World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Chapter 17, Problem 53A
Interpretation Introduction

Interpretation : The solubility of lead chromate in grams/L at 25°C needs to be determined.

Concept Introduction : The solubility product can be defined as the product of the equilibrium concentrations of the ions in a saturated solution of a salt.  In the expression of solubility product each concentration is raised to the power of their coefficient of ion as given in the balanced equation. For example; BaIO3 will show the equilibrium equation as:

  BaIO3(s)Ba+2(aq)+ IO3-2(aq)

The solubility expression will be:

  Ksp= [Ba+2] [ IO3-2]

The product of the concentrations of the ions at any moment in time for the given salt is called as ion product that is represented as Qsp . The expression for it is also same as Ksp but this is not at equilibrium condition.

Expert Solution & Answer
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Answer to Problem 53A

  1.7×10-4 g/L

Explanation of Solution

Given information:Chromate ion is used as a qualitative test for lead (II) ions as it forms bright yellow precipitate of lead chromate.

Ksp for PbCrO4 = 2.8×1013

  PbCrO4Pb2+  + CrO42

Thus, the Ksp expression should be:

  PbCrO4Pb2+  + CrO42Ksp=[Pb2+ ] [CrO42]Ksp=[s][s]=s22.8×10-13=s2s = 5.3×10-7M

Molar mass of PbCrO4 = 323.19 g/mol

Calculate the solubility in g/L:

   5.3×10-7mole1 L×323.19 g1 mole = 1.7×10-4 g/L

Conclusion

Thus, the solubility is 1.7×10-4 g/L .

Chapter 17 Solutions

World of Chemistry, 3rd edition

Ch. 17.2 - Prob. 4RQCh. 17.2 - Prob. 5RQCh. 17.3 - Prob. 1RQCh. 17.3 - Prob. 2RQCh. 17.3 - Prob. 3RQCh. 17.3 - Prob. 4RQCh. 17.3 - Prob. 5RQCh. 17.3 - Prob. 6RQCh. 17.3 - Prob. 7RQCh. 17 - Prob. 1ACh. 17 - Prob. 2ACh. 17 - Prob. 3ACh. 17 - Prob. 4ACh. 17 - Prob. 5ACh. 17 - Prob. 6ACh. 17 - Prob. 7ACh. 17 - Prob. 8ACh. 17 - Prob. 9ACh. 17 - Prob. 10ACh. 17 - Prob. 11ACh. 17 - Prob. 12ACh. 17 - Prob. 13ACh. 17 - Prob. 14ACh. 17 - Prob. 15ACh. 17 - Prob. 16ACh. 17 - Prob. 17ACh. 17 - Prob. 18ACh. 17 - Prob. 19ACh. 17 - Prob. 20ACh. 17 - Prob. 21ACh. 17 - Prob. 22ACh. 17 - Prob. 23ACh. 17 - Prob. 24ACh. 17 - Prob. 25ACh. 17 - Prob. 26ACh. 17 - Prob. 27ACh. 17 - Prob. 28ACh. 17 - Prob. 29ACh. 17 - Prob. 30ACh. 17 - Prob. 31ACh. 17 - Prob. 32ACh. 17 - Prob. 33ACh. 17 - Prob. 34ACh. 17 - Prob. 35ACh. 17 - Prob. 36ACh. 17 - Prob. 37ACh. 17 - Prob. 38ACh. 17 - Prob. 39ACh. 17 - Prob. 40ACh. 17 - Prob. 41ACh. 17 - Prob. 42ACh. 17 - Prob. 43ACh. 17 - Prob. 44ACh. 17 - Prob. 45ACh. 17 - Prob. 46ACh. 17 - Prob. 47ACh. 17 - Prob. 48ACh. 17 - Prob. 49ACh. 17 - Prob. 50ACh. 17 - Prob. 51ACh. 17 - Prob. 52ACh. 17 - Prob. 53ACh. 17 - Prob. 54ACh. 17 - Prob. 55ACh. 17 - Prob. 56ACh. 17 - Prob. 57ACh. 17 - Prob. 58ACh. 17 - Prob. 59ACh. 17 - Prob. 60ACh. 17 - Prob. 61ACh. 17 - Prob. 62ACh. 17 - Prob. 63ACh. 17 - Prob. 64ACh. 17 - Prob. 65ACh. 17 - Prob. 66ACh. 17 - Prob. 67ACh. 17 - Prob. 68ACh. 17 - Prob. 69ACh. 17 - Prob. 70ACh. 17 - Prob. 71ACh. 17 - Prob. 72ACh. 17 - Prob. 73ACh. 17 - Prob. 74ACh. 17 - Prob. 75ACh. 17 - Prob. 1STPCh. 17 - Prob. 2STPCh. 17 - Prob. 3STPCh. 17 - Prob. 4STPCh. 17 - Prob. 5STPCh. 17 - Prob. 6STPCh. 17 - Prob. 7STP
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