The pH and pOH of 0 .0146 M HNO 3 ( a q ) has to be calculated. Concept Introduction: Negative logarithm of the concentration of hydronium ion is the pH of solution. This can be given in equation form as follows; pH = − log [ H 3 O + ]

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Answer 1

Question

Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $pH$ of solution. This can be given in equation form as follows;

$pH=−log[H3O+]$

(a)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ is $1.84$ and $12.16$ respectively.

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion. Therefore, $0.0146 M$ of nitric acid gives $0.0146 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.0146 M$ .

The $pH$ of the nitric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.0146)=1.84$

Hence, the $pH$ of nitric acid solution is $1.84$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−1.84=12.16$

Hence, the $pOH$ of nitric acid solution is $12.16$ .

(b)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ is $0.96$ and $13.04$ respectively.

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HCl(aq)→H+(aq)+Cl−(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, $0.11 M$ of hydrochloric acid gives $0.11 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.11 M$ .

The $pH$ of the hydrochloric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.11)=0.96$

Hence, the $pH$ of hydrochloric acid solution is $0.96$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−0.96=13.04$

Hence, the $pOH$ of hydrochloric acid solution is $13.04$ .

(c)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ is $12.26$ and $1.74$ respectively.

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;

$Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, $0.0092 M$ of barium hydroxide gives $0.0184 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0184 M$ .

The $pOH$ of the barium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0184)=1.74$

Hence, the $pOH$ of barium hydroxide solution is $1.74$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−1.74=12.26$

Hence, the $pH$ of barium hydroxide solution is $12.26$ .

(d)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ after dilution from $2.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(d)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ is $10.85$ and $3.15$ respectively.

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(0.175 M)(2.00 mL)=Cfinal(50.0 mL)Cfinal=(0.175 M)(2.00 mL)500.0 mL=0.0007 M$

Therefore, $0.0007 M$ of potassium hydroxide gives $0.0007 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0007 M$ .

The $pOH$ of the potassium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0007)=3.15$

Hence, the $pOH$ of potassium hydroxide solution is $3.15$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.15=10.85$

Hence, the $pH$ of potassium hydroxide solution is $10.85$ .

(e)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $13.6 mg$ of $NaOH$ dissolved in $0.350 L$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

(e)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $NaOH(aq)$ is $10.99$ and $3.01$ respectively.

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

$Molarity=number of moles of soluteVolume of solution=13.6 mg×1 g1000 mg39.99 g⋅mol−1×0.350 L=9.72×10−4 M$

Therefore, $9.72×10−4 M$ of sodium hydroxide gives $9.72×10−4 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $9.72×10−4 M$ .

The $pOH$ of the sodium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(9.72×10−4)=3.01$

Hence, the $pOH$ of sodium hydroxide solution is $3.01$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.01=10.99$

Hence, the $pH$ of sodium hydroxide solution is $10.99$ .

(f)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ after dilution from $75.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(f)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ is $4.28$ and $9.72$ respectively.

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(3.5×10-4 M)(75.00 mL)=Cfinal(500.0 mL)Cfinal=(3.5×10-4 M)(75.00 mL)500.0 mL=0.525×10-4 M$

Therefore, $0.525×10-4 M$ of hydrobromic gives $0.525×10-4 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.525×10-4 M$ .

The $pH$ of the hydrobromic acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.525×10-4)=4.28$

Hence, the $pH$ of hydrobromic acid solution is $4.28$ .

The $pOH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pOH=14−pH=14−4.28=9.72$

Hence, the $pOH$ of hydrobromic acid solution is $9.72$ .

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Answer 2

Question

Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $pH$ of solution. This can be given in equation form as follows;

$pH=−log[H3O+]$

(a)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ is $1.84$ and $12.16$ respectively.

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion. Therefore, $0.0146 M$ of nitric acid gives $0.0146 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.0146 M$ .

The $pH$ of the nitric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.0146)=1.84$

Hence, the $pH$ of nitric acid solution is $1.84$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−1.84=12.16$

Hence, the $pOH$ of nitric acid solution is $12.16$ .

(b)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ is $0.96$ and $13.04$ respectively.

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HCl(aq)→H+(aq)+Cl−(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, $0.11 M$ of hydrochloric acid gives $0.11 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.11 M$ .

The $pH$ of the hydrochloric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.11)=0.96$

Hence, the $pH$ of hydrochloric acid solution is $0.96$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−0.96=13.04$

Hence, the $pOH$ of hydrochloric acid solution is $13.04$ .

(c)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ is $12.26$ and $1.74$ respectively.

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;

$Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, $0.0092 M$ of barium hydroxide gives $0.0184 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0184 M$ .

The $pOH$ of the barium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0184)=1.74$

Hence, the $pOH$ of barium hydroxide solution is $1.74$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−1.74=12.26$

Hence, the $pH$ of barium hydroxide solution is $12.26$ .

(d)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ after dilution from $2.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(d)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ is $10.85$ and $3.15$ respectively.

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(0.175 M)(2.00 mL)=Cfinal(50.0 mL)Cfinal=(0.175 M)(2.00 mL)500.0 mL=0.0007 M$

Therefore, $0.0007 M$ of potassium hydroxide gives $0.0007 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0007 M$ .

The $pOH$ of the potassium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0007)=3.15$

Hence, the $pOH$ of potassium hydroxide solution is $3.15$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.15=10.85$

Hence, the $pH$ of potassium hydroxide solution is $10.85$ .

(e)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $13.6 mg$ of $NaOH$ dissolved in $0.350 L$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

(e)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $NaOH(aq)$ is $10.99$ and $3.01$ respectively.

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

$Molarity=number of moles of soluteVolume of solution=13.6 mg×1 g1000 mg39.99 g⋅mol−1×0.350 L=9.72×10−4 M$

Therefore, $9.72×10−4 M$ of sodium hydroxide gives $9.72×10−4 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $9.72×10−4 M$ .

The $pOH$ of the sodium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(9.72×10−4)=3.01$

Hence, the $pOH$ of sodium hydroxide solution is $3.01$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.01=10.99$

Hence, the $pH$ of sodium hydroxide solution is $10.99$ .

(f)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ after dilution from $75.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(f)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ is $4.28$ and $9.72$ respectively.

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(3.5×10-4 M)(75.00 mL)=Cfinal(500.0 mL)Cfinal=(3.5×10-4 M)(75.00 mL)500.0 mL=0.525×10-4 M$

Therefore, $0.525×10-4 M$ of hydrobromic gives $0.525×10-4 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.525×10-4 M$ .

The $pH$ of the hydrobromic acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.525×10-4)=4.28$

Hence, the $pH$ of hydrobromic acid solution is $4.28$ .

The $pOH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pOH=14−pH=14−4.28=9.72$

Hence, the $pOH$ of hydrobromic acid solution is $9.72$ .

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Answer 3

Question

Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $pH$ of solution. This can be given in equation form as follows;

$pH=−log[H3O+]$

(a)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ is $1.84$ and $12.16$ respectively.

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion. Therefore, $0.0146 M$ of nitric acid gives $0.0146 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.0146 M$ .

The $pH$ of the nitric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.0146)=1.84$

Hence, the $pH$ of nitric acid solution is $1.84$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−1.84=12.16$

Hence, the $pOH$ of nitric acid solution is $12.16$ .

(b)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ is $0.96$ and $13.04$ respectively.

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HCl(aq)→H+(aq)+Cl−(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, $0.11 M$ of hydrochloric acid gives $0.11 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.11 M$ .

The $pH$ of the hydrochloric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.11)=0.96$

Hence, the $pH$ of hydrochloric acid solution is $0.96$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−0.96=13.04$

Hence, the $pOH$ of hydrochloric acid solution is $13.04$ .

(c)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ is $12.26$ and $1.74$ respectively.

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;

$Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, $0.0092 M$ of barium hydroxide gives $0.0184 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0184 M$ .

The $pOH$ of the barium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0184)=1.74$

Hence, the $pOH$ of barium hydroxide solution is $1.74$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−1.74=12.26$

Hence, the $pH$ of barium hydroxide solution is $12.26$ .

(d)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ after dilution from $2.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(d)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ is $10.85$ and $3.15$ respectively.

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(0.175 M)(2.00 mL)=Cfinal(50.0 mL)Cfinal=(0.175 M)(2.00 mL)500.0 mL=0.0007 M$

Therefore, $0.0007 M$ of potassium hydroxide gives $0.0007 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0007 M$ .

The $pOH$ of the potassium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0007)=3.15$

Hence, the $pOH$ of potassium hydroxide solution is $3.15$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.15=10.85$

Hence, the $pH$ of potassium hydroxide solution is $10.85$ .

(e)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $13.6 mg$ of $NaOH$ dissolved in $0.350 L$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

(e)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $NaOH(aq)$ is $10.99$ and $3.01$ respectively.

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

$Molarity=number of moles of soluteVolume of solution=13.6 mg×1 g1000 mg39.99 g⋅mol−1×0.350 L=9.72×10−4 M$

Therefore, $9.72×10−4 M$ of sodium hydroxide gives $9.72×10−4 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $9.72×10−4 M$ .

The $pOH$ of the sodium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(9.72×10−4)=3.01$

Hence, the $pOH$ of sodium hydroxide solution is $3.01$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.01=10.99$

Hence, the $pH$ of sodium hydroxide solution is $10.99$ .

(f)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ after dilution from $75.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(f)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ is $4.28$ and $9.72$ respectively.

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(3.5×10-4 M)(75.00 mL)=Cfinal(500.0 mL)Cfinal=(3.5×10-4 M)(75.00 mL)500.0 mL=0.525×10-4 M$

Therefore, $0.525×10-4 M$ of hydrobromic gives $0.525×10-4 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.525×10-4 M$ .

The $pH$ of the hydrobromic acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.525×10-4)=4.28$

Hence, the $pH$ of hydrobromic acid solution is $4.28$ .

The $pOH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pOH=14−pH=14−4.28=9.72$

Hence, the $pOH$ of hydrobromic acid solution is $9.72$ .

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Answer 4

Question

Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $pH$ of solution. This can be given in equation form as follows;

$pH=−log[H3O+]$

(a)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ is $1.84$ and $12.16$ respectively.

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion. Therefore, $0.0146 M$ of nitric acid gives $0.0146 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.0146 M$ .

The $pH$ of the nitric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.0146)=1.84$

Hence, the $pH$ of nitric acid solution is $1.84$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−1.84=12.16$

Hence, the $pOH$ of nitric acid solution is $12.16$ .

(b)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ is $0.96$ and $13.04$ respectively.

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HCl(aq)→H+(aq)+Cl−(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, $0.11 M$ of hydrochloric acid gives $0.11 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.11 M$ .

The $pH$ of the hydrochloric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.11)=0.96$

Hence, the $pH$ of hydrochloric acid solution is $0.96$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−0.96=13.04$

Hence, the $pOH$ of hydrochloric acid solution is $13.04$ .

(c)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ is $12.26$ and $1.74$ respectively.

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;

$Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, $0.0092 M$ of barium hydroxide gives $0.0184 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0184 M$ .

The $pOH$ of the barium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0184)=1.74$

Hence, the $pOH$ of barium hydroxide solution is $1.74$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−1.74=12.26$

Hence, the $pH$ of barium hydroxide solution is $12.26$ .

(d)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ after dilution from $2.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(d)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ is $10.85$ and $3.15$ respectively.

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(0.175 M)(2.00 mL)=Cfinal(50.0 mL)Cfinal=(0.175 M)(2.00 mL)500.0 mL=0.0007 M$

Therefore, $0.0007 M$ of potassium hydroxide gives $0.0007 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0007 M$ .

The $pOH$ of the potassium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0007)=3.15$

Hence, the $pOH$ of potassium hydroxide solution is $3.15$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.15=10.85$

Hence, the $pH$ of potassium hydroxide solution is $10.85$ .

(e)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $13.6 mg$ of $NaOH$ dissolved in $0.350 L$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

(e)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $NaOH(aq)$ is $10.99$ and $3.01$ respectively.

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

$Molarity=number of moles of soluteVolume of solution=13.6 mg×1 g1000 mg39.99 g⋅mol−1×0.350 L=9.72×10−4 M$

Therefore, $9.72×10−4 M$ of sodium hydroxide gives $9.72×10−4 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $9.72×10−4 M$ .

The $pOH$ of the sodium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(9.72×10−4)=3.01$

Hence, the $pOH$ of sodium hydroxide solution is $3.01$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.01=10.99$

Hence, the $pH$ of sodium hydroxide solution is $10.99$ .

(f)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ after dilution from $75.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(f)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ is $4.28$ and $9.72$ respectively.

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(3.5×10-4 M)(75.00 mL)=Cfinal(500.0 mL)Cfinal=(3.5×10-4 M)(75.00 mL)500.0 mL=0.525×10-4 M$

Therefore, $0.525×10-4 M$ of hydrobromic gives $0.525×10-4 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.525×10-4 M$ .

The $pH$ of the hydrobromic acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.525×10-4)=4.28$

Hence, the $pH$ of hydrobromic acid solution is $4.28$ .

The $pOH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pOH=14−pH=14−4.28=9.72$

Hence, the $pOH$ of hydrobromic acid solution is $9.72$ .

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Answer 5

Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $pH$ of solution. This can be given in equation form as follows;

$pH=−log[H3O+]$

(a)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ is $1.84$ and $12.16$ respectively.

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion. Therefore, $0.0146 M$ of nitric acid gives $0.0146 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.0146 M$ .

The $pH$ of the nitric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.0146)=1.84$

Hence, the $pH$ of nitric acid solution is $1.84$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−1.84=12.16$

Hence, the $pOH$ of nitric acid solution is $12.16$ .

(b)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ is $0.96$ and $13.04$ respectively.

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HCl(aq)→H+(aq)+Cl−(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, $0.11 M$ of hydrochloric acid gives $0.11 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.11 M$ .

The $pH$ of the hydrochloric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.11)=0.96$

Hence, the $pH$ of hydrochloric acid solution is $0.96$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−0.96=13.04$

Hence, the $pOH$ of hydrochloric acid solution is $13.04$ .

(c)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ is $12.26$ and $1.74$ respectively.

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;

$Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, $0.0092 M$ of barium hydroxide gives $0.0184 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0184 M$ .

The $pOH$ of the barium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0184)=1.74$

Hence, the $pOH$ of barium hydroxide solution is $1.74$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−1.74=12.26$

Hence, the $pH$ of barium hydroxide solution is $12.26$ .

(d)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ after dilution from $2.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(d)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ is $10.85$ and $3.15$ respectively.

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(0.175 M)(2.00 mL)=Cfinal(50.0 mL)Cfinal=(0.175 M)(2.00 mL)500.0 mL=0.0007 M$

Therefore, $0.0007 M$ of potassium hydroxide gives $0.0007 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0007 M$ .

The $pOH$ of the potassium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0007)=3.15$

Hence, the $pOH$ of potassium hydroxide solution is $3.15$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.15=10.85$

Hence, the $pH$ of potassium hydroxide solution is $10.85$ .

(e)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $13.6 mg$ of $NaOH$ dissolved in $0.350 L$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

(e)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $NaOH(aq)$ is $10.99$ and $3.01$ respectively.

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

$Molarity=number of moles of soluteVolume of solution=13.6 mg×1 g1000 mg39.99 g⋅mol−1×0.350 L=9.72×10−4 M$

Therefore, $9.72×10−4 M$ of sodium hydroxide gives $9.72×10−4 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $9.72×10−4 M$ .

The $pOH$ of the sodium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(9.72×10−4)=3.01$

Hence, the $pOH$ of sodium hydroxide solution is $3.01$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.01=10.99$

Hence, the $pH$ of sodium hydroxide solution is $10.99$ .

(f)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ after dilution from $75.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(f)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ is $4.28$ and $9.72$ respectively.

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(3.5×10-4 M)(75.00 mL)=Cfinal(500.0 mL)Cfinal=(3.5×10-4 M)(75.00 mL)500.0 mL=0.525×10-4 M$

Therefore, $0.525×10-4 M$ of hydrobromic gives $0.525×10-4 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.525×10-4 M$ .

The $pH$ of the hydrobromic acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.525×10-4)=4.28$

Hence, the $pH$ of hydrobromic acid solution is $4.28$ .

The $pOH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pOH=14−pH=14−4.28=9.72$

Hence, the $pOH$ of hydrobromic acid solution is $9.72$ .

Answer 6

Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $pH$ of solution. This can be given in equation form as follows;

$pH=−log[H3O+]$

(a)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ is $1.84$ and $12.16$ respectively.

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion. Therefore, $0.0146 M$ of nitric acid gives $0.0146 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.0146 M$ .

The $pH$ of the nitric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.0146)=1.84$

Hence, the $pH$ of nitric acid solution is $1.84$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−1.84=12.16$

Hence, the $pOH$ of nitric acid solution is $12.16$ .

Answer 7

Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $pH$ of solution. This can be given in equation form as follows;

$pH=−log[H3O+]$

Answer 8

(a)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0146 M HNO3(aq)$ is $1.84$ and $12.16$ respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HNO3(aq)→H+(aq)+NO3−(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion. Therefore, $0.0146 M$ of nitric acid gives $0.0146 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.0146 M$ .

The $pH$ of the nitric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.0146)=1.84$

Hence, the $pH$ of nitric acid solution is $1.84$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−1.84=12.16$

Hence, the $pOH$ of nitric acid solution is $12.16$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ is $0.96$ and $13.04$ respectively.

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HCl(aq)→H+(aq)+Cl−(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, $0.11 M$ of hydrochloric acid gives $0.11 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.11 M$ .

The $pH$ of the hydrochloric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.11)=0.96$

Hence, the $pH$ of hydrochloric acid solution is $0.96$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−0.96=13.04$

Hence, the $pOH$ of hydrochloric acid solution is $13.04$ .

Answer 14

(b)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.11 M HCl(aq)$ is $0.96$ and $13.04$ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water. This can be represented as follows;

$HCl(aq)→H+(aq)+Cl−(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion. Therefore, $0.11 M$ of hydrochloric acid gives $0.11 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.11 M$ .

The $pH$ of the hydrochloric acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.11)=0.96$

Hence, the $pH$ of hydrochloric acid solution is $0.96$ .

The $pOH$ of the solution can be calculated using the formula of $pOH$ as follows:

$pH+pOH=14pOH=14−pH=14−0.96=13.04$

Hence, the $pOH$ of hydrochloric acid solution is $13.04$ .

Answer 20

(c)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ is $12.26$ and $1.74$ respectively.

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;

$Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, $0.0092 M$ of barium hydroxide gives $0.0184 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0184 M$ .

The $pOH$ of the barium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0184)=1.74$

Hence, the $pOH$ of barium hydroxide solution is $1.74$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−1.74=12.26$

Hence, the $pH$ of barium hydroxide solution is $12.26$ .

Answer 21

(c)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer 22

(c)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.0092 M Ba(OH)2(aq)$ is $12.26$ and $1.74$ respectively.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water. This can be represented as follows;

$Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion. Therefore, $0.0092 M$ of barium hydroxide gives $0.0184 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0184 M$ .

The $pOH$ of the barium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0184)=1.74$

Hence, the $pOH$ of barium hydroxide solution is $1.74$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−1.74=12.26$

Hence, the $pH$ of barium hydroxide solution is $12.26$ .

Answer 27

(d)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ after dilution from $2.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(d)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ is $10.85$ and $3.15$ respectively.

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(0.175 M)(2.00 mL)=Cfinal(50.0 mL)Cfinal=(0.175 M)(2.00 mL)500.0 mL=0.0007 M$

Therefore, $0.0007 M$ of potassium hydroxide gives $0.0007 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0007 M$ .

The $pOH$ of the potassium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0007)=3.15$

Hence, the $pOH$ of potassium hydroxide solution is $3.15$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.15=10.85$

Hence, the $pH$ of potassium hydroxide solution is $10.85$ .

Answer 28

(d)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ after dilution from $2.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer 29

(d)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $0.175 M KOH(aq)$ is $10.85$ and $3.15$ respectively.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(0.175 M)(2.00 mL)=Cfinal(50.0 mL)Cfinal=(0.175 M)(2.00 mL)500.0 mL=0.0007 M$

Therefore, $0.0007 M$ of potassium hydroxide gives $0.0007 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $0.0007 M$ .

The $pOH$ of the potassium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(0.0007)=3.15$

Hence, the $pOH$ of potassium hydroxide solution is $3.15$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.15=10.85$

Hence, the $pH$ of potassium hydroxide solution is $10.85$ .

Answer 34

(e)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $13.6 mg$ of $NaOH$ dissolved in $0.350 L$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

(e)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $NaOH(aq)$ is $10.99$ and $3.01$ respectively.

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

$Molarity=number of moles of soluteVolume of solution=13.6 mg×1 g1000 mg39.99 g⋅mol−1×0.350 L=9.72×10−4 M$

Therefore, $9.72×10−4 M$ of sodium hydroxide gives $9.72×10−4 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $9.72×10−4 M$ .

The $pOH$ of the sodium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(9.72×10−4)=3.01$

Hence, the $pOH$ of sodium hydroxide solution is $3.01$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.01=10.99$

Hence, the $pH$ of sodium hydroxide solution is $10.99$ .

Answer 35

(e)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $13.6 mg$ of $NaOH$ dissolved in $0.350 L$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

Answer 36

(e)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $NaOH(aq)$ is $10.99$ and $3.01$ respectively.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

$Molarity=number of moles of soluteVolume of solution=13.6 mg×1 g1000 mg39.99 g⋅mol−1×0.350 L=9.72×10−4 M$

Therefore, $9.72×10−4 M$ of sodium hydroxide gives $9.72×10−4 M$ of hydroxide ions. Hence, the concentration of hydroxide ion is $9.72×10−4 M$ .

The $pOH$ of the sodium hydroxide solution is calculated as shown below;

$pOH=−log[OH−]=−log(9.72×10−4)=3.01$

Hence, the $pOH$ of sodium hydroxide solution is $3.01$ .

The $pH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pH=14−pOH=14−3.01=10.99$

Hence, the $pH$ of sodium hydroxide solution is $10.99$ .

Answer 41

(f)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ after dilution from $75.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

(f)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ is $4.28$ and $9.72$ respectively.

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(3.5×10-4 M)(75.00 mL)=Cfinal(500.0 mL)Cfinal=(3.5×10-4 M)(75.00 mL)500.0 mL=0.525×10-4 M$

Therefore, $0.525×10-4 M$ of hydrobromic gives $0.525×10-4 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.525×10-4 M$ .

The $pH$ of the hydrobromic acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.525×10-4)=4.28$

Hence, the $pH$ of hydrobromic acid solution is $4.28$ .

The $pOH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pOH=14−pH=14−4.28=9.72$

Hence, the $pOH$ of hydrobromic acid solution is $9.72$ .

Answer 42

(f)

Interpretation Introduction

Interpretation:

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ after dilution from $75.00 mL$ to $0.500 L$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer 43

(f)

Expert Solution

Answer to Problem 6B.5E

The $pH$ and $pOH$ of $3.5×10-4 M HBr(aq)$ is $4.28$ and $9.72$ respectively.

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

$CinitialVinitial=CfinalVfinal(3.5×10-4 M)(75.00 mL)=Cfinal(500.0 mL)Cfinal=(3.5×10-4 M)(75.00 mL)500.0 mL=0.525×10-4 M$

Therefore, $0.525×10-4 M$ of hydrobromic gives $0.525×10-4 M$ of hydronium ions. Hence, the concentration of hydronium ion is $0.525×10-4 M$ .

The $pH$ of the hydrobromic acid solution is calculated as shown below;

$pH=−log[H3O+]=−log(0.525×10-4)=4.28$

Hence, the $pH$ of hydrobromic acid solution is $4.28$ .

The $pOH$ of the solution can be calculated using the formula of $pH$ as follows:

$pH+pOH=14pOH=14−pH=14−4.28=9.72$

Hence, the $pOH$ of hydrobromic acid solution is $9.72$ .

Answer 48

Ch. 6 - Prob. 6A.1AST Ch. 6 - Prob. 6A.1BST Ch. 6 - Prob. 6A.2AST Ch. 6 - Prob. 6A.2BST Ch. 6 - Prob. 6A.3AST Ch. 6 - Prob. 6A.3BST Ch. 6 - Prob. 6A.1E Ch. 6 - Prob. 6A.2E Ch. 6 - Prob. 6A.3E Ch. 6 - Prob. 6A.4E

The pH and pOH of 0 .0146 M HNO 3 ( a q ) has to be calculated. Concept Introduction: Negative logarithm of the concentration of hydronium ion is the pH of solution. This can be given in equation form as follows; pH = − log [ H 3 O + ]

Answer to Problem 6B.5E

Explanation of Solution

Answer to Problem 6B.5E

Explanation of Solution

Answer to Problem 6B.5E

Explanation of Solution

Answer to Problem 6B.5E

Explanation of Solution

Answer to Problem 6B.5E

Explanation of Solution

Answer to Problem 6B.5E

Explanation of Solution

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