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Select the step(s) that will compose a rationale for the cation Pb+2 being absent in an unknown (but Ag+ is present). Question 3Select one or more: No white precipitate formed when 6 M HCl was added to the unknown solution in step 1-A. All of the white precipitate from step 1-A dissolved in hot water. A yellow precipitate did not form when K2CrO4 was added in step 1-C. The presence of a light blue decantate in step 1-A. A white precipitate did not form in step 2-B. The white precipitate from step 1-B dissolved in 6 M NH3 and then reformed when 6 M HNO3 was added. The white solid did not turn black upon addition of NaOH and SnCl2. A reddish brown precipitate did not form after adding K4Fe(CN)6. A lack of dark blue colored solution after addition of 15 M NH3.

          Select the step(s) that will compose a rationale for the cation Pb+2 being absent in an unknown (but Ag+ is present).
Question 3Select one or more:
No white precipitate formed when 6 M HCl was added to the unknown solution in step 1-A. 
All of the white precipitate from step 1-A dissolved in hot water.
A yellow precipitate did not form when K2CrO4 was added in step 1-C.  
The presence of a light blue decantate in step 1-A.
A white precipitate did not form in step 2-B.
The white precipitate from step 1-B dissolved in 6 M NH3 and then reformed when 6 M HNO3 was added.
The white solid did not turn black upon addition of NaOH and SnCl2.
A reddish brown precipitate did not form after adding K4Fe(CN)6.
A lack of dark blue colored solution after addition of 15 M NH3.

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Added by Michelle G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Sri K

11/07/2022

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Select the step(s) that will compose a rationale for the cation Pb+2 being absent in an unknown (but Ag+ is present). Question 3Select one or more: No white precipitate formed when 6 M HCl was added to the unknown solution in step 1-A. All of the white precipitate from step 1-A dissolved in hot water. A yellow precipitate did not form when K2CrO4 was added in step 1-C. The presence of a light blue decantate in step 1-A. A white precipitate did not form in step 2-B. The white precipitate from step 1-B dissolved in 6 M NH3 and then reformed when 6 M HNO3 was added. The white solid did not turn black upon addition of NaOH and SnCl2. A reddish brown precipitate did not form after adding K4Fe(CN)6. A lack of dark blue colored solution after addition of 15 M NH3.

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Transcript

00:01 Dear students, this is a question from salt analysis.

00:03 Silver, lead and mercuracine, these are group 1 radicals.

00:08 These are group 1 cations.

00:12 And this is a test for question on the test for group 1 cations.

00:17 The first question is a white basalt formed when concentrated h .l is added to the solution.

00:23 This is due to the formation of the chlorides.

00:26 Ag plus reacts with hcl to form agcl.

00:33 So let us write here, cl minus.

00:36 So it forms a gcl which is white.

00:40 Similarly, pb2 plus reacts with chloride ions from hcl to form pbcl to similarly mercury iron reacts with chloride ions to form mercurice chloride.

00:59 All these are white.

01:04 Now, the second step is water is added to the white precipitate and the mixture is heated.

01:11 Sodium dichromate is added to the supernatal liquid, but no yellow precipitate is formed.

01:16 We know lead chloride is more soluble in hot water compared to agcl and mercurose chloride.

01:23 Therefore when we are heating by addition of water, then what happens? laid chloride goes into solution forming lead ions.

01:32 Lead chloride goes into solution forming lead ions, then when we are adding sodium dichromate to this hot solution, then lead chromate is getting formed.

01:44 Pb cero4...

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