Question

All amino acids have at least two functional groups with acidic or basic properties. In alanine the carboxylic acid group has ${\mathbf{\text{K}}}_{\mathbf{\text{a}}}{\mathbf{\text{= 4.5×10}}}^{\mathbf{\text{- 3}}}$and the amino group has ${\mathbf{\text{K}}}_{\mathbf{\text{b}}}{\mathbf{\text{= 7.4×10}}}^{\mathbf{\text{- 5}}}$. Three ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with $\mathbf{\left[}{\mathbf{\text{H}}}^{\mathbf{\text{+}}}\mathbf{\right]}\mathbf{\text{= 1.0M}}$in a solution with $\left[{\text{OH}}^{\text{-}}\right]\mathbf{\text{= 1.0M}}$?

Short answer

The carboxylic and amine ends will predominate the since the${\text{H}}^{\text{+}}$ is in excess and they will form the protonated form of alanine. protons in the molecule.

Step-by-step solution

To determine three ions of alanine

First, to be able to know which of the three ions of alanine will predominate in the various solutions, we must first look at the structure of alanine, and it's ions.

To Observe that the first proton

Observe that the first proton that is removed from alanine is from the carboxylic acid group which has a ${\text{K}}_{\text{b}}{\text{= 7.4×10}}^{\text{- 5}}$.The second proton that is removed comes from the amine group which has the${\text{K}}_{\text{b}}{\text{= 7.4×10}}^{\text{- 5}}$ To know the${\text{K}}_{\text{a}}$ of this ion, we must use the${\text{K}}_{\text{2}}$ equation.

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}} \\ \text{=}\frac{{\text{1.0×10}}^{\text{- 14}}}{{\text{7.4×10}}^{\text{- - 5}}} \\ {\text{= 1.4×10}}^{\text{- 10}} \end{gathered}$$

 To predominate in a solution

Now that we know the${\text{K}}_{\text{a}}$of the two ions, we can say that inlocalid="1664879169265" $1.0\text{M}$of the localid="1664879174453" ${\text{H}}^{\text{+}}$,the carboxylic and amine ends will predominate the since theis in excess and they will form the protonated form of alanine. protons in the molecule.

The predominated structures are shown below:

Hence the carboxylic and amine ends will predominate. Since the${\text{H}}^{\text{+}}$is in excess and they will form the protonated form of alanine. protons in the molecule.