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Question

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Explanation / Answer

Analysis of ions

1. The boxes for CuI2 and FeI3 are left blank because these compounds are unstable and they tend to decompose upon standing rapidly.

2. If we were to tell the white solid is PbCl2, we would heat this in boiling water. PbCl2 solid is soluble in hot water and this would confirm that the solid is PbCl2.

3. Without using NaI, we would add dilute HCl to the solution containing Ag+ and Pb2+ ions,

Both of these precipitate out as white solid.

Ag+ + Cl- ---> AgCl (ppt)

Pb2+ + 2Cl- ---> PbCl2 (ppt)

4. Minimum [Cl-] needed to precipitate Pb2+ = sq.rt.(Ksp/[Pb2+])

                                                                       = sq.rt.(2 x 10^-5/0.02) = 0.0316 M

Volume of 6 M HCl required = 0.0316 M x 2 ml/6 M = 0.010 ml

5. Adding excess HCl forms complex ion PbCl4^2- which is soluble, therefore not all Pb2+ would precipitate out of solution,

PbCl2(s) + 2Cl- --> PbCl4^2- (soluble)

6. minimum [S^2-] needed to precipitate ZnS = Ksp/[Zn2+]

                                                                        = 2 x 10^-25/0.01 = 2 x 10^-23 M

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