8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature. At equilibrium after partial dissociation, its pH is found to be 5.05. What is the acid ionization constant, K a , for hypobromous acid at this temperature?

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Answer 1

Textbook Question

Chapter 8, Problem 104P

8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature.

Chapter 8, Problem 104P, 8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature. At

At equilibrium after partial dissociation, its pH is found to be 5.05. What is the acid ionization constant, K_a, for hypobromous acid at this temperature?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$HA(aq)⇌H+(aq)+A−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][A−][HA]$

Here, $[H+]$ is concentration of hydrogen ion, $[A−]$ is concentration of conjugate base and $[HA]$ is concentration of acid.

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$HOBr(aq)⇌H+(aq)+OBr−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][OBr−][HOBr]$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$HOBr(aq)⇌H+(aq)+OBr−(aq) [HOBr] [H+] [OBr−]i 0.040 0 0 C −X +X +X E 0.040−X +X +X$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$pH=−log[H+]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$pH=−log[H+][H+]=10−5.05=8.9125×10−6 mol L−1X=[H+]=8.9125×10−6 mol L−1$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$X=[OBr−]=8.9125×10−6 mol L−1$

Now, we need to calculate the concentration of $[HOBr]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[HOBr]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$Ka=[H+][OBr−][HOBr]=[8.9125×10−6][8.9125×10−6]0.040Ka=1.9858×10−9$

Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

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Answer 2

Textbook Question

Chapter 8, Problem 104P

8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature.

Chapter 8, Problem 104P, 8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature. At

At equilibrium after partial dissociation, its pH is found to be 5.05. What is the acid ionization constant, K_a, for hypobromous acid at this temperature?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$HA(aq)⇌H+(aq)+A−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][A−][HA]$

Here, $[H+]$ is concentration of hydrogen ion, $[A−]$ is concentration of conjugate base and $[HA]$ is concentration of acid.

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$HOBr(aq)⇌H+(aq)+OBr−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][OBr−][HOBr]$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$HOBr(aq)⇌H+(aq)+OBr−(aq) [HOBr] [H+] [OBr−]i 0.040 0 0 C −X +X +X E 0.040−X +X +X$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$pH=−log[H+]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$pH=−log[H+][H+]=10−5.05=8.9125×10−6 mol L−1X=[H+]=8.9125×10−6 mol L−1$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$X=[OBr−]=8.9125×10−6 mol L−1$

Now, we need to calculate the concentration of $[HOBr]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[HOBr]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$Ka=[H+][OBr−][HOBr]=[8.9125×10−6][8.9125×10−6]0.040Ka=1.9858×10−9$

Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

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Answer 3

Textbook Question

Chapter 8, Problem 104P

8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature.

Chapter 8, Problem 104P, 8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature. At

At equilibrium after partial dissociation, its pH is found to be 5.05. What is the acid ionization constant, K_a, for hypobromous acid at this temperature?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$HA(aq)⇌H+(aq)+A−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][A−][HA]$

Here, $[H+]$ is concentration of hydrogen ion, $[A−]$ is concentration of conjugate base and $[HA]$ is concentration of acid.

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$HOBr(aq)⇌H+(aq)+OBr−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][OBr−][HOBr]$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$HOBr(aq)⇌H+(aq)+OBr−(aq) [HOBr] [H+] [OBr−]i 0.040 0 0 C −X +X +X E 0.040−X +X +X$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$pH=−log[H+]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$pH=−log[H+][H+]=10−5.05=8.9125×10−6 mol L−1X=[H+]=8.9125×10−6 mol L−1$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$X=[OBr−]=8.9125×10−6 mol L−1$

Now, we need to calculate the concentration of $[HOBr]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[HOBr]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$Ka=[H+][OBr−][HOBr]=[8.9125×10−6][8.9125×10−6]0.040Ka=1.9858×10−9$

Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

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Answer 4

Textbook Question

Chapter 8, Problem 104P

8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature.

Chapter 8, Problem 104P, 8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature. At

At equilibrium after partial dissociation, its pH is found to be 5.05. What is the acid ionization constant, K_a, for hypobromous acid at this temperature?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$HA(aq)⇌H+(aq)+A−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][A−][HA]$

Here, $[H+]$ is concentration of hydrogen ion, $[A−]$ is concentration of conjugate base and $[HA]$ is concentration of acid.

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$HOBr(aq)⇌H+(aq)+OBr−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][OBr−][HOBr]$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$HOBr(aq)⇌H+(aq)+OBr−(aq) [HOBr] [H+] [OBr−]i 0.040 0 0 C −X +X +X E 0.040−X +X +X$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$pH=−log[H+]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$pH=−log[H+][H+]=10−5.05=8.9125×10−6 mol L−1X=[H+]=8.9125×10−6 mol L−1$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$X=[OBr−]=8.9125×10−6 mol L−1$

Now, we need to calculate the concentration of $[HOBr]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[HOBr]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$Ka=[H+][OBr−][HOBr]=[8.9125×10−6][8.9125×10−6]0.040Ka=1.9858×10−9$

Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$HA(aq)⇌H+(aq)+A−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][A−][HA]$

Here, $[H+]$ is concentration of hydrogen ion, $[A−]$ is concentration of conjugate base and $[HA]$ is concentration of acid.

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$HOBr(aq)⇌H+(aq)+OBr−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][OBr−][HOBr]$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$HOBr(aq)⇌H+(aq)+OBr−(aq) [HOBr] [H+] [OBr−]i 0.040 0 0 C −X +X +X E 0.040−X +X +X$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$pH=−log[H+]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$pH=−log[H+][H+]=10−5.05=8.9125×10−6 mol L−1X=[H+]=8.9125×10−6 mol L−1$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$X=[OBr−]=8.9125×10−6 mol L−1$

Now, we need to calculate the concentration of $[HOBr]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[HOBr]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$Ka=[H+][OBr−][HOBr]=[8.9125×10−6][8.9125×10−6]0.040Ka=1.9858×10−9$

Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$HA(aq)⇌H+(aq)+A−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][A−][HA]$

Here, $[H+]$ is concentration of hydrogen ion, $[A−]$ is concentration of conjugate base and $[HA]$ is concentration of acid.

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$HOBr(aq)⇌H+(aq)+OBr−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][OBr−][HOBr]$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$HOBr(aq)⇌H+(aq)+OBr−(aq) [HOBr] [H+] [OBr−]i 0.040 0 0 C −X +X +X E 0.040−X +X +X$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$pH=−log[H+]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$pH=−log[H+][H+]=10−5.05=8.9125×10−6 mol L−1X=[H+]=8.9125×10−6 mol L−1$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$X=[OBr−]=8.9125×10−6 mol L−1$

Now, we need to calculate the concentration of $[HOBr]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[HOBr]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$Ka=[H+][OBr−][HOBr]=[8.9125×10−6][8.9125×10−6]0.040Ka=1.9858×10−9$

Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The acid dissociation constant of hypobromous acid is to be calculated.

Concept Introduction:

Weak acids do not dissociate completely. Let HA be a weak acid. The dissociation of the weak acid can be represented by the chemical equation,

$HA(aq)⇌H+(aq)+A−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][A−][HA]$

Here, $[H+]$ is concentration of hydrogen ion, $[A−]$ is concentration of conjugate base and $[HA]$ is concentration of acid.

Answer 12

Answer to Problem 104P

The acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Answer 13

Explanation of Solution

Hypobromous acid is a weak acid. Hence, it do not dissociate completely. The dissociation of the given weak acid can be represented by the chemical equation,

$HOBr(aq)⇌H+(aq)+OBr−(aq)$

The equation for acid dissociation constant can be written from this chemical equation.

$Ka=[H+][OBr−][HOBr]$

The concentrations of each of the ions at equilibrium can be obtained from the ICE table. Where ICE represents the Initial, Change and Equilibrium concentrations of the weak acid.

$HOBr(aq)⇌H+(aq)+OBr−(aq) [HOBr] [H+] [OBr−]i 0.040 0 0 C −X +X +X E 0.040−X +X +X$

The hydrogen ion concentration can be obtained from the given pH. The pH is defined as the negative logarithm of the hydrogen ion concentration.

$pH=−log[H+]$

The pH of the weak acid solution at equilibrium is 5.05. Thus, we can calculate the concentration of the hydrogen ion.

$pH=−log[H+][H+]=10−5.05=8.9125×10−6 mol L−1X=[H+]=8.9125×10−6 mol L−1$

We calculated the “x” which is the concentration of hydrogen ion. The concentration of the anion is also “x”. Thus,

$X=[OBr−]=8.9125×10−6 mol L−1$

Now, we need to calculate the concentration of $[HOBr]$ = 0.040 − x. The value of x is very negligible hence, it can be ignored, and put just 0.040. Thus, the concentration of hypobromous acid is.

$[HOBr]$ = 0.040.

The concentrations of the anion, hydrogen ion and hypobromous acid are used in the equation used for acid dissociation constant.

$Ka=[H+][OBr−][HOBr]=[8.9125×10−6][8.9125×10−6]0.040Ka=1.9858×10−9$

Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Answer 14

Conclusion

Weak acids do not dissociate completely. Each weak acid has a specific dissociation constant. Here, ICE table is made from the given chemical equation. Thus, the acid dissociation constant of hypobromous acid is $Ka=1.9858×10−9$ .

Answer 15

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8-112 Consider an initial 0.040 M hypobromous acid (HOBr) solution at a certain temperature. At equilibrium after partial dissociation, its pH is found to be 5.05. What is the acid ionization constant, K a , for hypobromous acid at this temperature?

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Answer to Problem 104P

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