The K a and the percent ionization of 0 .22 M HC 4 H 5 O 3 is to be stated. Concept introduction: A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

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Answer 1

Question

Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The $Ka$ and the percent ionization of $0.22 M HC4H5O3$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Expert Solution & Answer

Answer to Problem 69E

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $[H+]$ is given below.

$pH=−log[H+]$ …(1)

The $pH$ of $0.22 M HC4H5O3$ is $2.12$ .

Substitute the value of $pH$ in equation (1).

$2.12=−log[H+][H+]=antilog(−2.12)=7.6×10−3 M$

The equation for the dissociation $HC4H5O3$ is shown below.

$HC4H5O3(aq)→C4H5O3−(aq)+H+(aq)$

The dissociation constant, $Ka$ for the above reaction is given below.

$Ka=[C4H5O3−][H+][HC4H5O3]$ …(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$[C4H5O3−]=[H+]$

The value of $[H+]$ is $7.6×10−3 M$ .

Therefore, the value of $[C4H5O3−]$ is $7.6×10−3 M$ .

The concentration of $HC4H5O3$ , $[HC4H5O3]$ is $0.22 M$ .

Substitute the values of $[H+]$ , $[C4H5O3−]$ , and $[HC4H5O3]$ in equation (2).

$Ka=7.6×10−3×7.6×10−30.22=2.6×10−4$

Therefore, the value of $Ka$ is $2.6×10−4$ .

The formula to calculate percent ionization is given below.

$Percent ionization=Concentration of ionized acidInitial concentration of acid×100%$ …(3)

The initial concentration is $0.22 M$ .

The concentration of ionized acid, $[C4H5O3−]$ is $7.6×10−3 M$ .

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$Percent ionization=7.6×10−3 M0.22 M×100%=3.5%$

Therefore, the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Conclusion

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

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Answer 2

Question

Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The $Ka$ and the percent ionization of $0.22 M HC4H5O3$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Expert Solution & Answer

Answer to Problem 69E

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $[H+]$ is given below.

$pH=−log[H+]$ …(1)

The $pH$ of $0.22 M HC4H5O3$ is $2.12$ .

Substitute the value of $pH$ in equation (1).

$2.12=−log[H+][H+]=antilog(−2.12)=7.6×10−3 M$

The equation for the dissociation $HC4H5O3$ is shown below.

$HC4H5O3(aq)→C4H5O3−(aq)+H+(aq)$

The dissociation constant, $Ka$ for the above reaction is given below.

$Ka=[C4H5O3−][H+][HC4H5O3]$ …(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$[C4H5O3−]=[H+]$

The value of $[H+]$ is $7.6×10−3 M$ .

Therefore, the value of $[C4H5O3−]$ is $7.6×10−3 M$ .

The concentration of $HC4H5O3$ , $[HC4H5O3]$ is $0.22 M$ .

Substitute the values of $[H+]$ , $[C4H5O3−]$ , and $[HC4H5O3]$ in equation (2).

$Ka=7.6×10−3×7.6×10−30.22=2.6×10−4$

Therefore, the value of $Ka$ is $2.6×10−4$ .

The formula to calculate percent ionization is given below.

$Percent ionization=Concentration of ionized acidInitial concentration of acid×100%$ …(3)

The initial concentration is $0.22 M$ .

The concentration of ionized acid, $[C4H5O3−]$ is $7.6×10−3 M$ .

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$Percent ionization=7.6×10−3 M0.22 M×100%=3.5%$

Therefore, the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Conclusion

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

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Answer 3

Question

Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The $Ka$ and the percent ionization of $0.22 M HC4H5O3$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Expert Solution & Answer

Answer to Problem 69E

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $[H+]$ is given below.

$pH=−log[H+]$ …(1)

The $pH$ of $0.22 M HC4H5O3$ is $2.12$ .

Substitute the value of $pH$ in equation (1).

$2.12=−log[H+][H+]=antilog(−2.12)=7.6×10−3 M$

The equation for the dissociation $HC4H5O3$ is shown below.

$HC4H5O3(aq)→C4H5O3−(aq)+H+(aq)$

The dissociation constant, $Ka$ for the above reaction is given below.

$Ka=[C4H5O3−][H+][HC4H5O3]$ …(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$[C4H5O3−]=[H+]$

The value of $[H+]$ is $7.6×10−3 M$ .

Therefore, the value of $[C4H5O3−]$ is $7.6×10−3 M$ .

The concentration of $HC4H5O3$ , $[HC4H5O3]$ is $0.22 M$ .

Substitute the values of $[H+]$ , $[C4H5O3−]$ , and $[HC4H5O3]$ in equation (2).

$Ka=7.6×10−3×7.6×10−30.22=2.6×10−4$

Therefore, the value of $Ka$ is $2.6×10−4$ .

The formula to calculate percent ionization is given below.

$Percent ionization=Concentration of ionized acidInitial concentration of acid×100%$ …(3)

The initial concentration is $0.22 M$ .

The concentration of ionized acid, $[C4H5O3−]$ is $7.6×10−3 M$ .

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$Percent ionization=7.6×10−3 M0.22 M×100%=3.5%$

Therefore, the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Conclusion

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

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Answer 4

Question

Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The $Ka$ and the percent ionization of $0.22 M HC4H5O3$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Expert Solution & Answer

Answer to Problem 69E

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $[H+]$ is given below.

$pH=−log[H+]$ …(1)

The $pH$ of $0.22 M HC4H5O3$ is $2.12$ .

Substitute the value of $pH$ in equation (1).

$2.12=−log[H+][H+]=antilog(−2.12)=7.6×10−3 M$

The equation for the dissociation $HC4H5O3$ is shown below.

$HC4H5O3(aq)→C4H5O3−(aq)+H+(aq)$

The dissociation constant, $Ka$ for the above reaction is given below.

$Ka=[C4H5O3−][H+][HC4H5O3]$ …(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$[C4H5O3−]=[H+]$

The value of $[H+]$ is $7.6×10−3 M$ .

Therefore, the value of $[C4H5O3−]$ is $7.6×10−3 M$ .

The concentration of $HC4H5O3$ , $[HC4H5O3]$ is $0.22 M$ .

Substitute the values of $[H+]$ , $[C4H5O3−]$ , and $[HC4H5O3]$ in equation (2).

$Ka=7.6×10−3×7.6×10−30.22=2.6×10−4$

Therefore, the value of $Ka$ is $2.6×10−4$ .

The formula to calculate percent ionization is given below.

$Percent ionization=Concentration of ionized acidInitial concentration of acid×100%$ …(3)

The initial concentration is $0.22 M$ .

The concentration of ionized acid, $[C4H5O3−]$ is $7.6×10−3 M$ .

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$Percent ionization=7.6×10−3 M0.22 M×100%=3.5%$

Therefore, the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Conclusion

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

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Answer 5

Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The $Ka$ and the percent ionization of $0.22 M HC4H5O3$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Expert Solution & Answer

Answer to Problem 69E

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $[H+]$ is given below.

$pH=−log[H+]$ …(1)

The $pH$ of $0.22 M HC4H5O3$ is $2.12$ .

Substitute the value of $pH$ in equation (1).

$2.12=−log[H+][H+]=antilog(−2.12)=7.6×10−3 M$

The equation for the dissociation $HC4H5O3$ is shown below.

$HC4H5O3(aq)→C4H5O3−(aq)+H+(aq)$

The dissociation constant, $Ka$ for the above reaction is given below.

$Ka=[C4H5O3−][H+][HC4H5O3]$ …(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$[C4H5O3−]=[H+]$

The value of $[H+]$ is $7.6×10−3 M$ .

Therefore, the value of $[C4H5O3−]$ is $7.6×10−3 M$ .

The concentration of $HC4H5O3$ , $[HC4H5O3]$ is $0.22 M$ .

Substitute the values of $[H+]$ , $[C4H5O3−]$ , and $[HC4H5O3]$ in equation (2).

$Ka=7.6×10−3×7.6×10−30.22=2.6×10−4$

Therefore, the value of $Ka$ is $2.6×10−4$ .

The formula to calculate percent ionization is given below.

$Percent ionization=Concentration of ionized acidInitial concentration of acid×100%$ …(3)

The initial concentration is $0.22 M$ .

The concentration of ionized acid, $[C4H5O3−]$ is $7.6×10−3 M$ .

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$Percent ionization=7.6×10−3 M0.22 M×100%=3.5%$

Therefore, the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Conclusion

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Answer 6

Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The $Ka$ and the percent ionization of $0.22 M HC4H5O3$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Expert Solution & Answer

Answer to Problem 69E

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $[H+]$ is given below.

$pH=−log[H+]$ …(1)

The $pH$ of $0.22 M HC4H5O3$ is $2.12$ .

Substitute the value of $pH$ in equation (1).

$2.12=−log[H+][H+]=antilog(−2.12)=7.6×10−3 M$

The equation for the dissociation $HC4H5O3$ is shown below.

$HC4H5O3(aq)→C4H5O3−(aq)+H+(aq)$

The dissociation constant, $Ka$ for the above reaction is given below.

$Ka=[C4H5O3−][H+][HC4H5O3]$ …(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$[C4H5O3−]=[H+]$

The value of $[H+]$ is $7.6×10−3 M$ .

Therefore, the value of $[C4H5O3−]$ is $7.6×10−3 M$ .

The concentration of $HC4H5O3$ , $[HC4H5O3]$ is $0.22 M$ .

Substitute the values of $[H+]$ , $[C4H5O3−]$ , and $[HC4H5O3]$ in equation (2).

$Ka=7.6×10−3×7.6×10−30.22=2.6×10−4$

Therefore, the value of $Ka$ is $2.6×10−4$ .

The formula to calculate percent ionization is given below.

$Percent ionization=Concentration of ionized acidInitial concentration of acid×100%$ …(3)

The initial concentration is $0.22 M$ .

The concentration of ionized acid, $[C4H5O3−]$ is $7.6×10−3 M$ .

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$Percent ionization=7.6×10−3 M0.22 M×100%=3.5%$

Therefore, the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Conclusion

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Answer 7

Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The $Ka$ and the percent ionization of $0.22 M HC4H5O3$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Answer 8

Expert Solution & Answer

Answer to Problem 69E

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $[H+]$ is given below.

$pH=−log[H+]$ …(1)

The $pH$ of $0.22 M HC4H5O3$ is $2.12$ .

Substitute the value of $pH$ in equation (1).

$2.12=−log[H+][H+]=antilog(−2.12)=7.6×10−3 M$

The equation for the dissociation $HC4H5O3$ is shown below.

$HC4H5O3(aq)→C4H5O3−(aq)+H+(aq)$

The dissociation constant, $Ka$ for the above reaction is given below.

$Ka=[C4H5O3−][H+][HC4H5O3]$ …(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$[C4H5O3−]=[H+]$

The value of $[H+]$ is $7.6×10−3 M$ .

Therefore, the value of $[C4H5O3−]$ is $7.6×10−3 M$ .

The concentration of $HC4H5O3$ , $[HC4H5O3]$ is $0.22 M$ .

Substitute the values of $[H+]$ , $[C4H5O3−]$ , and $[HC4H5O3]$ in equation (2).

$Ka=7.6×10−3×7.6×10−30.22=2.6×10−4$

Therefore, the value of $Ka$ is $2.6×10−4$ .

The formula to calculate percent ionization is given below.

$Percent ionization=Concentration of ionized acidInitial concentration of acid×100%$ …(3)

The initial concentration is $0.22 M$ .

The concentration of ionized acid, $[C4H5O3−]$ is $7.6×10−3 M$ .

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$Percent ionization=7.6×10−3 M0.22 M×100%=3.5%$

Therefore, the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Answer 12

Conclusion

The value of $Ka$ of $0.22 M HC4H5O3$ is $2.6×10−4$ and the percent ionization of $0.22 M HC4H5O3$ is $3.5%$ .

Answer 13

Ch. 18 - Prob. 1E Ch. 18 - Prob. 2E Ch. 18 - Prob. 3E Ch. 18 - Prob. 4E Ch. 18 - Prob. 5E Ch. 18 - Prob. 6E Ch. 18 - Prob. 7E Ch. 18 - Prob. 8E Ch. 18 - Prob. 9E Ch. 18 - Prob. 10E

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Answer to Problem 69E

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