Chapter 2, Problem 14P

(a)

Summary Introduction

To determine: The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 2:1.

Introduction:

The Henderson-Hasselbalch equation depicts the measurement of acidity as $\text{pH}$ (potential of hydrogen) in biochemical reactions. The isoelectric point and $\text{pH}$ of buffers are calculated by this reaction. The $\text{pH}$ is used to calculate the ${\text{pK}}_{\text{a}}$ of molecules. The “ionized and unionized” concentrations of chemicals are estimated with the help of this equation. The ${\text{pK}}_{\text{a}}$ is used to calculate the $\text{pH}$ of molecules.

Explanation of Solution

Explanation:

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation:

$\text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

$\text{p}{K}_{\text{a}}$ is the acid dissociation constant.

$\begin{array}{c}\text{pH}=4.76+\text{log}\left(\frac{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{K}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\right)\\ =4.76+\text{log}\left(\frac{\left[\text{2}\right]}{\left[\text{1}\right]}\right)\\ =4.76+(0.30)\\ =5.06\end{array}$

Conclusion

Conclusion:

The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of is $\underset{\_}{5.06}.$

(b)

Summary Introduction

T determine: The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 1:3.

Introduction:

The Henderson-Hasselbalch equation depicts the measurement of acidity as $\text{pH}$ (potential of hydrogen) in biochemical reactions. The isoelectric point and $\text{pH}$ of buffers are calculated by this reaction. The $\text{pH}$ is used to calculate the ${\text{pK}}_{\text{a}}$ of molecules. The “ionized and unionized” concentrations of chemicals are estimated with the help of this equation. The ${\text{pK}}_{\text{a}}$ is used to calculate the $\text{pH}$ of molecules.

Explanation of Solution

Explanation:

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation:

$\text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

$\text{p}{K}_{\text{a}}$ is the acid dissociation constant.

$\begin{array}{c}\text{pH}=4.76+\text{log}\left(\frac{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{K}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\right)\\ =4.76+\text{log}\left(\frac{\left[\text{1}\right]}{\left[\text{3}\right]}\right)\\ =4.76+(-0.48)\\ =4.28\end{array}$

Conclusion

Conclusion:

The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 1:3 is $\underset{\_}{4.28}.$

(c)

Summary Introduction

To determine: The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 5:1.

Introduction:

The Henderson-Hasselbalch equation depicts the measurement of acidity as $\text{pH}$ (potential of hydrogen) in biochemical reactions. The isoelectric point and $\text{pH}$ of buffers are calculated by this reaction. The $\text{pH}$ is used to calculate the ${\text{pK}}_{\text{a}}$ of molecules. The “ionized and unionized” concentrations of chemicals are estimated with the help of this equation. The ${\text{pK}}_{\text{a}}$ is used to calculate the $\text{pH}$ of molecules.

Explanation of Solution

Explanation:

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation:

$\text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

$\text{p}{K}_{\text{a}}$ is the acid dissociation constant.

$\begin{array}{c}\text{pH}=4.76+\text{log}\left(\frac{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{K}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\right)\\ =4.76+\text{log}\left(\frac{\left[\text{5}\right]}{\left[\text{1}\right]}\right)\\ =4.76+(0.69)\\ =5.46\end{array}$

Conclusion

Conclusion:

The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 5:1 is $\underset{\_}{5.46}.$

(d)

Summary Introduction

To determine: The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 1:1.

Introduction:

The Henderson-Hasselbalch equation depicts the measurement of acidity as $\text{pH}$ (potential of hydrogen) in biochemical reactions. The isoelectric point and $\text{pH}$ of buffers are calculated by this reaction. The $\text{pH}$ is used to calculate the ${\text{pK}}_{\text{a}}$ of molecules. The “ionized and unionized” concentrations of chemicals are estimated with the help of this equation. The ${\text{pK}}_{\text{a}}$ is used to calculate the $\text{pH}$ of molecules.

Explanation of Solution

Explanation:

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation:

$\text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

$\text{p}{K}_{\text{a}}$ is the acid dissociation constant.

$\begin{array}{c}\text{pH}=4.76+\text{log}\left(\frac{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{K}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\right)\\ =4.76+\text{log}\left(\frac{\left[\text{1}\right]}{\left[\text{1}\right]}\right)\\ =4.76+(0)\\ =4.76\end{array}$

Conclusion

Conclusion:

The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 1:1 is $\underset{\_}{4.76}.$

(e)

Summary Introduction

To determine: The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 1:10.

Introduction:

The Henderson-Hasselbalch equation depicts the measurement of acidity as $\text{pH}$ (potential of hydrogen) in biochemical reactions. The isoelectric point and $\text{pH}$ of buffers are calculated by this reaction. The $\text{pH}$ is used to calculate the ${\text{pK}}_{\text{a}}$ of molecules. The “ionized and unionized” concentrations of chemicals are estimated with the help of this equation. The ${\text{pK}}_{\text{a}}$ is used to calculate the $\text{pH}$ of molecules.

Explanation of Solution

Explanation:

The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation:

$\text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}\right)$

The $\text{p}{K}_{\text{a}}$ is the acid dissociation constant.

$\begin{array}{c}\text{pH}=4.76+\text{log}\left(\frac{\left[{\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{K}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\right)\\ =4.76+\text{log}\left(\frac{\left[\text{1}\right]}{\left[\text{10}\right]}\right)\\ =4.76+(-0.1)\\ =3.76\end{array}$

Conclusion

Conclusion:

The pH of a dilute solution that contains a molar ratio of potassium acetate to acetic acid of 1:10 is $\underset{\_}{3.76}.$