which are Pb, Hg, and Ag ions will form precipitates in the presence of chloride anion in an acidic aqueous solution. This is a method called selective precipitation. The supernatant in the part has little cations in it. If add some anions that has a lower K value, then the solution will from the precipitation. In the Part two, the solution forms some precipitation because PbCrO4 has a smaller Ksp value. In the part three, the known solutions
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Unformatted text preview: perform some precipitation because HgOH has a small Ksp value. In the other hand, the unknown solution did not form out the precipitation, so there are no Hg2 ions in the unknown solution. In the part four, the solution forms out the precipitation because the AgNo3 has a smaller Ksp value. The result reach my expectation. Also, I found there are Pb and Ag ions in the unknown solutions. Conclusion: The cations in the unknown #2 solution are the Pb and Ag ions. Questions: 1. A. NH4OH B. HCl C. K2CrO4 D. has different color. 2. (Cl-) + (Ag+) ----- AgCl (s) AgCl + (NH4+)---- Ag(NH3)2 + 3. Part 1: Pb+Cl---PbCl Ag+ Cl--- AgCl Hg + ClHgCl Part2: Pb + CrO4 --- PbCrO4 Part 3: Ag + NH4--- Ag(NH3)2+ Hg+ Cl + NH$--- HgNH2Cl5 Part 4: Ag(NH3) ---- AgCl...
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Exp #10 - perform some precipitation because HgOH has a...

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