Chapter 2, Problem 20P

(a)

Summary Introduction

To determine: The $\text{pH}$ range, in which the glycine can be used as an effective buffer due to its amino group.

Introduction:

The simplest amino acid is glycine. It is one of the “proteinogenic amino acids” and does not have a chiral carbon. The glycine is used in buffers because of its minimum repulsion in both hydrophobic and hydrophilic environments.

Explanation of Solution

Explanation:

The solution that is capable of resisting the drastic change in the levels of pH of the solution is called a buffer solution. The buffer varies in their chemical composition and thus they have the different buffering capacity and have a range of adjusting the pH of the solutions.

The effectiveness of the buffers is listed as follows:

$\begin{array}{c}\text{pH}=p{K}_a\pm 1\\ =9.6\pm 1\end{array}$

Therefore, the $\text{pH}$ range of buffer is $8.6-10.6$.

Conclusion

Conclusion:

The $\text{pH}$ range of buffer is $\underset{\_}{8.6-10.6}.$

(b)

Summary Introduction

To determine: The fraction of glycine has its amino group in the $\left({\text{NH}}_3^{+}\right)$ form in a $\text{0}\text{.1 M}$ solution of glycine at $\text{pH}$ $9.0$.

Introduction:

The simplest amino acid is glycine. It is one of the “proteinogenic amino acids” and does not have a chiral carbon. The glycine is used in buffers because of its minimum repulsion in both hydrophobic and hydrophilic environments.

Explanation of Solution

Explanation:

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

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

The acid dissociation constant is $9.6$.

$\begin{array}{c}9.0=9.6+\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)\\ -0.6=\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)\end{array}$

$\begin{array}{c}\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)=-0.6\\ \left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)={10}^{-0.6}\\ =0.25\end{array}$

The percentage of glycine in form of ${\text{NH}}_{\text{2}}$ is calculated as follows:

$\begin{array}{c}{\text{P}}_{{\text{(NH}}_{\text{2}}\text{)}}=\frac{0.25\times 100}{[1+0.25]}\%\\ =20\%\end{array}$

The percentage of glycine in form of ${\text{(NH}}_3^{+})$ is calculated as follows:

$\begin{array}{c}{\text{P}}_{{\text{(NH}}_3^{+})}=100\%-{\text{P}}_{{\text{(NH}}_{\text{2}}\text{)}}\\ =100\%-20\%\\ =80\%\end{array}$

The fraction can be simplified in the following form:

$\frac{4}{5}$

Conclusion

Conclusion:

The fraction of glycine has its amino group in the $\left({\text{NH}}_3^{+}\right)$ form in a $\text{0}\text{.1 M}$ solution of glycine at $\text{pH}$ $9.0$ is $\underset{\_}{\frac{4}{5}}.$

(c)

Summary Introduction

To determine: The volume of $\text{5 M}$ $\text{KOH}$ that must be added to $\text{1}\text{.0 L}$ of $\text{0}\text{.1 M}$ glycine at $\text{pH}$ $9.0$ to bring its $\text{pH}$ exactly to $\text{10}\text{.0}$.

Introduction:

The simplest amino acid is glycine. It is one of the “proteinogenic amino acids” and does not have a chiral carbon. The glycine is used in buffers because of its minimum repulsion in both hydrophobic and hydrophilic environments.

Explanation of Solution

Explanation:

One mole of $\left[{\text{A}}^{-}\right]$ salt will be formed by one mole $\text{KOH}$. The number of moles present in glycine is calculated as follows:

$\begin{array}{c}\text{N}=\text{M}\times \text{L}\\ =0.1\times 1\\ =0.1\text{ moles}\end{array}$

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

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

The acid dissociation constant is $9.6$.

$\begin{array}{c}9.0=9.6+\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)\\ -0.6=\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)\end{array}$

$\begin{array}{c}\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)=-0.6\\ \left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)={10}^{-0.6}\\ =0.25\end{array}$

The sum of moles of $[{\text{NH}}_2]$ and $[{\text{NH}}_3^{+}]$ is $0.1$ and ratio of $[{\text{NH}}_2]/[{\text{NH}}_3^{+}]$ is $0.25$.

The values of $[{\text{NH}}_2]$ and $[{\text{NH}}_3^{+}]$ are $\text{0}\text{.08}\text{mol}$ and $\text{0}\text{.02}\text{mol}$.

The acid dissociation constant is $9.6$ and $\text{pH}$ is $1.0$.

$\begin{array}{c}10.0=9.6+\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)\\ 0.4=\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)\end{array}$

$\begin{array}{c}\text{log}\left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)=0.4\\ \left(\frac{\left[{\text{NH}}_{\text{2}}\right]}{\left[{\text{NH}}_3^{+}\right]}\right)={10}^{0.4}\\ =2.5\end{array}$

The sum of moles of $[{\text{NH}}_2]$ and $[{\text{NH}}_3^{+}]$ is $0.1$ and ratio of $[{\text{NH}}_2]/[{\text{NH}}_3^{+}]$ is $2.5$.

The values of $[{\text{NH}}_2]$ and $[{\text{NH}}_3^{+}]$ are $\text{0}\text{.03}\text{mol}$ and $\text{0}\text{.07}\text{mol}$.

The number of moles that should be added is calculated as follows:

$\begin{array}{c}\text{N}=0.07\text{mol}-0.02\text{mol}\\ =0.05\text{mol}\end{array}$

The volume of $\text{KOH}$ added is calculated as follows:

$\begin{array}{c}\text{V}=\frac{\text{number}\text{of}\text{moles}}{\text{Molarity}}\\ =\frac{\text{0}\text{.05}\text{moles}}{\text{5}\text{M}}\\ =10\text{mL}\end{array}$

Conclusion

Conclusion:

The volume of $\text{5 M}$ $\text{KOH}$ that must be added to $\text{1}\text{.0 L}$ of $\text{0}\text{.1 M}$ glycine at $\text{pH}$ $9.0$ to bring its $\text{pH}$ exactly to $\text{10}\text{.0}$ is $\underset{\_}{10.0mL}.$

(d)

Summary Introduction

To determine: The numerical relation between the $\text{pH}$ of the solution and the pKa of the amino group when $\text{99\%}$ of glycine is in its $\left({\text{NH}}_3^{+}\right)$ form.

Introduction:

The simplest amino acid is glycine. It is one of the “proteinogenic amino acids” and does not have a chiral carbon. The glycine is used in buffers because of its minimum repulsion in both hydrophobic and hydrophilic environments.

Explanation of Solution

Explanation:

The $99\%$ percent of glycine get deprotonated. Therefore, $\left[{\text{A}}^{-}\right]$ is $0.01$ and $\left[\text{HA}\right]$ is $0.99$. The $\text{pH}$ of the solution is calculated by Henderson-Hasselbalch equation:

$\text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)$

$\begin{array}{c}\text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)\\ \text{pH}={\text{pK}}_{\text{a}}+\text{log}\left(\frac{0.01}{0.99}\right)\\ \text{pH}={\text{pK}}_{\text{a}}-2.0\end{array}$

Conclusion

Conclusion:

The relation between the $\text{pH}$ of the solution and the pKa of the amino group when $\text{99\%}$ of glycine is in its $\left({\text{NH}}_3^{+}\right)$ form is $\underset{\_}{pH=p{K}_a-2.0}.$