Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 14, Problem 14.69QE

(a)

Interpretation Introduction

Interpretation:

The solubility product constant for BaCrO4 has t o be determined.

Concept Introduction:

Solubility product is the equilibrium constant for the reaction that occurs when an ionic compound is dissolved to produce its constituent ions. It is represented by Ksp. Consider AxBy to be an ionic compound. Its dissociation occurs as follows:

  AxByxAy++yAx

The expression for its Ksp is as follows:

  Ksp=[Ay+]x[Bx]y

(a)

Expert Solution
Check Mark

Answer to Problem 14.69QE

The solubility product constant of BaCrO4 is 1.2×1010.

Explanation of Solution

The dissociation reaction of BaCrO4 is as follows:

  BaCrO4Ba2++CrO42

The solubility of BaCrO4 is 1.1×105 M. Therefore the concentration of both Ag+ and CrO42 is 1.1×105 M.

The formula to calculate the molar solubility of BaCrO4 is as follows:

  Ksp=[Ba2+][CrO42]        (1)

Substitute 1.1×105 M for [Ba2+] and 1.1×105 M for [CrO42] in equation (1).

  Ksp=(1.1×105)(1.1×105)=1.2×1010

Therefore the solubility product constant of BaCrO4 is 1.2×1010.

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for CsMnO4 has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 14.69QE

The solubility product constant of CsMnO4 is 76×1018.

Explanation of Solution

The dissociation reaction of CsMnO4 is as follows:

  CsMnO4Cs++MnO4

The formula to calculate the molar solubility of CsMnO4 is as follows:

  Ksp=[Cs+][MnO4]        (2)

The solubility of CsMnO4 can be calculated as follows:

  Solubility of CsMnO4=(0.22 g100 mL)(1 mL103L)=0.0000022 g/L

The formula to convert solubility of CsMnO4 to moles per litre is as follows:

  Molar solubility(mol/L)=Solubility(g/L)Molar mass of CsMnO4        (3)

Substitute 0.0000022 g/L for solubility of CsMnO4 and 251.841 g/mol for the molar mass of CsMnO4 in equation (3).

  Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×109 M

The solubility of CsMnO4 is 8.7×109 M. Therefore the concentration of both Cs+ and MnO4 is 8.7×109 M.

Substitute 8.7×109 M for [Cs+] and 8.7×109 M for [MnO4] in equation (2).

  Ksp=(8.7×109)(8.7×109)=76×1018

Therefore the solubility product constant of CsMnO4 is 76×1018.

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for Ag3PO4 has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 14.69QE

The solubility product constant of Ag3PO4 is 9.67×1017.

Explanation of Solution

The dissociation reaction of Ag3PO4 is as follows:

  Ag3PO43Ag++PO43

The ICE table for the above reaction is as follows:

EquationAg3PO43Ag++PO43Initial(M)Solid00Change(M)4.4×1053(4.4×105)4.4×105Equilibrium(M)Solid13×1054.4×105

The formula to calculate the molar solubility of Ag3PO4 is as follows:

  Ksp=[Ag+]3[PO43]        (4)

Substitute 13×105 M for [Ag+] and 4.4×105 M for [PO43] in equation (4).

  Ksp=(13×105)3(4.4×105 M)=9.67×1017

Therefore the solubility product constant of Ag3PO4 is 9.67×1017.

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Chapter 14 Solutions

Chemistry: Principles and Practice

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