Question

The isoelectric point of an amino acid is the pH at which the molecule has no net charge. For glycine, that would be the pH at which virtually all glycine molecules are in the form${\mathbf{\text{+}}}^{\mathbf{\text{-}}}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}\mathbf{\text{H}}$. If we assume that the principal equilibrium is ${\mathbf{\text{2}}}^{\mathbf{\text{+}}}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{-}}}\mathbf{\rightleftharpoons }{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{-}}}{\mathbf{\text{+}}}^{\mathbf{\text{-}}}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}\mathbf{\text{H}}$ (i)

then at equilibrium

$\left[H_{2}NC{H}_{2}C{O}_2^{-}\right]\mathbf{=}\left[{}^{+}{H}_{3}NC{H}_{2}C{O}_{2}H\right]$ (ii)

Use this result and your answer to part c of Exercise 152 to calculate the pH at which Equation (ii) is true. This will be the isoelectric point of glycine.

Short answer

The pH of glycine at the isoelectric point is, pH = 6.07.

Step-by-step solution

To calculate the pH of the glycine at its isoelectric point,

To be able to calculate the pH of the glycine at its isoelectric point, first, we use the value of the equilibrium constant expression of exercise 146c which is $K_{eq}=7.3\times {10}^{-13}$This expression can also be written $K_{eq}=-K_a\left(-COOH{K}_a\left(-N{H}_3^{+}\right)\right).$Assuming that the following balanced equation:

_{${\text{2}}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{\text{-}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{\text{-}}{\text{+}}^{\text{-}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}$}

Is in principal equilibrium, we must also consider that the following relationship it be true:

$\left[H_{2}NC{H}_{2}C{O}_2^{-}\right]=\left[{}^{+}{H}_{3}NC{H}_{2}C{O}_{2}H\right]$

equilibrium constant equation

Then, we see the equilibrium constant equation to calculate the concentration of the hydrogen at the iso electric point.

Thus,

$$\begin{gathered} K_{eq}=\frac{{\left[H^{+}\right]}^2\left[H_{2}NC{H}_{2}C{O}_2^{-}\right]}{\left[{}^{+}{H}_{3}NC{H}_{2}C{O}_{2}H\right]}7.3\times {10}^{-13} \\ ={\left[H^{+}\right]}^2{\left[H^{+}\right]}^2 \\ =\sqrt{7.3\times {10}^{-13}} \\ =8.5\times {10}^{-7}\mathrm{M} \end{gathered}$$

The pH of glycine at the isoelectric point

Lastly, we use the concentration of $\left[{\mathrm{H}}^{+}\right]$to solve for the pH of glycine at its isoelectric point:

_{$pH=-\log \left[H^{+}\right]=-\log \left(8.5\times {10}^{-7}\right)=6.07$}

Hence The pH of glycine at the isoelectric point is $6.07$.