Chapter 5, Problem 5.58E

The formation of zwitterionic glycine, ${\text{CH}}_{\text{2}}\left({\text{NH}}_{\text{3}}^{\text{+}}\right)\left({\text{COO}}^{-}\right)$, from its fully acid form ${\text{CH}}_{\text{2}}\left({\text{NH}}_{\text{3}}^{\text{+}}\right)\left(\text{COOH}\right)$ and its fully basic form ${\text{CH}}_{\text{2}}\left({\text{NH}}_2\right)\left({\text{COO}}^{-}\right)$ is

${\text{CH}}_{\text{2}}\left({\text{NH}}_{\text{3}}^{\text{+}}\right)\left(\text{COOH}\right)+{\text{CH}}_{\text{2}}\left({\text{NH}}_2\right)\left({\text{COO}}^{-}\right)\rightleftharpoons 2{\text{CH}}_{\text{2}}\left({\text{NH}}_{\text{3}}^{\text{+}}\right)\left({\text{COO}}^{-}\right)$

Use the data in Table $5.1$ to determine the equilibrium constant for this reaction. What does your answer imply about the relative amount of the zwitterion in solution?