Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
Question
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Chapter 16, Problem 25E

(a)

Interpretation Introduction

Interpretation: The solubility of CaC2O4 and BiI3 is given. The solubility product of CaC2O4 and BiI3 is to be calculated.

Concept introduction: The solubility product Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated as,

Ksp=[A]x[B]y

(a)

Expert Solution
Check Mark

Answer to Problem 25E

Answer

The solubility product of CaC2O4 is 2.3×109_ .

Explanation of Solution

Explanation

To determine: The solubility product of CaC2O4 .

The concentration of Ca2+ is 4.8×105mol/L_ .

Given

Solubility of CaC2O4 is 4.8×105mol/L .

Since, solid CaC2O4 is placed in contact with water. Therefore, compound present before the reaction is CaC2O4 and H2O . The dissociation reaction of CaC2O4 is,

CaC2O4(s)Ca2+(aq)+C2O42(aq)

Since, CaC2O4 does not dissolved initially, hence,

[Ca2+]initial=[C2O42]initial=0

The concentration at equilibrium can be calculated from the measured solubility of CaC2O4 . If 4.8×105mol of CaC2O4 is dissolved in 1.0L of solution, the change in solubility will be equal to 4.8×105mol/L . The reaction is,

CaC2O4(s)Ca2+(aq)+C2O42(aq)

Therefore,

4.8×105mol/LCaC2O44.8×105mol/LCa2++4.8×105mol/LC2O4

The equilibrium concentration of [Ca2+] is written as,

[Ca2+]=[Ca2+]initial+changetoreachequilibrium

Substitute the value of [Ca2+]initial and change to reach equilibrium in the above equation.

[Ca2+]=[Ca2+]initial+changetoreachequilibrium=0+4.8×105mol/L=4.8×105mol/L_

The concentration of C2O42 is 4.8×105mol/L_ .

Given

Solubility of CaC2O4 is 4.8×105mol/L .

The equilibrium concentration of [C2O42] is written as,

[C2O42]=[C2O42]initial+changetoreachequilibrium

Substitute the value of [C2O42]initial and change to reach equilibrium in the above equation.

[C2O42]=[C2O42]initial+changetoreachequilibrium=0+4.8×105mol/L=4.8×105mol/L_

The solubility product of CaC2O4 is 2.3×109_ .

The concentration of Ca2+ is 4.8×105mol/L .

The concentration of C2O42 is 4.8×105mol/L .

Formula

The solubility product of CaC2O4 is calculated as,

Ksp=[Ca2+][C2O42]

Where,

  • Ksp is solubility product.
  • [Ca2+] is concentration of Ca2+ .
  • [C2O42] is concentration of C2O42

Substitute the values of [Ca2+] and [C2O42] in the above expression.

Ksp=[Ca2+][C2O42]=(4.8×105)(4.8×105)=2.3×109_

(b)

Interpretation Introduction

Interpretation: The solubility of CaC2O4 and BiI3 is given. The solubility product of CaC2O4 and BiI3 is to be calculated.

Concept introduction: The solubility product Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated as,

Ksp=[A]x[B]y

(b)

Expert Solution
Check Mark

Answer to Problem 25E

Answer

The solubility product of BiI3 is 8.20×1019_ .

Explanation of Solution

Explanation

To determine: The solubility product of BiI3 .

The concentration of Bi3+ is 1.32×105mol/L_ .

Given

Solubility of BiI3 is 1.32×105mol/L .

Since, solid BiI3 is placed in contact with water. Therefore, compound present before the reaction is BiI3 and H2O . The dissociation reaction of BiI3 is,

BiI3(s)Bi3+(aq)+3I(aq)

Since, BiI3 does not dissolved initially, hence,

[Bi3+]initial=[I]initial=0

The concentration at equilibrium can be calculated from the measured solubility of BiI3 . If 1.32×105mol/L of BiI3 is dissolved in 1.0L of solution, the change in solubility will be equal to 1.32×105mol/L . The reaction is,

BiI3(s)Bi3+(aq)+3I(aq)

Therefore,

1.32×105mol/LBiI31.32×105mol/LBi3++(3×1.32×105)mol/LI

The equilibrium concentration of Bi3+ is written as,

[Bi3+]=[Bi3+]initial+changetoreachequilibrium

Substitute the value of [Bi3+]initial and change to reach equilibrium in the above equation.

[Bi3+]=[Bi3+]initial+changetoreachequilibrium=0+1.32×105mol/L=1.32×105mol/L_

The concentration of I is 3.96×105mol/L_ .

Given

Solubility of BiI3 is 1.32×105mol/L .

The equilibrium concentration of I is written as,

[I]=[I]initial+changetoreachequilibrium

Substitute the value of [I]initial and change to reach equilibrium in the above equation.

[I]=[I]initial+changetoreachequilibrium=0+(3×1.32×105mol/L)=3.96×105mol/L_

The solubility product of BiI3 is 8.20×1019_ .

The concentration of Bi3+ is 1.32×105mol/L .

The concentration of I is 3.96×105mol/L .

Formula

The solubility product of BiI3 is calculated as,

Ksp=[Bi3+][I]3

Where,

  • Ksp is solubility product.
  • [Bi3+] is concentration of Bi3+ .
  • [I] is concentration of I

Substitute the values of [Bi3+] and [I] in the above expression.

Ksp=[Bi3+][I]3=(1.32×105)(3.96×105)3=8.20×1019_

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Chapter 16 Solutions

Chemistry

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