Chapter 21, Problem 153AE

Interpretation Introduction

Interpretation:The pH of lysine at isoelectric point needs to be determined.

Concept introduction:

The pH at which molecule does not carry any net electric charge is said to be its isoelectric point. In this case, it is electrically neutral and the amino acids do not migrate in an electric field.

Answer to Problem 153AE

The pH of lysine at isoelectric point is 6.07

Explanation of Solution

The equilibrium constant of reaction K_ey is known and the value is as follows:

  $K_{eq}=\text{7}{\text{.3 x 10}}^{\text{-13}}$

This can be depicted as:

  $K_{eq}=-K_a(-COOH)K_a(-N{H}_3{}^{+})$

Considering the equation

  $\begin{array}{l}\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}{}^{\text{-}}\right]\text{=}\left[\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\right]\\ 2\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}{}^{-}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}{}^{\text{-}}+\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\end{array}$

The equilibrium constant can be represented as follows:

  ${\text{K}}_{eq}=\frac{\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}{}^{\text{-}}\right]\left[\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\right]}{{\left[\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}{}^{-}\right]}^2}$

Since,

  $\begin{array}{l}{\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}\rightleftharpoons {\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{-}+{\text{H}}^{+}\\ {}^{+}{\text{NH}}_3{\text{CH}}_{\text{2}}\text{COOH}\rightleftharpoons {\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COOH}+{\text{H}}^{+}\\ \overline{\underset{\_}{{}^{+}{\text{NH}}_3{\text{CH}}_{\text{2}}\text{COOH}\rightleftharpoons {\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{-}+2{\text{H}}^{+}}}\end{array}$

Thus,

  ${\text{K}}_{eq}=\frac{{\left[{\text{H}}^{+}\right]}^2\left[\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\right]}{\left[\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}H\right]}$

Since,

  $\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}{}^{\text{-}}\right]\text{=}\left[\text{H}{}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\right]$

Thus,

  $\begin{array}{l}\text{7}{\text{.3 x 10}}^{\text{-13}}={\left[{\text{H}}^{+}\right]}^2\\ \left[{\text{H}}^{+}\right]=\sqrt{\text{7}{\text{.3 x 10}}^{\text{-13}}}\\ \text{ =}\sqrt{{\text{73 x 10}}^{\text{-14}}}\\ \text{ =8}{\text{.5 x 10}}^{\text{-7}}\text{ M}\end{array}$

To compute the pH at isoelectric point is done below

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

  $\left[{\text{H}}^{+}\right]\text{=8}{\text{.5 x 10}}^{\text{-7}}\text{ M}$

  $\begin{array}{l}\text{pH = - log }\left[{\text{H}}^{+}\right]\\ \text{ = - log (8}{\text{.5 x 10}}^{\text{-7}})\\ \text{ = 6}\text{.07}\end{array}$

Hence, the pH of glycine at isoelectric point is 6.07.

Conclusion

The pH of glycine at isoelectric point is 6.07.