Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 17, Problem 45QP

Write balanced equations and solubility product expressions for the solubility equilibria of the following compounds: ( a )  CuBr,  ( b )  ZnC 2 O 4 ( c )  Ag 2 CrO 4 ( d )  Hg 2 Cl 2 ( e )  AuCl 3 ( f )  Mn 3 ( PO 4 ) 2 .

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

Thebalanced equation and the solubility product constant of the given compounds are to be calculated.

Concept introduction:

The amount of solute dissolved in a given volume of the solvent to form a saturated solution at a given temperatureis termed as the solubility of the solute in the solvent at that temperature.

The solubility product of a sparingly-soluble salt is given as the product of the concentration of the ions raised to the power equal to the number of times the ion occurs in the equation, after the dissociation of the electrolyte.

The number of moles of the solute dissolved per litre of the solution is called molar solubility.

For a general reaction: AB(s)A+(aq)+B(aq)

The solubility product can be calculated by the expression as: Ksp= [A+][B]

Here, Ksp is the solubility product constant and sp standsfor solubility product.

Answer to Problem 45QP

Solution:

(a)

Balanced equation: CuBr(s)Cu+(aq)+Br(aq)

Solubility product: Ksp= [Cu+][Br]

(b)

Balanced equation: ZnC2O4(s)Zn2+(aq)+C2O42(aq)

Solubility product: Ksp= [Zn2+][C2O42]

(c)

Balanced equation: Ag2CrO4(s)2Ag+(aq)+ CrO42(aq)

Solubility product: Ksp= [Ag+]2[CrO42]

(d)

Balanced equation: Hg2Cl2(s)Hg22+(aq)+2Cl(aq)

Solubility product: Ksp=[Hg22+][Cl]2

(e)

Balanced equation: AuCl3(s)Au3+(aq)+3Cl(aq)

Solubility product: Ksp= [Au3+][Cl]3

(f)

Balanced equation: Mn3(PO4)2(s)3Mn2+(aq)+2PO43(aq)

Solubility product: Ksp= [Mn2+]3[PO43]2

Explanation of Solution

a) CuBr(s)

The balanced equation for the dissociation of CuBr(s) is as follows:

CuBr(s)Cu+(aq)+Br(aq)

Let s be the molar solubility.

The initial change equilibrium table for the ionisation of CuBr(s) is as follows:

CuBr(s)Cu+(aq)+Br(aq)Initial(M)00Change(M)s+s+sEquilibrium(M)ss

The equilibrium expression for the reaction is written as follows:

Ksp= [Cu+][Br]

Here, Ksp is the solubility product constant, [Cu+] is the concentration of the copper ion, and [Br] is the concentration of the bromine ion.

Substitute the values of [Cu+] and [Br] in the above expression,

Ksp= (s)(s)=s2

Hence, the solubility product constant of CuBr(s) is Ksp= s2.

b) ZnC2O4(s)

The balanced equation for the dissociation of ZnC2O4(s) is as follows:

ZnC2O4(s)Zn2+(aq)+C2O42(aq)

Let s be the molar solubility.

The initial change equilibrium table for the ionisation of ZnC2O4(s) is as follows:

ZnC2O4(s)Zn2+(aq)+ C2O42(aq)Initial(M)00Change(M)s+s+sEquilibrium(M)ss

The equilibrium expression for the reaction is written as follows:

Ksp= [Zn2+][C2O42]

Here, Ksp is the solubility product constant, [Zn2+] is the concentration of the zinc ion, and [C2O42] is the concentration of the oxalate ion.

Substitute the values of [Zn2+] and [C2O42] in the above expression,

Ksp= (s)(s)=s2

Hence, the solubility product constant of ZnC2O4(s) is Ksp= s2.

c) Ag2CrO4(s)

The balanced equation for the dissociation of Ag2CrO4(s) is as follows:

Ag2CrO42Ag+(aq)+CrO42(aq)

Let s be the molar solubility.

The initial change equilibrium table for theionization of Ag2CrO4(s) is as follows:

Ag2CrO4(s)2Ag+(aq)+CrO42(aq)Initial(M)00Change(M)s+2s+sEquilibrium(M)2ss

The equilibrium expression for the reaction is written as follows:

Ksp= [Ag+]2[CrO42]

Here, Ksp is the solubility product constant, [Ag+] is the concentration of the silver ion, and [CrO42] is the concentration of the chromate ion.

Substitute the values of [Ag+] and [CrO42] in the above expression,

Ksp= (2s)2(s)=4s3

Hence, the solubility product constant of Ag2CrO4(s) is Ksp= 4s3.

d) Hg2Cl2(s)

The balanced equation for the dissociation of Hg2Cl2(s) is as follows:

Hg2Cl2(s)Hg22+(aq)+2Cl(aq)

Let s be the molar solubility.

The initial change equilibrium table for the ionisation of Hg2Cl2(s) is as follows:

Hg2Cl2(s)Hg22+(aq)+ 2Cl(aq)Initial(M)00Change(M)s+s+2sEquilibrium(M)s2s

The equilibrium expression for the reaction is written as follows:

Ksp= [Hg22+][Cl]2

Here, Ksp is the solubility product constant, [Hg22+] is the concentration of Dimercury(2+), and [Cl] is the concentration of the chlorine ion.

Substitute the values of [Hg22+] and [Cl] in the above expression,

Ksp= (s)(2s)2=4s3

Hence, the solubility product constant of Hg2Cl2(s) is Ksp= 4s3.

e) AuCl3(s)

The balanced equation for the dissociation of AuCl3(s) is as follows:

AuCl3(s)Au3+(aq)+3Cl(aq)

The ICE table for the ionisation for AuCl3(s) is as follows:

Let s be the molar solubility.

AuCl3(s)Au3+(aq)+ 3Cl(aq)Initial(M)00Change(M)s+s+3sEquilibrium(M)s3s

The equilibrium expression for the reaction is written as follows:

Ksp= [Au3+][Cl]3

Here, Ksp is the solubility product constant, [Au3+] is the concentration of Gold(3+), and [Cl] is the concentration of the chlorine ion.

Substitute the values of [Au3+] and [Cl] in the above expression,

Ksp= (s)(3s)3=27s4

Hence, the solubility product constant of AuCl3(s) is Ksp= 27s4.

f) Mn3(PO4)2(s)

The balanced equation for the dissociation of Mn3(PO4)2(s) is as follows:

Mn3(PO4)2(s)3Mn2+(aq)+2PO43(aq)

Let s be the molar solubility.

The initial change equilibrium table for the ionisation of Mn3(PO4)2(s) is as follows:

Mn3(PO4)2(s)3Mn2+(aq)+ 2PO42(aq)Initial(M)00Change(M)s+3s+2sEquilibrium(M)3s2s

The equilibrium expression for the reaction is written as follows:

Ksp= [Mn2+]3[PO43]2

Here, Ksp is the solubility product constant, [Mn2+] is the concentration of the manganese (2+) ion, and [PO43] is the concentration of the phosphate (3+) ion.

Substitute the values of [Mn2+] and [PO43] in the above expression,

Ksp=(3s)3(2s)2=108s5

Hence, the solubility product constant of Mn3(PO4)2(s) is Ksp= 108s5.

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Chapter 17 Solutions

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