The solubility product constant for BaCrO 4 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 K sp . Consider A x B y to be an ionic compound. Its dissociation occurs as follows: A x B y ⇌ x A y + + y A x − The expression for its K sp is as follows: K sp = [ A y + ] x [ B x − ] y

Question

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Answer 1

Question

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:

$AxBy⇌xAy++yAx−$

The expression for its $Ksp$ is as follows:

$Ksp=[Ay+]x[Bx−]y$

(a)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Explanation of Solution

The dissociation reaction of $BaCrO4$ is as follows:

$BaCrO4⇌Ba2++CrO42−$

The solubility of $BaCrO4$ is $1.1×10−5 M$ . Therefore the concentration of both $Ag+$ and $CrO42−$ is $1.1×10−5 M$ .

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

$Ksp=[Ba2+][CrO42−]$ (1)

Substitute $1.1×10−5 M$ for $[Ba2+]$ and $1.1×10−5 M$ for $[CrO42−]$ in equation (1).

$Ksp=(1.1×10−5)(1.1×10−5)=1.2×10−10$

Therefore the solubility product constant of $BaCrO4$ is $1.2×10−10$ .

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $CsMnO4$ is $76×10−18$ .

Explanation of Solution

The dissociation reaction of $CsMnO4$ is as follows:

$CsMnO4⇌Cs++MnO4−$

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:

$Solubility of CsMnO4=(0.22 g100 mL)(1 mL10−3L)=0.0000022 g/L$

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).

$Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×10−9 M$

The solubility of $CsMnO4$ is $8.7×10−9 M$ . Therefore the concentration of both $Cs+$ and $MnO4−$ is $8.7×10−9 M$ .

Substitute $8.7×10−9 M$ for $[Cs+]$ and $8.7×10−9 M$ for $[MnO4−]$ in equation (2).

$Ksp=(8.7×10−9)(8.7×10−9)=76×10−18$

Therefore the solubility product constant of $CsMnO4$ is $76×10−18$ .

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Explanation of Solution

The dissociation reaction of $Ag3PO4$ is as follows:

$Ag3PO4⇌3Ag++PO43−$

The ICE table for the above reaction is as follows:

$EquationAg3PO4⇌3Ag++PO43−Initial(M)Solid00Change(M)−4.4×10−53(4.4×10−5)4.4×10−5Equilibrium(M)Solid13×10−54.4×10−5$

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

$Ksp=[Ag+]3[PO43−]$ (4)

Substitute $13×10−5 M$ for $[Ag+]$ and $4.4×10−5 M$ for $[PO43−]$ in equation (4).

$Ksp=(13×10−5)3(4.4×10−5 M)=9.67×10−17$

Therefore the solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

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Help me solve this problem. Thank you in advance.

22.7 Predict the monoalkylated products of the following reactions with benzene. (a) AlCl3 Ya (b) AlCl3 (c) H3PO4 (d) 22.8 Think-Pair-Share AICI3 The reaction below is a common electrophilic aromatic substitution. SO3 H₂SO4 SO₂H (a) Draw the reaction mechanism for this reaction using HSO,+ as the electrophile. (b) Sketch the reaction coordinate diagram, where the product is lower in energy than the starting reactant. (c) Which step in the reaction mechanism is highest in energy? Explain. (d) Which of the following reaction conditions could be used in an electrophilic aro- matic substitution with benzene to provide substituted phenyl derivatives? (i) AICI3 HNO3 H₂SO4 K2Cr2O7 (iii) H₂SO4 (iv) H₂PO₁

Is an acid-base reaction the only type of reaction that would cause leavening products to rise?

Answer 2

Question

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:

$AxBy⇌xAy++yAx−$

The expression for its $Ksp$ is as follows:

$Ksp=[Ay+]x[Bx−]y$

(a)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Explanation of Solution

The dissociation reaction of $BaCrO4$ is as follows:

$BaCrO4⇌Ba2++CrO42−$

The solubility of $BaCrO4$ is $1.1×10−5 M$ . Therefore the concentration of both $Ag+$ and $CrO42−$ is $1.1×10−5 M$ .

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

$Ksp=[Ba2+][CrO42−]$ (1)

Substitute $1.1×10−5 M$ for $[Ba2+]$ and $1.1×10−5 M$ for $[CrO42−]$ in equation (1).

$Ksp=(1.1×10−5)(1.1×10−5)=1.2×10−10$

Therefore the solubility product constant of $BaCrO4$ is $1.2×10−10$ .

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $CsMnO4$ is $76×10−18$ .

Explanation of Solution

The dissociation reaction of $CsMnO4$ is as follows:

$CsMnO4⇌Cs++MnO4−$

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:

$Solubility of CsMnO4=(0.22 g100 mL)(1 mL10−3L)=0.0000022 g/L$

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).

$Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×10−9 M$

The solubility of $CsMnO4$ is $8.7×10−9 M$ . Therefore the concentration of both $Cs+$ and $MnO4−$ is $8.7×10−9 M$ .

Substitute $8.7×10−9 M$ for $[Cs+]$ and $8.7×10−9 M$ for $[MnO4−]$ in equation (2).

$Ksp=(8.7×10−9)(8.7×10−9)=76×10−18$

Therefore the solubility product constant of $CsMnO4$ is $76×10−18$ .

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Explanation of Solution

The dissociation reaction of $Ag3PO4$ is as follows:

$Ag3PO4⇌3Ag++PO43−$

The ICE table for the above reaction is as follows:

$EquationAg3PO4⇌3Ag++PO43−Initial(M)Solid00Change(M)−4.4×10−53(4.4×10−5)4.4×10−5Equilibrium(M)Solid13×10−54.4×10−5$

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

$Ksp=[Ag+]3[PO43−]$ (4)

Substitute $13×10−5 M$ for $[Ag+]$ and $4.4×10−5 M$ for $[PO43−]$ in equation (4).

$Ksp=(13×10−5)3(4.4×10−5 M)=9.67×10−17$

Therefore the solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

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Answer 3

Question

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:

$AxBy⇌xAy++yAx−$

The expression for its $Ksp$ is as follows:

$Ksp=[Ay+]x[Bx−]y$

(a)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Explanation of Solution

The dissociation reaction of $BaCrO4$ is as follows:

$BaCrO4⇌Ba2++CrO42−$

The solubility of $BaCrO4$ is $1.1×10−5 M$ . Therefore the concentration of both $Ag+$ and $CrO42−$ is $1.1×10−5 M$ .

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

$Ksp=[Ba2+][CrO42−]$ (1)

Substitute $1.1×10−5 M$ for $[Ba2+]$ and $1.1×10−5 M$ for $[CrO42−]$ in equation (1).

$Ksp=(1.1×10−5)(1.1×10−5)=1.2×10−10$

Therefore the solubility product constant of $BaCrO4$ is $1.2×10−10$ .

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $CsMnO4$ is $76×10−18$ .

Explanation of Solution

The dissociation reaction of $CsMnO4$ is as follows:

$CsMnO4⇌Cs++MnO4−$

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:

$Solubility of CsMnO4=(0.22 g100 mL)(1 mL10−3L)=0.0000022 g/L$

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).

$Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×10−9 M$

The solubility of $CsMnO4$ is $8.7×10−9 M$ . Therefore the concentration of both $Cs+$ and $MnO4−$ is $8.7×10−9 M$ .

Substitute $8.7×10−9 M$ for $[Cs+]$ and $8.7×10−9 M$ for $[MnO4−]$ in equation (2).

$Ksp=(8.7×10−9)(8.7×10−9)=76×10−18$

Therefore the solubility product constant of $CsMnO4$ is $76×10−18$ .

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Explanation of Solution

The dissociation reaction of $Ag3PO4$ is as follows:

$Ag3PO4⇌3Ag++PO43−$

The ICE table for the above reaction is as follows:

$EquationAg3PO4⇌3Ag++PO43−Initial(M)Solid00Change(M)−4.4×10−53(4.4×10−5)4.4×10−5Equilibrium(M)Solid13×10−54.4×10−5$

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

$Ksp=[Ag+]3[PO43−]$ (4)

Substitute $13×10−5 M$ for $[Ag+]$ and $4.4×10−5 M$ for $[PO43−]$ in equation (4).

$Ksp=(13×10−5)3(4.4×10−5 M)=9.67×10−17$

Therefore the solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

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Answer 4

Question

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:

$AxBy⇌xAy++yAx−$

The expression for its $Ksp$ is as follows:

$Ksp=[Ay+]x[Bx−]y$

(a)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Explanation of Solution

The dissociation reaction of $BaCrO4$ is as follows:

$BaCrO4⇌Ba2++CrO42−$

The solubility of $BaCrO4$ is $1.1×10−5 M$ . Therefore the concentration of both $Ag+$ and $CrO42−$ is $1.1×10−5 M$ .

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

$Ksp=[Ba2+][CrO42−]$ (1)

Substitute $1.1×10−5 M$ for $[Ba2+]$ and $1.1×10−5 M$ for $[CrO42−]$ in equation (1).

$Ksp=(1.1×10−5)(1.1×10−5)=1.2×10−10$

Therefore the solubility product constant of $BaCrO4$ is $1.2×10−10$ .

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $CsMnO4$ is $76×10−18$ .

Explanation of Solution

The dissociation reaction of $CsMnO4$ is as follows:

$CsMnO4⇌Cs++MnO4−$

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:

$Solubility of CsMnO4=(0.22 g100 mL)(1 mL10−3L)=0.0000022 g/L$

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).

$Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×10−9 M$

The solubility of $CsMnO4$ is $8.7×10−9 M$ . Therefore the concentration of both $Cs+$ and $MnO4−$ is $8.7×10−9 M$ .

Substitute $8.7×10−9 M$ for $[Cs+]$ and $8.7×10−9 M$ for $[MnO4−]$ in equation (2).

$Ksp=(8.7×10−9)(8.7×10−9)=76×10−18$

Therefore the solubility product constant of $CsMnO4$ is $76×10−18$ .

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Explanation of Solution

The dissociation reaction of $Ag3PO4$ is as follows:

$Ag3PO4⇌3Ag++PO43−$

The ICE table for the above reaction is as follows:

$EquationAg3PO4⇌3Ag++PO43−Initial(M)Solid00Change(M)−4.4×10−53(4.4×10−5)4.4×10−5Equilibrium(M)Solid13×10−54.4×10−5$

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

$Ksp=[Ag+]3[PO43−]$ (4)

Substitute $13×10−5 M$ for $[Ag+]$ and $4.4×10−5 M$ for $[PO43−]$ in equation (4).

$Ksp=(13×10−5)3(4.4×10−5 M)=9.67×10−17$

Therefore the solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

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Answer 5

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:

$AxBy⇌xAy++yAx−$

The expression for its $Ksp$ is as follows:

$Ksp=[Ay+]x[Bx−]y$

(a)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Explanation of Solution

The dissociation reaction of $BaCrO4$ is as follows:

$BaCrO4⇌Ba2++CrO42−$

The solubility of $BaCrO4$ is $1.1×10−5 M$ . Therefore the concentration of both $Ag+$ and $CrO42−$ is $1.1×10−5 M$ .

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

$Ksp=[Ba2+][CrO42−]$ (1)

Substitute $1.1×10−5 M$ for $[Ba2+]$ and $1.1×10−5 M$ for $[CrO42−]$ in equation (1).

$Ksp=(1.1×10−5)(1.1×10−5)=1.2×10−10$

Therefore the solubility product constant of $BaCrO4$ is $1.2×10−10$ .

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $CsMnO4$ is $76×10−18$ .

Explanation of Solution

The dissociation reaction of $CsMnO4$ is as follows:

$CsMnO4⇌Cs++MnO4−$

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:

$Solubility of CsMnO4=(0.22 g100 mL)(1 mL10−3L)=0.0000022 g/L$

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).

$Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×10−9 M$

The solubility of $CsMnO4$ is $8.7×10−9 M$ . Therefore the concentration of both $Cs+$ and $MnO4−$ is $8.7×10−9 M$ .

Substitute $8.7×10−9 M$ for $[Cs+]$ and $8.7×10−9 M$ for $[MnO4−]$ in equation (2).

$Ksp=(8.7×10−9)(8.7×10−9)=76×10−18$

Therefore the solubility product constant of $CsMnO4$ is $76×10−18$ .

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Explanation of Solution

The dissociation reaction of $Ag3PO4$ is as follows:

$Ag3PO4⇌3Ag++PO43−$

The ICE table for the above reaction is as follows:

$EquationAg3PO4⇌3Ag++PO43−Initial(M)Solid00Change(M)−4.4×10−53(4.4×10−5)4.4×10−5Equilibrium(M)Solid13×10−54.4×10−5$

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

$Ksp=[Ag+]3[PO43−]$ (4)

Substitute $13×10−5 M$ for $[Ag+]$ and $4.4×10−5 M$ for $[PO43−]$ in equation (4).

$Ksp=(13×10−5)3(4.4×10−5 M)=9.67×10−17$

Therefore the solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Answer 6

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:

$AxBy⇌xAy++yAx−$

The expression for its $Ksp$ is as follows:

$Ksp=[Ay+]x[Bx−]y$

(a)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Explanation of Solution

The dissociation reaction of $BaCrO4$ is as follows:

$BaCrO4⇌Ba2++CrO42−$

The solubility of $BaCrO4$ is $1.1×10−5 M$ . Therefore the concentration of both $Ag+$ and $CrO42−$ is $1.1×10−5 M$ .

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

$Ksp=[Ba2+][CrO42−]$ (1)

Substitute $1.1×10−5 M$ for $[Ba2+]$ and $1.1×10−5 M$ for $[CrO42−]$ in equation (1).

$Ksp=(1.1×10−5)(1.1×10−5)=1.2×10−10$

Therefore the solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Answer 7

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:

$AxBy⇌xAy++yAx−$

The expression for its $Ksp$ is as follows:

$Ksp=[Ay+]x[Bx−]y$

Answer 8

(a)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The dissociation reaction of $BaCrO4$ is as follows:

$BaCrO4⇌Ba2++CrO42−$

The solubility of $BaCrO4$ is $1.1×10−5 M$ . Therefore the concentration of both $Ag+$ and $CrO42−$ is $1.1×10−5 M$ .

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

$Ksp=[Ba2+][CrO42−]$ (1)

Substitute $1.1×10−5 M$ for $[Ba2+]$ and $1.1×10−5 M$ for $[CrO42−]$ in equation (1).

$Ksp=(1.1×10−5)(1.1×10−5)=1.2×10−10$

Therefore the solubility product constant of $BaCrO4$ is $1.2×10−10$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $CsMnO4$ is $76×10−18$ .

Explanation of Solution

The dissociation reaction of $CsMnO4$ is as follows:

$CsMnO4⇌Cs++MnO4−$

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:

$Solubility of CsMnO4=(0.22 g100 mL)(1 mL10−3L)=0.0000022 g/L$

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).

$Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×10−9 M$

The solubility of $CsMnO4$ is $8.7×10−9 M$ . Therefore the concentration of both $Cs+$ and $MnO4−$ is $8.7×10−9 M$ .

Substitute $8.7×10−9 M$ for $[Cs+]$ and $8.7×10−9 M$ for $[MnO4−]$ in equation (2).

$Ksp=(8.7×10−9)(8.7×10−9)=76×10−18$

Therefore the solubility product constant of $CsMnO4$ is $76×10−18$ .

Answer 14

(b)

Interpretation Introduction

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $CsMnO4$ is $76×10−18$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The dissociation reaction of $CsMnO4$ is as follows:

$CsMnO4⇌Cs++MnO4−$

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:

$Solubility of CsMnO4=(0.22 g100 mL)(1 mL10−3L)=0.0000022 g/L$

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).

$Molar solubility(mol/L)=(0.0000022 g1 L)(1 mol251.841 g)=8.7×10−9 M$

The solubility of $CsMnO4$ is $8.7×10−9 M$ . Therefore the concentration of both $Cs+$ and $MnO4−$ is $8.7×10−9 M$ .

Substitute $8.7×10−9 M$ for $[Cs+]$ and $8.7×10−9 M$ for $[MnO4−]$ in equation (2).

$Ksp=(8.7×10−9)(8.7×10−9)=76×10−18$

Therefore the solubility product constant of $CsMnO4$ is $76×10−18$ .

Answer 20

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Explanation of Solution

The dissociation reaction of $Ag3PO4$ is as follows:

$Ag3PO4⇌3Ag++PO43−$

The ICE table for the above reaction is as follows:

$EquationAg3PO4⇌3Ag++PO43−Initial(M)Solid00Change(M)−4.4×10−53(4.4×10−5)4.4×10−5Equilibrium(M)Solid13×10−54.4×10−5$

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

$Ksp=[Ag+]3[PO43−]$ (4)

Substitute $13×10−5 M$ for $[Ag+]$ and $4.4×10−5 M$ for $[PO43−]$ in equation (4).

$Ksp=(13×10−5)3(4.4×10−5 M)=9.67×10−17$

Therefore the solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Answer 21

(c)

Interpretation Introduction

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer 22

(c)

Expert Solution

Answer to Problem 14.69QE

The solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The dissociation reaction of $Ag3PO4$ is as follows:

$Ag3PO4⇌3Ag++PO43−$

The ICE table for the above reaction is as follows:

$EquationAg3PO4⇌3Ag++PO43−Initial(M)Solid00Change(M)−4.4×10−53(4.4×10−5)4.4×10−5Equilibrium(M)Solid13×10−54.4×10−5$

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

$Ksp=[Ag+]3[PO43−]$ (4)

Substitute $13×10−5 M$ for $[Ag+]$ and $4.4×10−5 M$ for $[PO43−]$ in equation (4).

$Ksp=(13×10−5)3(4.4×10−5 M)=9.67×10−17$

Therefore the solubility product constant of $Ag3PO4$ is $9.67×10−17$ .

Answer 27

Ch. 14 - Prob. 14.1QE Ch. 14 - Describe a nonchemical system that is in...Ch. 14 - Describe a nonchemical system that is not in...Ch. 14 - Prob. 14.4QE Ch. 14 - Prob. 14.5QE Ch. 14 - Prob. 14.6QE Ch. 14 - Prob. 14.7QE Ch. 14 - Prob. 14.8QE Ch. 14 - Prob. 14.9QE Ch. 14 - Prob. 14.10QE

Concept explainers

Videos

Answer to Problem 14.69QE

Explanation of Solution

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 14.69QE

Explanation of Solution

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 14.69QE

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

Concept explainers

Videos

Answer to Problem 14.69QE

Explanation of Solution

Interpretation: The solubility product constant for CsMnO4<m:math><m:mrow><m:msub><m:mrow><m:mtext>CsMnO</m:mtext></m:mrow><m:mtext>4</m:mtext></m:msub></m:mrow></m:math> has to be determined. Concept Introduction: Refer to part (a).

Answer to Problem 14.69QE

Explanation of Solution

Interpretation: The solubility product constant for Ag3PO4<m:math><m:mrow><m:msub><m:mrow><m:mtext>Ag</m:mtext></m:mrow><m:mn>3</m:mn></m:msub><m:msub><m:mrow><m:mtext>PO</m:mtext></m:mrow><m:mtext>4</m:mtext></m:msub></m:mrow></m:math> has to be determined. Concept Introduction: Refer to part (a).

Answer to Problem 14.69QE

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

Interpretation:

The solubility product constant for $CsMnO4$ has to be determined.

Concept Introduction:

Refer to part (a).

Interpretation:

The solubility product constant for $Ag3PO4$ has to be determined.

Concept Introduction:

Refer to part (a).