The solubility in grams per liter of A g B r ( s ) in 0.200 M K B r a n d 0.200 M N H 3 should be calculated. Concept Introduction: Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature. Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity). Consider a general reaction: M n X m (s) ⇌ n M m+ (aq)+m X n+ (aq) The relation between solubility product and molar solubility is as follows: K sp = [ M m+ ] n [ X n- ] m Here The solubility product of salt is K sp . The molar solubility of M m+ ion is [ M m+ ] The molar solubility of X n- ion is [ X n- ]

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Answer 1

Question

Chapter 22, Problem 22.26P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr(s)$ in $0.200 M KBr and0.200 M NH3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

$MnXm(s)⇌nMm+(aq)+mXn+(aq)$

The relation between solubility product and molar solubility is as follows:

$Ksp=[Mm+]n[Xn-]m$

Here

The solubility product of salt is $Ksp$ .

The molar solubility of $Mm+$ ion is $[Mm+]$

The molar solubility of $Xn-$ ion is $[Xn-]$

Expert Solution & Answer

Answer to Problem 22.26P

The solubility in grams per liter of $AgBr(s)$ is given as $4.1×10−4g L−1$ .

Explanation of Solution

The two applicable equations are given below:

$AgBr(s)⇌Ag+(aq)+Br−(aq) Ksp=5.4×10−13 M2 (1)Ag+(aq)+2NH3⇌[Ag(NH3)]2+(aq) Kf=2.0×107 M−2 (2)$

Addition of the both equation will give,

$AgBr(s)+2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq) (3) Kc=Ksp.Kf=5.4×10−13 M2×2.0×107 M−2=1.1×10−5$

Since, $Kc>>Ksp$ , equation (1) can be neglected and (3) can be used to form the concentration table.

$AgBr(s)+ 2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq)Initial 0.200 M 0 M 0.200MChange −2x +x +xEquilibrium 0.200 M−2x x 0.200 M+x$

The expression for equilibrium constant is,

$Kc=x(0.20 M+x)(0.20 M−2x)2=1.1×10−5x(0.20 M+x)(0.20 M)2=1.1×10−5 since x is so smallx=2.2×10−6$

The concentration of $Ag+$ can be calculated as follows:

$Kf=[Ag(NH3)]2+[Ag+][NH3]2=2.0×107 M−2 [Ag+]=[Ag(NH3)]2+2.0×107 M−2[NH3]2=2.2×10−6 M2.0×107 M−2[0.2−(2×2.2×10−6) M]2[Ag+]=2.75×10−12 M$

The total solubility of $AgBr(s)$ can be calculated as follows:

$s=[Ag+]+[Ag(NH3)]2+=2.75×10−12 M+2.2×10−6 Ms=2.2×10−6 M$

Therefore, the solubility of $AgBr(s)$ in grams per liter is given below:

$s= 2.2×10−6 mol.L−1(187.77g1 mol AgBr)s=4.1×10−4g L−1$

The solubility in grams per liter of $AgBr(s)$ is calculated to be $4.1×10−4g L−1$ .

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Answer 2

Question

Chapter 22, Problem 22.26P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr(s)$ in $0.200 M KBr and0.200 M NH3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

$MnXm(s)⇌nMm+(aq)+mXn+(aq)$

The relation between solubility product and molar solubility is as follows:

$Ksp=[Mm+]n[Xn-]m$

Here

The solubility product of salt is $Ksp$ .

The molar solubility of $Mm+$ ion is $[Mm+]$

The molar solubility of $Xn-$ ion is $[Xn-]$

Expert Solution & Answer

Answer to Problem 22.26P

The solubility in grams per liter of $AgBr(s)$ is given as $4.1×10−4g L−1$ .

Explanation of Solution

The two applicable equations are given below:

$AgBr(s)⇌Ag+(aq)+Br−(aq) Ksp=5.4×10−13 M2 (1)Ag+(aq)+2NH3⇌[Ag(NH3)]2+(aq) Kf=2.0×107 M−2 (2)$

Addition of the both equation will give,

$AgBr(s)+2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq) (3) Kc=Ksp.Kf=5.4×10−13 M2×2.0×107 M−2=1.1×10−5$

Since, $Kc>>Ksp$ , equation (1) can be neglected and (3) can be used to form the concentration table.

$AgBr(s)+ 2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq)Initial 0.200 M 0 M 0.200MChange −2x +x +xEquilibrium 0.200 M−2x x 0.200 M+x$

The expression for equilibrium constant is,

$Kc=x(0.20 M+x)(0.20 M−2x)2=1.1×10−5x(0.20 M+x)(0.20 M)2=1.1×10−5 since x is so smallx=2.2×10−6$

The concentration of $Ag+$ can be calculated as follows:

$Kf=[Ag(NH3)]2+[Ag+][NH3]2=2.0×107 M−2 [Ag+]=[Ag(NH3)]2+2.0×107 M−2[NH3]2=2.2×10−6 M2.0×107 M−2[0.2−(2×2.2×10−6) M]2[Ag+]=2.75×10−12 M$

The total solubility of $AgBr(s)$ can be calculated as follows:

$s=[Ag+]+[Ag(NH3)]2+=2.75×10−12 M+2.2×10−6 Ms=2.2×10−6 M$

Therefore, the solubility of $AgBr(s)$ in grams per liter is given below:

$s= 2.2×10−6 mol.L−1(187.77g1 mol AgBr)s=4.1×10−4g L−1$

The solubility in grams per liter of $AgBr(s)$ is calculated to be $4.1×10−4g L−1$ .

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Answer 3

Question

Chapter 22, Problem 22.26P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr(s)$ in $0.200 M KBr and0.200 M NH3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

$MnXm(s)⇌nMm+(aq)+mXn+(aq)$

The relation between solubility product and molar solubility is as follows:

$Ksp=[Mm+]n[Xn-]m$

Here

The solubility product of salt is $Ksp$ .

The molar solubility of $Mm+$ ion is $[Mm+]$

The molar solubility of $Xn-$ ion is $[Xn-]$

Expert Solution & Answer

Answer to Problem 22.26P

The solubility in grams per liter of $AgBr(s)$ is given as $4.1×10−4g L−1$ .

Explanation of Solution

The two applicable equations are given below:

$AgBr(s)⇌Ag+(aq)+Br−(aq) Ksp=5.4×10−13 M2 (1)Ag+(aq)+2NH3⇌[Ag(NH3)]2+(aq) Kf=2.0×107 M−2 (2)$

Addition of the both equation will give,

$AgBr(s)+2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq) (3) Kc=Ksp.Kf=5.4×10−13 M2×2.0×107 M−2=1.1×10−5$

Since, $Kc>>Ksp$ , equation (1) can be neglected and (3) can be used to form the concentration table.

$AgBr(s)+ 2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq)Initial 0.200 M 0 M 0.200MChange −2x +x +xEquilibrium 0.200 M−2x x 0.200 M+x$

The expression for equilibrium constant is,

$Kc=x(0.20 M+x)(0.20 M−2x)2=1.1×10−5x(0.20 M+x)(0.20 M)2=1.1×10−5 since x is so smallx=2.2×10−6$

The concentration of $Ag+$ can be calculated as follows:

$Kf=[Ag(NH3)]2+[Ag+][NH3]2=2.0×107 M−2 [Ag+]=[Ag(NH3)]2+2.0×107 M−2[NH3]2=2.2×10−6 M2.0×107 M−2[0.2−(2×2.2×10−6) M]2[Ag+]=2.75×10−12 M$

The total solubility of $AgBr(s)$ can be calculated as follows:

$s=[Ag+]+[Ag(NH3)]2+=2.75×10−12 M+2.2×10−6 Ms=2.2×10−6 M$

Therefore, the solubility of $AgBr(s)$ in grams per liter is given below:

$s= 2.2×10−6 mol.L−1(187.77g1 mol AgBr)s=4.1×10−4g L−1$

The solubility in grams per liter of $AgBr(s)$ is calculated to be $4.1×10−4g L−1$ .

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Answer 4

Question

Chapter 22, Problem 22.26P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr(s)$ in $0.200 M KBr and0.200 M NH3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

$MnXm(s)⇌nMm+(aq)+mXn+(aq)$

The relation between solubility product and molar solubility is as follows:

$Ksp=[Mm+]n[Xn-]m$

Here

The solubility product of salt is $Ksp$ .

The molar solubility of $Mm+$ ion is $[Mm+]$

The molar solubility of $Xn-$ ion is $[Xn-]$

Expert Solution & Answer

Answer to Problem 22.26P

The solubility in grams per liter of $AgBr(s)$ is given as $4.1×10−4g L−1$ .

Explanation of Solution

The two applicable equations are given below:

$AgBr(s)⇌Ag+(aq)+Br−(aq) Ksp=5.4×10−13 M2 (1)Ag+(aq)+2NH3⇌[Ag(NH3)]2+(aq) Kf=2.0×107 M−2 (2)$

Addition of the both equation will give,

$AgBr(s)+2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq) (3) Kc=Ksp.Kf=5.4×10−13 M2×2.0×107 M−2=1.1×10−5$

Since, $Kc>>Ksp$ , equation (1) can be neglected and (3) can be used to form the concentration table.

$AgBr(s)+ 2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq)Initial 0.200 M 0 M 0.200MChange −2x +x +xEquilibrium 0.200 M−2x x 0.200 M+x$

The expression for equilibrium constant is,

$Kc=x(0.20 M+x)(0.20 M−2x)2=1.1×10−5x(0.20 M+x)(0.20 M)2=1.1×10−5 since x is so smallx=2.2×10−6$

The concentration of $Ag+$ can be calculated as follows:

$Kf=[Ag(NH3)]2+[Ag+][NH3]2=2.0×107 M−2 [Ag+]=[Ag(NH3)]2+2.0×107 M−2[NH3]2=2.2×10−6 M2.0×107 M−2[0.2−(2×2.2×10−6) M]2[Ag+]=2.75×10−12 M$

The total solubility of $AgBr(s)$ can be calculated as follows:

$s=[Ag+]+[Ag(NH3)]2+=2.75×10−12 M+2.2×10−6 Ms=2.2×10−6 M$

Therefore, the solubility of $AgBr(s)$ in grams per liter is given below:

$s= 2.2×10−6 mol.L−1(187.77g1 mol AgBr)s=4.1×10−4g L−1$

The solubility in grams per liter of $AgBr(s)$ is calculated to be $4.1×10−4g L−1$ .

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Answer 5

Chapter 22, Problem 22.26P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr(s)$ in $0.200 M KBr and0.200 M NH3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

$MnXm(s)⇌nMm+(aq)+mXn+(aq)$

The relation between solubility product and molar solubility is as follows:

$Ksp=[Mm+]n[Xn-]m$

Here

The solubility product of salt is $Ksp$ .

The molar solubility of $Mm+$ ion is $[Mm+]$

The molar solubility of $Xn-$ ion is $[Xn-]$

Expert Solution & Answer

Answer to Problem 22.26P

The solubility in grams per liter of $AgBr(s)$ is given as $4.1×10−4g L−1$ .

Explanation of Solution

The two applicable equations are given below:

$AgBr(s)⇌Ag+(aq)+Br−(aq) Ksp=5.4×10−13 M2 (1)Ag+(aq)+2NH3⇌[Ag(NH3)]2+(aq) Kf=2.0×107 M−2 (2)$

Addition of the both equation will give,

$AgBr(s)+2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq) (3) Kc=Ksp.Kf=5.4×10−13 M2×2.0×107 M−2=1.1×10−5$

Since, $Kc>>Ksp$ , equation (1) can be neglected and (3) can be used to form the concentration table.

$AgBr(s)+ 2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq)Initial 0.200 M 0 M 0.200MChange −2x +x +xEquilibrium 0.200 M−2x x 0.200 M+x$

The expression for equilibrium constant is,

$Kc=x(0.20 M+x)(0.20 M−2x)2=1.1×10−5x(0.20 M+x)(0.20 M)2=1.1×10−5 since x is so smallx=2.2×10−6$

The concentration of $Ag+$ can be calculated as follows:

$Kf=[Ag(NH3)]2+[Ag+][NH3]2=2.0×107 M−2 [Ag+]=[Ag(NH3)]2+2.0×107 M−2[NH3]2=2.2×10−6 M2.0×107 M−2[0.2−(2×2.2×10−6) M]2[Ag+]=2.75×10−12 M$

The total solubility of $AgBr(s)$ can be calculated as follows:

$s=[Ag+]+[Ag(NH3)]2+=2.75×10−12 M+2.2×10−6 Ms=2.2×10−6 M$

Therefore, the solubility of $AgBr(s)$ in grams per liter is given below:

$s= 2.2×10−6 mol.L−1(187.77g1 mol AgBr)s=4.1×10−4g L−1$

The solubility in grams per liter of $AgBr(s)$ is calculated to be $4.1×10−4g L−1$ .

Answer 6

Chapter 22, Problem 22.26P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr(s)$ in $0.200 M KBr and0.200 M NH3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

$MnXm(s)⇌nMm+(aq)+mXn+(aq)$

The relation between solubility product and molar solubility is as follows:

$Ksp=[Mm+]n[Xn-]m$

Here

The solubility product of salt is $Ksp$ .

The molar solubility of $Mm+$ ion is $[Mm+]$

The molar solubility of $Xn-$ ion is $[Xn-]$

Expert Solution & Answer

Answer to Problem 22.26P

The solubility in grams per liter of $AgBr(s)$ is given as $4.1×10−4g L−1$ .

Explanation of Solution

The two applicable equations are given below:

$AgBr(s)⇌Ag+(aq)+Br−(aq) Ksp=5.4×10−13 M2 (1)Ag+(aq)+2NH3⇌[Ag(NH3)]2+(aq) Kf=2.0×107 M−2 (2)$

Addition of the both equation will give,

$AgBr(s)+2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq) (3) Kc=Ksp.Kf=5.4×10−13 M2×2.0×107 M−2=1.1×10−5$

Since, $Kc>>Ksp$ , equation (1) can be neglected and (3) can be used to form the concentration table.

$AgBr(s)+ 2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq)Initial 0.200 M 0 M 0.200MChange −2x +x +xEquilibrium 0.200 M−2x x 0.200 M+x$

The expression for equilibrium constant is,

$Kc=x(0.20 M+x)(0.20 M−2x)2=1.1×10−5x(0.20 M+x)(0.20 M)2=1.1×10−5 since x is so smallx=2.2×10−6$

The concentration of $Ag+$ can be calculated as follows:

$Kf=[Ag(NH3)]2+[Ag+][NH3]2=2.0×107 M−2 [Ag+]=[Ag(NH3)]2+2.0×107 M−2[NH3]2=2.2×10−6 M2.0×107 M−2[0.2−(2×2.2×10−6) M]2[Ag+]=2.75×10−12 M$

The total solubility of $AgBr(s)$ can be calculated as follows:

$s=[Ag+]+[Ag(NH3)]2+=2.75×10−12 M+2.2×10−6 Ms=2.2×10−6 M$

Therefore, the solubility of $AgBr(s)$ in grams per liter is given below:

$s= 2.2×10−6 mol.L−1(187.77g1 mol AgBr)s=4.1×10−4g L−1$

The solubility in grams per liter of $AgBr(s)$ is calculated to be $4.1×10−4g L−1$ .

Answer 7

Chapter 22, Problem 22.26P

Interpretation Introduction

Interpretation:

The solubility in grams per liter of $AgBr(s)$ in $0.200 M KBr and0.200 M NH3$ should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

$MnXm(s)⇌nMm+(aq)+mXn+(aq)$

The relation between solubility product and molar solubility is as follows:

$Ksp=[Mm+]n[Xn-]m$

Here

The solubility product of salt is $Ksp$ .

The molar solubility of $Mm+$ ion is $[Mm+]$

The molar solubility of $Xn-$ ion is $[Xn-]$

Answer 8

Expert Solution & Answer

Answer to Problem 22.26P

The solubility in grams per liter of $AgBr(s)$ is given as $4.1×10−4g L−1$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The two applicable equations are given below:

$AgBr(s)⇌Ag+(aq)+Br−(aq) Ksp=5.4×10−13 M2 (1)Ag+(aq)+2NH3⇌[Ag(NH3)]2+(aq) Kf=2.0×107 M−2 (2)$

Addition of the both equation will give,

$AgBr(s)+2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq) (3) Kc=Ksp.Kf=5.4×10−13 M2×2.0×107 M−2=1.1×10−5$

Since, $Kc>>Ksp$ , equation (1) can be neglected and (3) can be used to form the concentration table.

$AgBr(s)+ 2NH3⇌[Ag(NH3)]2+(aq)+Br−(aq)Initial 0.200 M 0 M 0.200MChange −2x +x +xEquilibrium 0.200 M−2x x 0.200 M+x$

The expression for equilibrium constant is,

$Kc=x(0.20 M+x)(0.20 M−2x)2=1.1×10−5x(0.20 M+x)(0.20 M)2=1.1×10−5 since x is so smallx=2.2×10−6$

The concentration of $Ag+$ can be calculated as follows:

$Kf=[Ag(NH3)]2+[Ag+][NH3]2=2.0×107 M−2 [Ag+]=[Ag(NH3)]2+2.0×107 M−2[NH3]2=2.2×10−6 M2.0×107 M−2[0.2−(2×2.2×10−6) M]2[Ag+]=2.75×10−12 M$

The total solubility of $AgBr(s)$ can be calculated as follows:

$s=[Ag+]+[Ag(NH3)]2+=2.75×10−12 M+2.2×10−6 Ms=2.2×10−6 M$

Therefore, the solubility of $AgBr(s)$ in grams per liter is given below:

$s= 2.2×10−6 mol.L−1(187.77g1 mol AgBr)s=4.1×10−4g L−1$

The solubility in grams per liter of $AgBr(s)$ is calculated to be $4.1×10−4g L−1$ .

Answer 12

Ch. 22 - Prob. 22.1P Ch. 22 - Prob. 22.2P Ch. 22 - Prob. 22.3P Ch. 22 - Prob. 22.4P Ch. 22 - Prob. 22.5P Ch. 22 - Prob. 22.6P Ch. 22 - Prob. 22.7P Ch. 22 - Prob. 22.8P Ch. 22 - Prob. 22.9P Ch. 22 - Prob. 22.10P

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