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lab rationale - (aq was added and a white precipitate was...

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Ions Present: Pb 2+ and Hg 2 2 Ions Absent: Ag + Pb 2+ is present To an unknown solution, 6 M HCl (aq) was added and a white precipitate was formed. The white precipitate was PbCl 2(s) because it is insoluble according to the chloride solubility rule. Upon heating the mixture in the hot water bath, some white precipitate was dissolved. PbCl 2(s) is not soluble at room temperature, but when the mixture is heated it becomes Pb 2+ (aq) , therefore Pb 2+ (aq) maybe be present in the solution. When K 2 CrO 4(aq) was added, a yellow precipitate was formed. PbCrO 4(s) is the yellow precipitate. Pb 2+ (aq) is present in the original unknown sample. Hg 2 2+ is present To an unknown solution, 6 M HCl
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Unformatted text preview: (aq) was added and a white precipitate was formed. The white precipitate was Hg 2 Cl 2(s) because it is insoluble according to the chloride solubility rule. Upon the addition of 6M NH 3 the white precipitate turned to a gray- black precipitate. Hg 2 2+ (aq) was present in the original unknown sample. Ag + is absent In Step 1-E, 6 M HNO 3(aq) was added to the decantate from step 1- D1 and a colorless solution remained. Ag + (aq) was absent in the original unknown sample. A white precipitate, AgCl (s) , would have been observed if Ag + (aq) had been in the original unknown sample....
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