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Complete the table. Assume that all concentrations are equilibrium concentrations in

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 89P Chapter 15

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 89P

Problem 89P

Complete the table. Assume that all concentrations are equilibrium concentrations in moles per liter, M.

Compound

[Cation]

[Anion]

Ksp

SrCO3

2.4 × 10−5

2.4 × 10−5

_____

SrF2

1.0 × 10−3

_____

4.0 × 10−9

Ag2CO3

_____

1.3 × 10−4

8.8 × 10−12

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to complete the given table.

Step1:

Solubility product of any substance is the product of the concentrations of the dissolved ions raised to the power of their stoichiometric coefficients.

Consider the ionization of a substance AB,

                                AB  A+ + B-

Then, the solubility product, Ksp = [A+][B-]

\Step2:

Compound

[Cation]

[Anion]

Ksp

SrCO3

2.4 x 10-5

2.4 x 10-5

5.76 x 10-10

SrF2

1.0 x 10-3

2.0 x 10-3

4.0 x 10-9

Ag2CO3

2.6 x 10-4

1.3 x 10-4

8.8 x 10-12

  1. The ionization of SrCO3 takes place in the following manner:

                                SrCO3  Sr2+ + CO32-

Solubility product for the reaction, Ksp = [Sr2+][CO32-]

                                          = (2.4 x 10-5)(2.4 x 10-5)

                                          = 5.76 x 10-10

    b)        The ionization of SrF2 takes place in the following manner:

                                SrF2  Sr2+ + 2F-

Solubility product for the reaction, Ksp = [Sr2+][F-]2

                        Therefore, [F-]2 = Ksp / [Sr2+]

                                          = (4.0 x 10-9) / (1.0 x 10-3)

                                          = 4.0 x 10-6

                                      [F-] =

                                          = 2.0 x 10-3

     c)        The ionization of Ag2CO3  takes place in the following manner:

                                Ag2CO3  2Ag+ + CO32-

Solubility product for the reaction, Ksp = [Ag+]2 [CO32-]

                        Therefore, [Ag+]2 = Ksp / [CO32-]

                                          = (8.8 x 10-12) / (1.3 x 10-4)

                                          = 6.769 x 10-8

                                      [Ag+] =

                                          = 2.6 x 10-4

                                        -----------------------

                        

Step 2 of 3

Chapter 15, Problem 89P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Complete the table. Assume that all concentrations are equilibrium concentrations in