Foundations of College Chemistry, Binder Ready Version
Foundations of College Chemistry, Binder Ready Version
15th Edition
ISBN: 9781119083900
Author: Morris Hein, Susan Arena, Cary Willard
Publisher: WILEY
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Chapter 16, Problem 22RQ

(a)

Interpretation Introduction

Interpretation:

Among Mn(OH)2 and Ag2CrO4 that has higher molar solubility of Mn(OH)2 and Ag2CrO4 has to be determined.

Concept Introduction:

The equilibrium constant used for the partially soluble salt in water is termed as solubility product constant Ksp. Consider solubility of AB in water and equilibrium equation of AB is as follows:

  AB(s)A+(aq)+B(aq)

The expression for Ksp is as follows:

  Ksp=[A+][B][AB(s)]

Generally, the concentration of solid is taken as constant. Therefore the expression for Ksp of AB is as follows:

  Ksp=[A+][B]

(a)

Expert Solution
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Explanation of Solution

The equilibrium reaction for Mn(OH)2 is as follows:

  Mn(OH)2(s)Mn2+(aq)+2OH(aq)

In expression for Ksp molar concentration of solid is considered constant. Therefore, expression for Mn(OH)2 is as follows:

  Ksp=[Mn2+][OH]2        (1)

Ionization of solid Mn(OH)2 produces one mole of Mn2+ and two moles of OH. Therefore concentration of OH is two times of concentration of Mn2+ and that is concentration of Mn2+ is s and concentration of OH is 2s.

Substitute s for [Mn2+] and 2s for [OH] in the equation (1).

  Ksp=(s)(2s)2Ksp43=s

Substitute 2×1013 for Ksp in above equation.

  s=2×101343=3.68×105

The equilibrium reaction for Ag2CrO4 is as follows:

  Ag2CrO4(s)2Ag+(aq)+CrO42(aq)

In expression for Ksp molar concentration of solid is considered constant. Therefore, expression for Ag2CrO4 is as follows:

  Ksp=[Ag+]2[CrO42]        (2)

Ionization of solid Ag2CrO4 produces one mole of CrO42 and two moles of Ag+. Therefore concentration of Ag+ is two times of concentration of CrO42 and that is concentration of CrO42 is s and concentration of Ag+ is 2s.

Substitute 2s for [Ag+] and s for [CrO42] in the equation (2).

  Ksp=(2s)2(s)Ksp43=s

Substitute 1.9×1012 for Ksp in above equation.

  s=1.9×101243=7.80×105

Value of molar solubility is greater in Ag2CrO4 than Mn(OH)2.

(b)

Interpretation Introduction

Interpretation:

Among BaCrO4 and Ag2CrO4 that has higher molar solubility of Mn(OH)2 and Ag2CrO4 has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The equilibrium reaction for BaCrO4 is as follows:

  BaCrO4(s)Ba2+(aq)+CrO42(aq)

In expression for Ksp molar concentration of solid is considered constant. Therefore, expression for BaCrO4 is as follows:

  Ksp=[Ba2+][CrO42]        (3)

Ionization of solid BaCrO4 produces Ba2+ and CrO42 in equal amounts that is has equal molar concentrations. Value of concentration of Ba2+ and CrO42 is s.

Substitute s for [Ba2+] and s for [CrO42] in the equation (1).

  Ksp=(s)(s)Ksp=s2

Substitute 8.5×1011 for Ksp in above equation.

  s=8.5×1011=9.21×106

The equilibrium reaction for Ag2CrO4 is as follows:

  Ag2CrO4(s)2Ag+(aq)+CrO42(aq)

In expression for Ksp molar concentration of solid is considered constant. Therefore, expression for Ag2CrO4 is as follows:

  Ksp=[Ag+]2[CrO42]        (2)

Ionization of solid Ag2CrO4 produces one mole of CrO42 and two moles of Ag+. Therefore concentration of Ag+ is two times of concentration of CrO42 and that is concentration of CrO42 is s and concentration of Ag+ is 2s.

Substitute 2s for [Ag+] and s for [CrO42] in the equation (2).

  Ksp=(2s)2(s)Ksp43=s

Substitute 1.9×1012 for Ksp in above equation.

  s=1.9×101243=7.80×105

Value of molar solubility is greater in Ag2CrO4 than BaCrO4.

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Chapter 16 Solutions

Foundations of College Chemistry, Binder Ready Version

Ch. 16.6 - Prob. 16.11PCh. 16.6 - Prob. 16.12PCh. 16.7 - Prob. 16.13PCh. 16.7 - Prob. 16.14PCh. 16.7 - Prob. 16.15PCh. 16.8 - Prob. 16.16PCh. 16 - Prob. 1RQCh. 16 - Prob. 2RQCh. 16 - Prob. 3RQCh. 16 - Prob. 4RQCh. 16 - Prob. 5RQCh. 16 - Prob. 6RQCh. 16 - Prob. 7RQCh. 16 - Prob. 8RQCh. 16 - Prob. 9RQCh. 16 - Prob. 10RQCh. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 14RQCh. 16 - Prob. 15RQCh. 16 - Prob. 16RQCh. 16 - Prob. 17RQCh. 16 - Prob. 18RQCh. 16 - Prob. 19RQCh. 16 - Prob. 20RQCh. 16 - Prob. 21RQCh. 16 - Prob. 22RQCh. 16 - Prob. 23RQCh. 16 - Prob. 24RQCh. 16 - Prob. 25RQCh. 16 - Prob. 26RQCh. 16 - Prob. 27RQCh. 16 - Prob. 1PECh. 16 - Prob. 2PECh. 16 - Prob. 3PECh. 16 - Prob. 4PECh. 16 - Prob. 5PECh. 16 - Prob. 6PECh. 16 - Prob. 7PECh. 16 - Prob. 8PECh. 16 - Prob. 9PECh. 16 - Prob. 10PECh. 16 - Prob. 11PECh. 16 - Prob. 12PECh. 16 - Prob. 13PECh. 16 - Prob. 14PECh. 16 - Prob. 15PECh. 16 - Prob. 16PECh. 16 - Prob. 17PECh. 16 - Prob. 18PECh. 16 - Prob. 19PECh. 16 - Prob. 20PECh. 16 - Prob. 21PECh. 16 - Prob. 22PECh. 16 - Prob. 23PECh. 16 - Prob. 24PECh. 16 - Prob. 25PECh. 16 - Prob. 26PECh. 16 - Prob. 27PECh. 16 - Prob. 28PECh. 16 - Prob. 29PECh. 16 - Prob. 30PECh. 16 - Prob. 31PECh. 16 - Prob. 32PECh. 16 - Prob. 33PECh. 16 - Prob. 34PECh. 16 - Prob. 35PECh. 16 - Prob. 36PECh. 16 - Prob. 37PECh. 16 - Prob. 38PECh. 16 - Prob. 39PECh. 16 - Prob. 40PECh. 16 - Prob. 41PECh. 16 - Prob. 42PECh. 16 - Prob. 43PECh. 16 - Prob. 44PECh. 16 - Prob. 45PECh. 16 - Prob. 46PECh. 16 - Prob. 47PECh. 16 - Prob. 48PECh. 16 - Prob. 49AECh. 16 - Prob. 50AECh. 16 - Prob. 51AECh. 16 - Prob. 52AECh. 16 - Prob. 53AECh. 16 - Prob. 54AECh. 16 - Prob. 55AECh. 16 - Prob. 56AECh. 16 - Prob. 57AECh. 16 - Prob. 58AECh. 16 - Prob. 59AECh. 16 - Prob. 60AECh. 16 - Prob. 61AECh. 16 - Prob. 62AECh. 16 - Prob. 63AECh. 16 - Prob. 64AECh. 16 - Prob. 65AECh. 16 - Prob. 66AECh. 16 - Prob. 67AECh. 16 - Prob. 68AECh. 16 - Prob. 69AECh. 16 - Prob. 70AECh. 16 - Prob. 71AECh. 16 - Prob. 72AECh. 16 - Prob. 73AECh. 16 - Prob. 74AECh. 16 - Prob. 75AECh. 16 - Prob. 76AECh. 16 - Prob. 77AECh. 16 - Prob. 78AECh. 16 - Prob. 79AECh. 16 - Prob. 80AECh. 16 - Prob. 81AECh. 16 - Prob. 83AECh. 16 - Prob. 84AECh. 16 - Prob. 85AECh. 16 - Prob. 86CECh. 16 - Prob. 87CECh. 16 - Prob. 88CECh. 16 - Prob. 89CE
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