(a) Interpretation: The pH and pOH values for 0.00562 M HClO 4 solution are to be calculated. Concept Introduction: The term pH referred to the power of hydrogen. The pH of a solution is mathematically equal to the negative logarithm of concentration of H + ion in solution. The pH of a solution is represented as, pH = − log [ H + ] Where, [ H + ] represents the concentration of H + ions in the solution. The pOH of a solution is used to measure the alkalinity of the solution. The pOH of a solution is mathematically equal to the negative logarithm of concentration of OH − ion in solution. The pOH of a solution is represented as, pOH = − log [ OH − ] Where, [ OH − ] represents the concentration of OH − ions in the solution.

Question

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Answer 1

Question

Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are $2.2503$ and $11.7497$ respectively.

Explanation of Solution

The concentration of the $HClO4$ solution is $0.00562 M$ .

The dissociation of $HClO4$ in aqueous medium is represented as,

$HClO4(aq)→H+(aq)+ClO4−(aq)$

One mole of $HClO4$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HClO4$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.00562 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.00562 M)=2.2503$

The $pH$ of the solution is $2.2503$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−2.2503=11.7497$

The $pOH$ of the solution is $11.7497$ .

Interpretation Introduction

(b)

Interpretation:

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are $10.5999$ and $3.4001$ respectively.

Explanation of Solution

The concentration of the $KOH$ solution is $3.98×10−4 M$ .

The dissociation of $KOH$ in aqueous medium is represented as,

$KOH(aq)→K+(aq)+OH−(aq)$

One mole of $KOH$ produces one mole of $OH−$ ions in the solution, Hence, the concentration of the $KOH$ solution and the concentration of $OH−$ ions in the solution is equal.

Therefore, the concentration of $OH−$ ions in the solution is $3.98×10−4 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(3.98×10−4 M)=3.4001$

The $pOH$ of the solution is $3.4001$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−3.4001=10.5999$

The $pH$ of the solution is $10.5999$ .

Interpretation Introduction

(c)

Interpretation:

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are $1.1079$ and $12.8921$ respectively.

Explanation of Solution

The concentration of the $HNO3$ solution is $0.078 M$ .

The dissociation of $HNO3$ in aqueous medium is represented as,

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of $HNO3$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HNO3$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.078 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.078 M)=1.1079$

The $pH$ of the solution is $1.1079$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−1.1079=12.8921$

The $pOH$ of the solution is $12.8921$ .

Interpretation Introduction

(d)

Interpretation:

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are $8.974$ and $5.026$ respectively.

Explanation of Solution

The concentration of the $Ca(OH)2$ solution is $4.71×10−6 M$ .

The dissociation of $Ca(OH)2$ in aqueous medium is represented as,

$Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)$

One mole of $Ca(OH)2$ produces two moles of $OH−$ ions in the solution, Hence, the concentration of $OH−$ ions in the solution is twice as the concentration of the $Ca(OH)2$ solution. The relation between the number of concentration of $Ca(OH)2$ and $OH−$ in the solution is given as,

$M1=2M2$

Where,

$M1$ represents the concentration of $OH−$ in the solution.
$M2$ represents the concentration of the $Ca(OH)2$ solution.

Substitute the value of $M1$ and $M2$ in the above equation.

$M1=(2)(4.71×10−6 M)=9.42×10−6 M$

Therefore, the concentration of $OH−$ ions in the solution is $9.42×10−6 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(9.42×10−6 M)=5.026$

The $pOH$ of the solution is $5.026$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−5.026=8.974$

The $pH$ of the solution is $8.974$ .

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Answer 2

Question

Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are $2.2503$ and $11.7497$ respectively.

Explanation of Solution

The concentration of the $HClO4$ solution is $0.00562 M$ .

The dissociation of $HClO4$ in aqueous medium is represented as,

$HClO4(aq)→H+(aq)+ClO4−(aq)$

One mole of $HClO4$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HClO4$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.00562 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.00562 M)=2.2503$

The $pH$ of the solution is $2.2503$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−2.2503=11.7497$

The $pOH$ of the solution is $11.7497$ .

Interpretation Introduction

(b)

Interpretation:

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are $10.5999$ and $3.4001$ respectively.

Explanation of Solution

The concentration of the $KOH$ solution is $3.98×10−4 M$ .

The dissociation of $KOH$ in aqueous medium is represented as,

$KOH(aq)→K+(aq)+OH−(aq)$

One mole of $KOH$ produces one mole of $OH−$ ions in the solution, Hence, the concentration of the $KOH$ solution and the concentration of $OH−$ ions in the solution is equal.

Therefore, the concentration of $OH−$ ions in the solution is $3.98×10−4 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(3.98×10−4 M)=3.4001$

The $pOH$ of the solution is $3.4001$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−3.4001=10.5999$

The $pH$ of the solution is $10.5999$ .

Interpretation Introduction

(c)

Interpretation:

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are $1.1079$ and $12.8921$ respectively.

Explanation of Solution

The concentration of the $HNO3$ solution is $0.078 M$ .

The dissociation of $HNO3$ in aqueous medium is represented as,

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of $HNO3$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HNO3$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.078 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.078 M)=1.1079$

The $pH$ of the solution is $1.1079$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−1.1079=12.8921$

The $pOH$ of the solution is $12.8921$ .

Interpretation Introduction

(d)

Interpretation:

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are $8.974$ and $5.026$ respectively.

Explanation of Solution

The concentration of the $Ca(OH)2$ solution is $4.71×10−6 M$ .

The dissociation of $Ca(OH)2$ in aqueous medium is represented as,

$Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)$

One mole of $Ca(OH)2$ produces two moles of $OH−$ ions in the solution, Hence, the concentration of $OH−$ ions in the solution is twice as the concentration of the $Ca(OH)2$ solution. The relation between the number of concentration of $Ca(OH)2$ and $OH−$ in the solution is given as,

$M1=2M2$

Where,

$M1$ represents the concentration of $OH−$ in the solution.
$M2$ represents the concentration of the $Ca(OH)2$ solution.

Substitute the value of $M1$ and $M2$ in the above equation.

$M1=(2)(4.71×10−6 M)=9.42×10−6 M$

Therefore, the concentration of $OH−$ ions in the solution is $9.42×10−6 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(9.42×10−6 M)=5.026$

The $pOH$ of the solution is $5.026$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−5.026=8.974$

The $pH$ of the solution is $8.974$ .

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Answer 3

Question

Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are $2.2503$ and $11.7497$ respectively.

Explanation of Solution

The concentration of the $HClO4$ solution is $0.00562 M$ .

The dissociation of $HClO4$ in aqueous medium is represented as,

$HClO4(aq)→H+(aq)+ClO4−(aq)$

One mole of $HClO4$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HClO4$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.00562 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.00562 M)=2.2503$

The $pH$ of the solution is $2.2503$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−2.2503=11.7497$

The $pOH$ of the solution is $11.7497$ .

Interpretation Introduction

(b)

Interpretation:

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are $10.5999$ and $3.4001$ respectively.

Explanation of Solution

The concentration of the $KOH$ solution is $3.98×10−4 M$ .

The dissociation of $KOH$ in aqueous medium is represented as,

$KOH(aq)→K+(aq)+OH−(aq)$

One mole of $KOH$ produces one mole of $OH−$ ions in the solution, Hence, the concentration of the $KOH$ solution and the concentration of $OH−$ ions in the solution is equal.

Therefore, the concentration of $OH−$ ions in the solution is $3.98×10−4 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(3.98×10−4 M)=3.4001$

The $pOH$ of the solution is $3.4001$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−3.4001=10.5999$

The $pH$ of the solution is $10.5999$ .

Interpretation Introduction

(c)

Interpretation:

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are $1.1079$ and $12.8921$ respectively.

Explanation of Solution

The concentration of the $HNO3$ solution is $0.078 M$ .

The dissociation of $HNO3$ in aqueous medium is represented as,

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of $HNO3$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HNO3$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.078 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.078 M)=1.1079$

The $pH$ of the solution is $1.1079$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−1.1079=12.8921$

The $pOH$ of the solution is $12.8921$ .

Interpretation Introduction

(d)

Interpretation:

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are $8.974$ and $5.026$ respectively.

Explanation of Solution

The concentration of the $Ca(OH)2$ solution is $4.71×10−6 M$ .

The dissociation of $Ca(OH)2$ in aqueous medium is represented as,

$Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)$

One mole of $Ca(OH)2$ produces two moles of $OH−$ ions in the solution, Hence, the concentration of $OH−$ ions in the solution is twice as the concentration of the $Ca(OH)2$ solution. The relation between the number of concentration of $Ca(OH)2$ and $OH−$ in the solution is given as,

$M1=2M2$

Where,

$M1$ represents the concentration of $OH−$ in the solution.
$M2$ represents the concentration of the $Ca(OH)2$ solution.

Substitute the value of $M1$ and $M2$ in the above equation.

$M1=(2)(4.71×10−6 M)=9.42×10−6 M$

Therefore, the concentration of $OH−$ ions in the solution is $9.42×10−6 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(9.42×10−6 M)=5.026$

The $pOH$ of the solution is $5.026$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−5.026=8.974$

The $pH$ of the solution is $8.974$ .

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Answer 4

Question

Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are $2.2503$ and $11.7497$ respectively.

Explanation of Solution

The concentration of the $HClO4$ solution is $0.00562 M$ .

The dissociation of $HClO4$ in aqueous medium is represented as,

$HClO4(aq)→H+(aq)+ClO4−(aq)$

One mole of $HClO4$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HClO4$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.00562 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.00562 M)=2.2503$

The $pH$ of the solution is $2.2503$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−2.2503=11.7497$

The $pOH$ of the solution is $11.7497$ .

Interpretation Introduction

(b)

Interpretation:

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are $10.5999$ and $3.4001$ respectively.

Explanation of Solution

The concentration of the $KOH$ solution is $3.98×10−4 M$ .

The dissociation of $KOH$ in aqueous medium is represented as,

$KOH(aq)→K+(aq)+OH−(aq)$

One mole of $KOH$ produces one mole of $OH−$ ions in the solution, Hence, the concentration of the $KOH$ solution and the concentration of $OH−$ ions in the solution is equal.

Therefore, the concentration of $OH−$ ions in the solution is $3.98×10−4 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(3.98×10−4 M)=3.4001$

The $pOH$ of the solution is $3.4001$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−3.4001=10.5999$

The $pH$ of the solution is $10.5999$ .

Interpretation Introduction

(c)

Interpretation:

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are $1.1079$ and $12.8921$ respectively.

Explanation of Solution

The concentration of the $HNO3$ solution is $0.078 M$ .

The dissociation of $HNO3$ in aqueous medium is represented as,

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of $HNO3$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HNO3$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.078 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.078 M)=1.1079$

The $pH$ of the solution is $1.1079$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−1.1079=12.8921$

The $pOH$ of the solution is $12.8921$ .

Interpretation Introduction

(d)

Interpretation:

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are $8.974$ and $5.026$ respectively.

Explanation of Solution

The concentration of the $Ca(OH)2$ solution is $4.71×10−6 M$ .

The dissociation of $Ca(OH)2$ in aqueous medium is represented as,

$Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)$

One mole of $Ca(OH)2$ produces two moles of $OH−$ ions in the solution, Hence, the concentration of $OH−$ ions in the solution is twice as the concentration of the $Ca(OH)2$ solution. The relation between the number of concentration of $Ca(OH)2$ and $OH−$ in the solution is given as,

$M1=2M2$

Where,

$M1$ represents the concentration of $OH−$ in the solution.
$M2$ represents the concentration of the $Ca(OH)2$ solution.

Substitute the value of $M1$ and $M2$ in the above equation.

$M1=(2)(4.71×10−6 M)=9.42×10−6 M$

Therefore, the concentration of $OH−$ ions in the solution is $9.42×10−6 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(9.42×10−6 M)=5.026$

The $pOH$ of the solution is $5.026$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−5.026=8.974$

The $pH$ of the solution is $8.974$ .

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Answer 5

Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are $2.2503$ and $11.7497$ respectively.

Explanation of Solution

The concentration of the $HClO4$ solution is $0.00562 M$ .

The dissociation of $HClO4$ in aqueous medium is represented as,

$HClO4(aq)→H+(aq)+ClO4−(aq)$

One mole of $HClO4$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HClO4$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.00562 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.00562 M)=2.2503$

The $pH$ of the solution is $2.2503$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−2.2503=11.7497$

The $pOH$ of the solution is $11.7497$ .

Interpretation Introduction

(b)

Interpretation:

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are $10.5999$ and $3.4001$ respectively.

Explanation of Solution

The concentration of the $KOH$ solution is $3.98×10−4 M$ .

The dissociation of $KOH$ in aqueous medium is represented as,

$KOH(aq)→K+(aq)+OH−(aq)$

One mole of $KOH$ produces one mole of $OH−$ ions in the solution, Hence, the concentration of the $KOH$ solution and the concentration of $OH−$ ions in the solution is equal.

Therefore, the concentration of $OH−$ ions in the solution is $3.98×10−4 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(3.98×10−4 M)=3.4001$

The $pOH$ of the solution is $3.4001$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−3.4001=10.5999$

The $pH$ of the solution is $10.5999$ .

Interpretation Introduction

(c)

Interpretation:

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are $1.1079$ and $12.8921$ respectively.

Explanation of Solution

The concentration of the $HNO3$ solution is $0.078 M$ .

The dissociation of $HNO3$ in aqueous medium is represented as,

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of $HNO3$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HNO3$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.078 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.078 M)=1.1079$

The $pH$ of the solution is $1.1079$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−1.1079=12.8921$

The $pOH$ of the solution is $12.8921$ .

Interpretation Introduction

(d)

Interpretation:

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are $8.974$ and $5.026$ respectively.

Explanation of Solution

The concentration of the $Ca(OH)2$ solution is $4.71×10−6 M$ .

The dissociation of $Ca(OH)2$ in aqueous medium is represented as,

$Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)$

One mole of $Ca(OH)2$ produces two moles of $OH−$ ions in the solution, Hence, the concentration of $OH−$ ions in the solution is twice as the concentration of the $Ca(OH)2$ solution. The relation between the number of concentration of $Ca(OH)2$ and $OH−$ in the solution is given as,

$M1=2M2$

Where,

$M1$ represents the concentration of $OH−$ in the solution.
$M2$ represents the concentration of the $Ca(OH)2$ solution.

Substitute the value of $M1$ and $M2$ in the above equation.

$M1=(2)(4.71×10−6 M)=9.42×10−6 M$

Therefore, the concentration of $OH−$ ions in the solution is $9.42×10−6 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(9.42×10−6 M)=5.026$

The $pOH$ of the solution is $5.026$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−5.026=8.974$

The $pH$ of the solution is $8.974$ .

Answer 6

Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are $2.2503$ and $11.7497$ respectively.

Explanation of Solution

The concentration of the $HClO4$ solution is $0.00562 M$ .

The dissociation of $HClO4$ in aqueous medium is represented as,

$HClO4(aq)→H+(aq)+ClO4−(aq)$

One mole of $HClO4$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HClO4$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.00562 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.00562 M)=2.2503$

The $pH$ of the solution is $2.2503$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−2.2503=11.7497$

The $pOH$ of the solution is $11.7497$ .

Answer 7

Chapter 17, Problem 20CR

Interpretation Introduction

(a)

Interpretation:

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Answer 8

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.00562 M HClO4$ solution are $2.2503$ and $11.7497$ respectively.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The concentration of the $HClO4$ solution is $0.00562 M$ .

The dissociation of $HClO4$ in aqueous medium is represented as,

$HClO4(aq)→H+(aq)+ClO4−(aq)$

One mole of $HClO4$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HClO4$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.00562 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.00562 M)=2.2503$

The $pH$ of the solution is $2.2503$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−2.2503=11.7497$

The $pOH$ of the solution is $11.7497$ .

Answer 13

Interpretation Introduction

(b)

Interpretation:

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are $10.5999$ and $3.4001$ respectively.

Explanation of Solution

The concentration of the $KOH$ solution is $3.98×10−4 M$ .

The dissociation of $KOH$ in aqueous medium is represented as,

$KOH(aq)→K+(aq)+OH−(aq)$

One mole of $KOH$ produces one mole of $OH−$ ions in the solution, Hence, the concentration of the $KOH$ solution and the concentration of $OH−$ ions in the solution is equal.

Therefore, the concentration of $OH−$ ions in the solution is $3.98×10−4 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(3.98×10−4 M)=3.4001$

The $pOH$ of the solution is $3.4001$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−3.4001=10.5999$

The $pH$ of the solution is $10.5999$ .

Answer 14

Interpretation Introduction

(b)

Interpretation:

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Answer 15

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $3.98×10−4 M KOH$ solution are $10.5999$ and $3.4001$ respectively.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The concentration of the $KOH$ solution is $3.98×10−4 M$ .

The dissociation of $KOH$ in aqueous medium is represented as,

$KOH(aq)→K+(aq)+OH−(aq)$

One mole of $KOH$ produces one mole of $OH−$ ions in the solution, Hence, the concentration of the $KOH$ solution and the concentration of $OH−$ ions in the solution is equal.

Therefore, the concentration of $OH−$ ions in the solution is $3.98×10−4 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(3.98×10−4 M)=3.4001$

The $pOH$ of the solution is $3.4001$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−3.4001=10.5999$

The $pH$ of the solution is $10.5999$ .

Answer 20

Interpretation Introduction

(c)

Interpretation:

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are $1.1079$ and $12.8921$ respectively.

Explanation of Solution

The concentration of the $HNO3$ solution is $0.078 M$ .

The dissociation of $HNO3$ in aqueous medium is represented as,

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of $HNO3$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HNO3$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.078 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.078 M)=1.1079$

The $pH$ of the solution is $1.1079$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−1.1079=12.8921$

The $pOH$ of the solution is $12.8921$ .

Answer 21

Interpretation Introduction

(c)

Interpretation:

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Answer 22

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $0.078 M HNO3$ solution are $1.1079$ and $12.8921$ respectively.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The concentration of the $HNO3$ solution is $0.078 M$ .

The dissociation of $HNO3$ in aqueous medium is represented as,

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of $HNO3$ produces one mole of $H+$ ions in the solution, Hence, the concentration of the $HNO3$ solution and the concentration of $H+$ ions in the solution is equal.

Therefore, the concentration of $H+$ ions in the solution is $0.078 M$ .

The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

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

$pH=−log(0.078 M)=1.1079$

The $pH$ of the solution is $1.1079$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pOH=14−pH$

Substitute the value of $pH$ in the above equation.

$pOH=14−1.1079=12.8921$

The $pOH$ of the solution is $12.8921$ .

Answer 27

Interpretation Introduction

(d)

Interpretation:

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are $8.974$ and $5.026$ respectively.

Explanation of Solution

The concentration of the $Ca(OH)2$ solution is $4.71×10−6 M$ .

The dissociation of $Ca(OH)2$ in aqueous medium is represented as,

$Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)$

One mole of $Ca(OH)2$ produces two moles of $OH−$ ions in the solution, Hence, the concentration of $OH−$ ions in the solution is twice as the concentration of the $Ca(OH)2$ solution. The relation between the number of concentration of $Ca(OH)2$ and $OH−$ in the solution is given as,

$M1=2M2$

Where,

$M1$ represents the concentration of $OH−$ in the solution.
$M2$ represents the concentration of the $Ca(OH)2$ solution.

Substitute the value of $M1$ and $M2$ in the above equation.

$M1=(2)(4.71×10−6 M)=9.42×10−6 M$

Therefore, the concentration of $OH−$ ions in the solution is $9.42×10−6 M$ .

The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

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

$pOH=−log(9.42×10−6 M)=5.026$

The $pOH$ of the solution is $5.026$ .

The relation between $pH$ and $pOH$ of a solution is given as,

$pH=14−pOH$

Substitute the value of $pOH$ in the above equation.

$pH=14−5.026=8.974$

The $pH$ of the solution is $8.974$ .

Answer 28

Interpretation Introduction

(d)

Interpretation:

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are to be calculated.

Concept Introduction:

The term $pH$ referred to the power of hydrogen. The $pH$ of a solution is mathematically equal to the negative logarithm of concentration of $H+$ ion in solution. The $pH$ of a solution is represented as,

$pH=−log[H+]$

Where,

$[H+]$ represents the concentration of $H+$ ions in the solution.

The $pOH$ of a solution is used to measure the alkalinity of the solution. The $pOH$ of a solution is mathematically equal to the negative logarithm of concentration of $OH−$ ion in solution. The $pOH$ of a solution is represented as,

$pOH=−log[OH−]$

Where,

$[OH−]$ represents the concentration of $OH−$ ions in the solution.

Answer 29

Expert Solution

Answer to Problem 20CR

The $pH$ and $pOH$ values for $4.71×10−6 M Ca(OH)2$ solution are $8.974$ and $5.026$ respectively.

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The concentration of the $Ca(OH)2$ solution is $4.71×10−6 M$ .

The dissociation of $Ca(OH)2$ in aqueous medium is represented as,

$Ca(OH)2(aq)→Ca2+(aq)+2OH−(aq)$

One mole of $Ca(OH)2$ produces two moles of $OH−$ ions in the solution, Hence, the concentration of $OH−$ ions in the solution is twice as the concentration of the $Ca(OH)2$ solution. The relation between the number of concentration of $Ca(OH)2$ and $OH−$ in the solution is given as,