Calculate the pH of each of the following solutions. a. 0.10 M CH 3 NH 3 Cl b. 0.050 M NaCN

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Answer 1

Textbook Question

Chapter 14, Problem 119E

Calculate the pH of each of the following solutions.

a. 0.10 M CH₃NH₃Cl

b. 0.050 M NaCN

(a)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.10 M$ $CH3NH3Cl$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.10 M$ $CH3NH3Cl$ is $5.82_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $Ka=[CH3NH2][H+][CH3NH3+]$

$CH3NH3+$ is a stronger acid than $H2O$ .

The dominant equilibrium reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

Where,

$Ka$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$Ka=[CH3NH2][H+][CH3NH3+]$ (1)

The $Ka$ value is $2.28×10-11 M_$ .

The value, $Kw=Ka⋅Kb$

The value of $Kb$ for methyl amine is $4.38×10−4$ .

The value of $Ka$ is calculated by the formula,

$Ka=KwKb$

Substitute the value of $Kb$ in the above expression.

$Ka=1.0×10−144.38×10−4=2.28×10-11_$

The $[H+]$ is $1.50×10-6 M_$ .

The change in concentration of $CH3NH3+$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)Inititial concentration0.100Change−x+x+xEquilibrium concentration0.1−xxx$

The equilibrium concentration of $[CH3NH3+]$ is $(0.1−x) M$ .

The equilibrium concentration of $[CH3NH2]$ is $x M$ .

The equilibrium concentration of $[H+]$ is $x M$ .

The calculated value of $Ka$ is $2.28×10−11$ .

Substitute the value of $Ka$ , $[CH3NH3+]$ , $[CH3NH2]$ and $[H+]$ in equation (1).

$2.28×10−11=[x][x][0.1−x]2.28×10−11=[x]2[0.1−x]$

Solve the above expression.

$2.28×10−11=[x]2[0.1−x][x]=1.50×10-6 M_$

Therefore, the $[H+]$ is $1.50×10-6 M_$ .

The $pH$ of the given solution is $5.82_$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above expression.

$pH=−log[1.50×10−6]=5.82_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.050 M$ $NaCN$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.050 M$ $NaCN$ is $10.945_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$

The dominant equilibrium reaction is,

$CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Kb=Concentration of productsConcentration of reactants$

Where,

$Kb$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$ (1)

The $Kb$ value is $1.61×10-5_$ .

The value, $Kw=Ka⋅Kb$

The value of $Ka$ for $HCN$ is $6.2×10−10$ .

The value of $Kb$ is calculated by the formula,

$Kb=KwKa$

Substitute the value of $Kb$ in the above expression.

$Kb=1.0×10−146.2×10−10=1.61×10-5_$

The $[OH−]$ is $8.8×10-4 M_$ .

The change in concentration of $CN−$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CN−(aq)⇌HCN(aq)+OH−(aq)Inititial concentration0.0500Change−x+x+xEquilibrium concentration0.05−xxx$

The equilibrium concentration of $[CN−]$ is $(0.05−x) M$ .

The equilibrium concentration of $[HCN]$ is $x M$ .

The equilibrium concentration of $[OH−]$ is $x M$ .

The calculated value of $Kb$ is $1.61×10−5$ .

Substitute the value of $Kb$ , $[CN−]$ , $[CN−]$ and $[OH−]$ in equation (1).

$1.61×10−5=[x][x][0.05−x]1.61×10−5=[x]2[0.05−x]$

Solve the above expression.

$1.61×10−5=[x]2[0.05−x][x]=8.8×10-4 M_$

Therefore, the $[OH−]$ is $8.8×10-4 M_$ .

The $pOH$ of the given solution is $3.055_$ .

The $pOH$ of a solution is calculated by the formula,

$pOH=−log[OH−]$

Substitute the value of $[OH−]$ in the above expression.

$pOH=−log[8.8×10−4]=3.055_$

The $pH$ of the given solution is $10.945_$ .

The sum, $pH+pOH=14$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$

Substitute the calculated value of $pOH$ in the above expression.

$pH=14−3.055=10.945_$

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Answer 2

Textbook Question

Chapter 14, Problem 119E

Calculate the pH of each of the following solutions.

a. 0.10 M CH₃NH₃Cl

b. 0.050 M NaCN

(a)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.10 M$ $CH3NH3Cl$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.10 M$ $CH3NH3Cl$ is $5.82_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $Ka=[CH3NH2][H+][CH3NH3+]$

$CH3NH3+$ is a stronger acid than $H2O$ .

The dominant equilibrium reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

Where,

$Ka$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$Ka=[CH3NH2][H+][CH3NH3+]$ (1)

The $Ka$ value is $2.28×10-11 M_$ .

The value, $Kw=Ka⋅Kb$

The value of $Kb$ for methyl amine is $4.38×10−4$ .

The value of $Ka$ is calculated by the formula,

$Ka=KwKb$

Substitute the value of $Kb$ in the above expression.

$Ka=1.0×10−144.38×10−4=2.28×10-11_$

The $[H+]$ is $1.50×10-6 M_$ .

The change in concentration of $CH3NH3+$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)Inititial concentration0.100Change−x+x+xEquilibrium concentration0.1−xxx$

The equilibrium concentration of $[CH3NH3+]$ is $(0.1−x) M$ .

The equilibrium concentration of $[CH3NH2]$ is $x M$ .

The equilibrium concentration of $[H+]$ is $x M$ .

The calculated value of $Ka$ is $2.28×10−11$ .

Substitute the value of $Ka$ , $[CH3NH3+]$ , $[CH3NH2]$ and $[H+]$ in equation (1).

$2.28×10−11=[x][x][0.1−x]2.28×10−11=[x]2[0.1−x]$

Solve the above expression.

$2.28×10−11=[x]2[0.1−x][x]=1.50×10-6 M_$

Therefore, the $[H+]$ is $1.50×10-6 M_$ .

The $pH$ of the given solution is $5.82_$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above expression.

$pH=−log[1.50×10−6]=5.82_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.050 M$ $NaCN$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.050 M$ $NaCN$ is $10.945_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$

The dominant equilibrium reaction is,

$CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Kb=Concentration of productsConcentration of reactants$

Where,

$Kb$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$ (1)

The $Kb$ value is $1.61×10-5_$ .

The value, $Kw=Ka⋅Kb$

The value of $Ka$ for $HCN$ is $6.2×10−10$ .

The value of $Kb$ is calculated by the formula,

$Kb=KwKa$

Substitute the value of $Kb$ in the above expression.

$Kb=1.0×10−146.2×10−10=1.61×10-5_$

The $[OH−]$ is $8.8×10-4 M_$ .

The change in concentration of $CN−$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CN−(aq)⇌HCN(aq)+OH−(aq)Inititial concentration0.0500Change−x+x+xEquilibrium concentration0.05−xxx$

The equilibrium concentration of $[CN−]$ is $(0.05−x) M$ .

The equilibrium concentration of $[HCN]$ is $x M$ .

The equilibrium concentration of $[OH−]$ is $x M$ .

The calculated value of $Kb$ is $1.61×10−5$ .

Substitute the value of $Kb$ , $[CN−]$ , $[CN−]$ and $[OH−]$ in equation (1).

$1.61×10−5=[x][x][0.05−x]1.61×10−5=[x]2[0.05−x]$

Solve the above expression.

$1.61×10−5=[x]2[0.05−x][x]=8.8×10-4 M_$

Therefore, the $[OH−]$ is $8.8×10-4 M_$ .

The $pOH$ of the given solution is $3.055_$ .

The $pOH$ of a solution is calculated by the formula,

$pOH=−log[OH−]$

Substitute the value of $[OH−]$ in the above expression.

$pOH=−log[8.8×10−4]=3.055_$

The $pH$ of the given solution is $10.945_$ .

The sum, $pH+pOH=14$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$

Substitute the calculated value of $pOH$ in the above expression.

$pH=14−3.055=10.945_$

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Answer 3

Textbook Question

Chapter 14, Problem 119E

Calculate the pH of each of the following solutions.

a. 0.10 M CH₃NH₃Cl

b. 0.050 M NaCN

(a)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.10 M$ $CH3NH3Cl$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.10 M$ $CH3NH3Cl$ is $5.82_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $Ka=[CH3NH2][H+][CH3NH3+]$

$CH3NH3+$ is a stronger acid than $H2O$ .

The dominant equilibrium reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

Where,

$Ka$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$Ka=[CH3NH2][H+][CH3NH3+]$ (1)

The $Ka$ value is $2.28×10-11 M_$ .

The value, $Kw=Ka⋅Kb$

The value of $Kb$ for methyl amine is $4.38×10−4$ .

The value of $Ka$ is calculated by the formula,

$Ka=KwKb$

Substitute the value of $Kb$ in the above expression.

$Ka=1.0×10−144.38×10−4=2.28×10-11_$

The $[H+]$ is $1.50×10-6 M_$ .

The change in concentration of $CH3NH3+$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)Inititial concentration0.100Change−x+x+xEquilibrium concentration0.1−xxx$

The equilibrium concentration of $[CH3NH3+]$ is $(0.1−x) M$ .

The equilibrium concentration of $[CH3NH2]$ is $x M$ .

The equilibrium concentration of $[H+]$ is $x M$ .

The calculated value of $Ka$ is $2.28×10−11$ .

Substitute the value of $Ka$ , $[CH3NH3+]$ , $[CH3NH2]$ and $[H+]$ in equation (1).

$2.28×10−11=[x][x][0.1−x]2.28×10−11=[x]2[0.1−x]$

Solve the above expression.

$2.28×10−11=[x]2[0.1−x][x]=1.50×10-6 M_$

Therefore, the $[H+]$ is $1.50×10-6 M_$ .

The $pH$ of the given solution is $5.82_$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above expression.

$pH=−log[1.50×10−6]=5.82_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.050 M$ $NaCN$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.050 M$ $NaCN$ is $10.945_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$

The dominant equilibrium reaction is,

$CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Kb=Concentration of productsConcentration of reactants$

Where,

$Kb$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$ (1)

The $Kb$ value is $1.61×10-5_$ .

The value, $Kw=Ka⋅Kb$

The value of $Ka$ for $HCN$ is $6.2×10−10$ .

The value of $Kb$ is calculated by the formula,

$Kb=KwKa$

Substitute the value of $Kb$ in the above expression.

$Kb=1.0×10−146.2×10−10=1.61×10-5_$

The $[OH−]$ is $8.8×10-4 M_$ .

The change in concentration of $CN−$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CN−(aq)⇌HCN(aq)+OH−(aq)Inititial concentration0.0500Change−x+x+xEquilibrium concentration0.05−xxx$

The equilibrium concentration of $[CN−]$ is $(0.05−x) M$ .

The equilibrium concentration of $[HCN]$ is $x M$ .

The equilibrium concentration of $[OH−]$ is $x M$ .

The calculated value of $Kb$ is $1.61×10−5$ .

Substitute the value of $Kb$ , $[CN−]$ , $[CN−]$ and $[OH−]$ in equation (1).

$1.61×10−5=[x][x][0.05−x]1.61×10−5=[x]2[0.05−x]$

Solve the above expression.

$1.61×10−5=[x]2[0.05−x][x]=8.8×10-4 M_$

Therefore, the $[OH−]$ is $8.8×10-4 M_$ .

The $pOH$ of the given solution is $3.055_$ .

The $pOH$ of a solution is calculated by the formula,

$pOH=−log[OH−]$

Substitute the value of $[OH−]$ in the above expression.

$pOH=−log[8.8×10−4]=3.055_$

The $pH$ of the given solution is $10.945_$ .

The sum, $pH+pOH=14$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$

Substitute the calculated value of $pOH$ in the above expression.

$pH=14−3.055=10.945_$

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Answer 4

Textbook Question

Chapter 14, Problem 119E

Calculate the pH of each of the following solutions.

a. 0.10 M CH₃NH₃Cl

b. 0.050 M NaCN

(a)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.10 M$ $CH3NH3Cl$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.10 M$ $CH3NH3Cl$ is $5.82_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $Ka=[CH3NH2][H+][CH3NH3+]$

$CH3NH3+$ is a stronger acid than $H2O$ .

The dominant equilibrium reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

Where,

$Ka$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$Ka=[CH3NH2][H+][CH3NH3+]$ (1)

The $Ka$ value is $2.28×10-11 M_$ .

The value, $Kw=Ka⋅Kb$

The value of $Kb$ for methyl amine is $4.38×10−4$ .

The value of $Ka$ is calculated by the formula,

$Ka=KwKb$

Substitute the value of $Kb$ in the above expression.

$Ka=1.0×10−144.38×10−4=2.28×10-11_$

The $[H+]$ is $1.50×10-6 M_$ .

The change in concentration of $CH3NH3+$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)Inititial concentration0.100Change−x+x+xEquilibrium concentration0.1−xxx$

The equilibrium concentration of $[CH3NH3+]$ is $(0.1−x) M$ .

The equilibrium concentration of $[CH3NH2]$ is $x M$ .

The equilibrium concentration of $[H+]$ is $x M$ .

The calculated value of $Ka$ is $2.28×10−11$ .

Substitute the value of $Ka$ , $[CH3NH3+]$ , $[CH3NH2]$ and $[H+]$ in equation (1).

$2.28×10−11=[x][x][0.1−x]2.28×10−11=[x]2[0.1−x]$

Solve the above expression.

$2.28×10−11=[x]2[0.1−x][x]=1.50×10-6 M_$

Therefore, the $[H+]$ is $1.50×10-6 M_$ .

The $pH$ of the given solution is $5.82_$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above expression.

$pH=−log[1.50×10−6]=5.82_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.050 M$ $NaCN$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.050 M$ $NaCN$ is $10.945_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$

The dominant equilibrium reaction is,

$CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Kb=Concentration of productsConcentration of reactants$

Where,

$Kb$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$ (1)

The $Kb$ value is $1.61×10-5_$ .

The value, $Kw=Ka⋅Kb$

The value of $Ka$ for $HCN$ is $6.2×10−10$ .

The value of $Kb$ is calculated by the formula,

$Kb=KwKa$

Substitute the value of $Kb$ in the above expression.

$Kb=1.0×10−146.2×10−10=1.61×10-5_$

The $[OH−]$ is $8.8×10-4 M_$ .

The change in concentration of $CN−$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CN−(aq)⇌HCN(aq)+OH−(aq)Inititial concentration0.0500Change−x+x+xEquilibrium concentration0.05−xxx$

The equilibrium concentration of $[CN−]$ is $(0.05−x) M$ .

The equilibrium concentration of $[HCN]$ is $x M$ .

The equilibrium concentration of $[OH−]$ is $x M$ .

The calculated value of $Kb$ is $1.61×10−5$ .

Substitute the value of $Kb$ , $[CN−]$ , $[CN−]$ and $[OH−]$ in equation (1).

$1.61×10−5=[x][x][0.05−x]1.61×10−5=[x]2[0.05−x]$

Solve the above expression.

$1.61×10−5=[x]2[0.05−x][x]=8.8×10-4 M_$

Therefore, the $[OH−]$ is $8.8×10-4 M_$ .

The $pOH$ of the given solution is $3.055_$ .

The $pOH$ of a solution is calculated by the formula,

$pOH=−log[OH−]$

Substitute the value of $[OH−]$ in the above expression.

$pOH=−log[8.8×10−4]=3.055_$

The $pH$ of the given solution is $10.945_$ .

The sum, $pH+pOH=14$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$

Substitute the calculated value of $pOH$ in the above expression.

$pH=14−3.055=10.945_$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.10 M$ $CH3NH3Cl$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.10 M$ $CH3NH3Cl$ is $5.82_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $Ka=[CH3NH2][H+][CH3NH3+]$

$CH3NH3+$ is a stronger acid than $H2O$ .

The dominant equilibrium reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

Where,

$Ka$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$Ka=[CH3NH2][H+][CH3NH3+]$ (1)

The $Ka$ value is $2.28×10-11 M_$ .

The value, $Kw=Ka⋅Kb$

The value of $Kb$ for methyl amine is $4.38×10−4$ .

The value of $Ka$ is calculated by the formula,

$Ka=KwKb$

Substitute the value of $Kb$ in the above expression.

$Ka=1.0×10−144.38×10−4=2.28×10-11_$

The $[H+]$ is $1.50×10-6 M_$ .

The change in concentration of $CH3NH3+$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)Inititial concentration0.100Change−x+x+xEquilibrium concentration0.1−xxx$

The equilibrium concentration of $[CH3NH3+]$ is $(0.1−x) M$ .

The equilibrium concentration of $[CH3NH2]$ is $x M$ .

The equilibrium concentration of $[H+]$ is $x M$ .

The calculated value of $Ka$ is $2.28×10−11$ .

Substitute the value of $Ka$ , $[CH3NH3+]$ , $[CH3NH2]$ and $[H+]$ in equation (1).

$2.28×10−11=[x][x][0.1−x]2.28×10−11=[x]2[0.1−x]$

Solve the above expression.

$2.28×10−11=[x]2[0.1−x][x]=1.50×10-6 M_$

Therefore, the $[H+]$ is $1.50×10-6 M_$ .

The $pH$ of the given solution is $5.82_$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above expression.

$pH=−log[1.50×10−6]=5.82_$

(b)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.050 M$ $NaCN$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.050 M$ $NaCN$ is $10.945_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$

The dominant equilibrium reaction is,

$CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Kb=Concentration of productsConcentration of reactants$

Where,

$Kb$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$ (1)

The $Kb$ value is $1.61×10-5_$ .

The value, $Kw=Ka⋅Kb$

The value of $Ka$ for $HCN$ is $6.2×10−10$ .

The value of $Kb$ is calculated by the formula,

$Kb=KwKa$

Substitute the value of $Kb$ in the above expression.

$Kb=1.0×10−146.2×10−10=1.61×10-5_$

The $[OH−]$ is $8.8×10-4 M_$ .

The change in concentration of $CN−$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CN−(aq)⇌HCN(aq)+OH−(aq)Inititial concentration0.0500Change−x+x+xEquilibrium concentration0.05−xxx$

The equilibrium concentration of $[CN−]$ is $(0.05−x) M$ .

The equilibrium concentration of $[HCN]$ is $x M$ .

The equilibrium concentration of $[OH−]$ is $x M$ .

The calculated value of $Kb$ is $1.61×10−5$ .

Substitute the value of $Kb$ , $[CN−]$ , $[CN−]$ and $[OH−]$ in equation (1).

$1.61×10−5=[x][x][0.05−x]1.61×10−5=[x]2[0.05−x]$

Solve the above expression.

$1.61×10−5=[x]2[0.05−x][x]=8.8×10-4 M_$

Therefore, the $[OH−]$ is $8.8×10-4 M_$ .

The $pOH$ of the given solution is $3.055_$ .

The $pOH$ of a solution is calculated by the formula,

$pOH=−log[OH−]$

Substitute the value of $[OH−]$ in the above expression.

$pOH=−log[8.8×10−4]=3.055_$

The $pH$ of the given solution is $10.945_$ .

The sum, $pH+pOH=14$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$

Substitute the calculated value of $pOH$ in the above expression.

$pH=14−3.055=10.945_$

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.10 M$ $CH3NH3Cl$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.10 M$ $CH3NH3Cl$ is $5.82_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $Ka=[CH3NH2][H+][CH3NH3+]$

$CH3NH3+$ is a stronger acid than $H2O$ .

The dominant equilibrium reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

Where,

$Ka$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$Ka=[CH3NH2][H+][CH3NH3+]$ (1)

The $Ka$ value is $2.28×10-11 M_$ .

The value, $Kw=Ka⋅Kb$

The value of $Kb$ for methyl amine is $4.38×10−4$ .

The value of $Ka$ is calculated by the formula,

$Ka=KwKb$

Substitute the value of $Kb$ in the above expression.

$Ka=1.0×10−144.38×10−4=2.28×10-11_$

The $[H+]$ is $1.50×10-6 M_$ .

The change in concentration of $CH3NH3+$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)Inititial concentration0.100Change−x+x+xEquilibrium concentration0.1−xxx$

The equilibrium concentration of $[CH3NH3+]$ is $(0.1−x) M$ .

The equilibrium concentration of $[CH3NH2]$ is $x M$ .

The equilibrium concentration of $[H+]$ is $x M$ .

The calculated value of $Ka$ is $2.28×10−11$ .

Substitute the value of $Ka$ , $[CH3NH3+]$ , $[CH3NH2]$ and $[H+]$ in equation (1).

$2.28×10−11=[x][x][0.1−x]2.28×10−11=[x]2[0.1−x]$

Solve the above expression.

$2.28×10−11=[x]2[0.1−x][x]=1.50×10-6 M_$

Therefore, the $[H+]$ is $1.50×10-6 M_$ .

The $pH$ of the given solution is $5.82_$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above expression.

$pH=−log[1.50×10−6]=5.82_$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.10 M$ $CH3NH3Cl$ .

Answer 13

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.10 M$ $CH3NH3Cl$ is $5.82_$ .

Answer 14

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is, $Ka=[CH3NH2][H+][CH3NH3+]$

$CH3NH3+$ is a stronger acid than $H2O$ .

The dominant equilibrium reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

Where,

$Ka$ is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

$Ka=[CH3NH2][H+][CH3NH3+]$ (1)

The $Ka$ value is $2.28×10-11 M_$ .

The value, $Kw=Ka⋅Kb$

The value of $Kb$ for methyl amine is $4.38×10−4$ .

The value of $Ka$ is calculated by the formula,

$Ka=KwKb$

Substitute the value of $Kb$ in the above expression.

$Ka=1.0×10−144.38×10−4=2.28×10-11_$

The $[H+]$ is $1.50×10-6 M_$ .

The change in concentration of $CH3NH3+$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CH3NH3+(aq)⇌CH3NH2(aq)+H+(aq)Inititial concentration0.100Change−x+x+xEquilibrium concentration0.1−xxx$

The equilibrium concentration of $[CH3NH3+]$ is $(0.1−x) M$ .

The equilibrium concentration of $[CH3NH2]$ is $x M$ .

The equilibrium concentration of $[H+]$ is $x M$ .

The calculated value of $Ka$ is $2.28×10−11$ .

Substitute the value of $Ka$ , $[CH3NH3+]$ , $[CH3NH2]$ and $[H+]$ in equation (1).

$2.28×10−11=[x][x][0.1−x]2.28×10−11=[x]2[0.1−x]$

Solve the above expression.

$2.28×10−11=[x]2[0.1−x][x]=1.50×10-6 M_$

Therefore, the $[H+]$ is $1.50×10-6 M_$ .

The $pH$ of the given solution is $5.82_$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above expression.

$pH=−log[1.50×10−6]=5.82_$

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.050 M$ $NaCN$ .

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.050 M$ $NaCN$ is $10.945_$ .

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$

The dominant equilibrium reaction is,

$CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Kb=Concentration of productsConcentration of reactants$

Where,

$Kb$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$ (1)

The $Kb$ value is $1.61×10-5_$ .

The value, $Kw=Ka⋅Kb$

The value of $Ka$ for $HCN$ is $6.2×10−10$ .

The value of $Kb$ is calculated by the formula,

$Kb=KwKa$

Substitute the value of $Kb$ in the above expression.

$Kb=1.0×10−146.2×10−10=1.61×10-5_$

The $[OH−]$ is $8.8×10-4 M_$ .

The change in concentration of $CN−$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CN−(aq)⇌HCN(aq)+OH−(aq)Inititial concentration0.0500Change−x+x+xEquilibrium concentration0.05−xxx$

The equilibrium concentration of $[CN−]$ is $(0.05−x) M$ .

The equilibrium concentration of $[HCN]$ is $x M$ .

The equilibrium concentration of $[OH−]$ is $x M$ .

The calculated value of $Kb$ is $1.61×10−5$ .

Substitute the value of $Kb$ , $[CN−]$ , $[CN−]$ and $[OH−]$ in equation (1).

$1.61×10−5=[x][x][0.05−x]1.61×10−5=[x]2[0.05−x]$

Solve the above expression.

$1.61×10−5=[x]2[0.05−x][x]=8.8×10-4 M_$

Therefore, the $[OH−]$ is $8.8×10-4 M_$ .

The $pOH$ of the given solution is $3.055_$ .

The $pOH$ of a solution is calculated by the formula,

$pOH=−log[OH−]$

Substitute the value of $[OH−]$ in the above expression.

$pOH=−log[8.8×10−4]=3.055_$

The $pH$ of the given solution is $10.945_$ .

The sum, $pH+pOH=14$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$

Substitute the calculated value of $pOH$ in the above expression.

$pH=14−3.055=10.945_$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation: The $pH$ value for each of the given solutions to be calculated.

Concept introduction: The $pH$ of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.

The $pOH$ of a solution is calculated by the formula, $pOH=−log[OH−]$

The sum, $pH+pOH=14$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Ka=Concentration of productsConcentration of reactants$

The value of $Kw$ is calculated by the formula,

$Kw=Ka⋅Kb$

To determine: The $pH$ value for each of the given solution of $0.050 M$ $NaCN$ .

Answer 20

Answer to Problem 119E

Answer

The $pH$ of the given solution of $0.050 M$ $NaCN$ is $10.945_$ .

Answer 21

Explanation of Solution

Explanation

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$

The dominant equilibrium reaction is,

$CN−(aq)+H2O(l)⇌HCN(aq)+OH−(aq)$

At equilibrium, the equilibrium constant expression is expressed by the formula,

$Kb=Concentration of productsConcentration of reactants$

Where,

$Kb$ is the base ionization constant.

The equilibrium constant expression for the given reaction is,

$Kb=[HCN][H+][CN−]$ (1)

The $Kb$ value is $1.61×10-5_$ .

The value, $Kw=Ka⋅Kb$

The value of $Ka$ for $HCN$ is $6.2×10−10$ .

The value of $Kb$ is calculated by the formula,

$Kb=KwKa$

Substitute the value of $Kb$ in the above expression.

$Kb=1.0×10−146.2×10−10=1.61×10-5_$

The $[OH−]$ is $8.8×10-4 M_$ .

The change in concentration of $CN−$ is assumed to be $x$ .

The liquid components do not affect the value of the rate constant.

The ICE table for the stated reaction is,

$CN−(aq)⇌HCN(aq)+OH−(aq)Inititial concentration0.0500Change−x+x+xEquilibrium concentration0.05−xxx$

The equilibrium concentration of $[CN−]$ is $(0.05−x) M$ .

The equilibrium concentration of $[HCN]$ is $x M$ .

The equilibrium concentration of $[OH−]$ is $x M$ .

The calculated value of $Kb$ is $1.61×10−5$ .

Substitute the value of $Kb$ , $[CN−]$ , $[CN−]$ and $[OH−]$ in equation (1).

$1.61×10−5=[x][x][0.05−x]1.61×10−5=[x]2[0.05−x]$

Solve the above expression.

$1.61×10−5=[x]2[0.05−x][x]=8.8×10-4 M_$

Therefore, the $[OH−]$ is $8.8×10-4 M_$ .

The $pOH$ of the given solution is $3.055_$ .

The $pOH$ of a solution is calculated by the formula,

$pOH=−log[OH−]$

Substitute the value of $[OH−]$ in the above expression.

$pOH=−log[8.8×10−4]=3.055_$

The $pH$ of the given solution is $10.945_$ .

The sum, $pH+pOH=14$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$

Substitute the calculated value of $pOH$ in the above expression.

$pH=14−3.055=10.945_$

Answer 22

Ch. 14 - Define each of the following: a. Arrhenius acid b....Ch. 14 - Define or illustrate the meaning of the following...Ch. 14 - Define or illustrate the meaning of the following...Ch. 14 - How is acid strength related to the value of Ka?...Ch. 14 - Two strategies are followed when solving for the...Ch. 14 - Two strategies are also followed when solving for...Ch. 14 - Table 13-4 lists the stepwise Ka values for some...Ch. 14 - For conjugate acidbase pairs, how are Ka and Kb...Ch. 14 - What is a salt? List some anions that behave as...Ch. 14 - For oxyacids, how does acid strength depend on a....

Calculate the pH of each of the following solutions. a. 0.10 M CH 3 NH 3 Cl b. 0.050 M NaCN

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Answer to Problem 119E

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