Chapter 22, Problem 159CP

Interpretation Introduction

Interpretation: The pH or isoelectric point of glycine at which the given equation is true is to be calculated.

Concept introduction: Amino acids are defined as organic compounds which contain amine $\left({\text{-NH}}_{\text{2}}\right)$ and carboxyl $\left(\text{-COOH}\right)$ functional groups, along with a side chain (R group) attached to each amino acid. The isoelectric point is defined as the pH of the molecule at which the molecule has no net electrical charge in statistical mean.

To determine: The pH or isoelectric point of glycine at which the given equation is true.

Answer to Problem 159CP

Answer

The pH or isoelectric point of glycine is $\underset{\_}{\left(6.08\right)}$.

Explanation of Solution

Explanation

The equilibrium constant for the given reaction is $\underset{\_}{6.88\times 10^{-13}}$.

The equilibrium constant $\left(K_{\text{eq}}\right)$ of the reaction is given from the part (c) of Exercise 136. According to the part (c) of exercise 136, the equilibrium constant $\left(K_{\text{eq}}\right)$ of the reaction is ${\text{K}}_{\text{eq}}=6.88\times {\text{10}}^{-13}$ which is calculated as,

Given

The value of $K_{\text{a}}\left(\text{COOH}\right)$ is $4.3\times {10}^{-3}$.

The value of ${\text{K}}_{\text{a}}\left({\text{NH}}_3^{+}\right)$ is $1.6\times {10}^{-3}$.

The equilibrium constant is calculated by using the expression,

$K_{\text{eq}}=K_{\text{a}}\left(\text{COOH}\right)\times K_{\text{a}}\left({\text{NH}}_3^{+}\right)$

Substitute the value of $K_{\text{a}}\left(\text{COOH}\right)$ and $K_{\text{a}}\left({\text{NH}}_3^{+}\right)$ in the above expression,

$\begin{array}{c}{K}_{\text{eq}}=K_{\text{a}}\left(\text{COOH}\right)\times K_{\text{a}}\left({\text{NH}}_3^{+}\right)\\ =4.3\times {10}^{-3}\times 1.6\times {10}^{-10}\\ =\underset{\_}{6.88\times 10^{-13}}\end{array}$

Hence, $K_{\text{eq}}=\underset{\_}{6.88\times 10^{-13}}$.

By using equilibrium equation, the concentration of hydrogen ion at its isoelectric point is $\underset{\_}{8.3\times 10^{-7}M}$.

The equilibrium constant for the given reaction is $\text{6}{\text{.88×10}}^{-\text{13}}$.

The concentration of hydrogen ion at its isoelectric point is calculated by the equilibrium equation as follows,

$K_{\text{eq}}=\frac{{\left[{\text{H}}^{+}\right]}^{\text{2}}\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{-}\right]}{\left[{}^{+}{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\right]}$

Substitute the value of $K_{\text{eq}}$ in the above expression, the concentration of hydrogen ion at its isoelectric point is calculated.

$\begin{array}{c}{K}_{\text{eq}}=\frac{{\left[{\text{H}}^{+}\right]}^{\text{2}}\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{\text{-}}\right]}{\left[{}^{+}{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\right]}\\ 6.88\times {\text{10}}^{-13}={\left[{\text{H}}^{+}\right]}^{\text{2}}\\ \left[{\text{H}}^{+}\right]=\sqrt{\text{6}\text{.88}\times {\text{10}}^{-13}}\\ =\underset{\_}{8.3\times 10^{-7}M}\end{array}$

Therefore, concentration of hydrogen ion is $\underset{\_}{8.5\times 10^{-7}M}$.

The pH or isoelectric point of glycine is $\underset{\_}{6.08}$.

Finally, the pH of glycine or isoelectric point of glycine is calculated by using the value of the hydrogen ion concentration using an expression,

$\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]$

Substitute the value of hydrogen ion concentration $\left[{\text{H}}^{\text{+}}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=-\text{log}\left[{\text{H}}^{+}\right]\\ =-\text{log}\left(\text{8}\text{.3}\times {\text{10}}^{-7}\right)\\ =\underset{\_}{6.08}\end{array}$

Therefore, the pH of glycine is $\underset{\_}{\left(6.08\right)}$.

Conclusion

Conclusion

The pH or isoelectric point of glycine at which the given equation is true is $\underset{\_}{\left(6.08\right)}$.