Question

The protonated form of glycine \(\left({ }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{COOH}\right)\) has \(K_{a 1}=4.47 \times 10^{-3}\), and \(K_{a 2}=1.66 \times 10^{-10} \cdot\) (a) Write the chemical equations for the proton transfer equilibria. (b) What is the dominant form of glycine in solution at \(\mathrm{pH}=2, \mathrm{pH}=5\), and \(\mathrm{pH}=12\) ?

Short answer

The dominant form of glycine at pH 2 is the protonated form \( { }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{COOH} \. At pH 5, it is the zwitterionic form \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH} \) and at pH 12 it is the negatively charged form \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COO}^{-} \) due to the respective acidic and basic conditions.

Step-by-step solution

Write the equilibrium equation for the first proton transfer

For the first proton transfer, the protonated form of glycine, \( { }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{COOH} \) will lose a proton to form \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH} \) and release a hydronium ion \( \mathrm{H}_3 \mathrm{O}^{+} \). The equilibrium equation is \( { }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{COOH} \longleftrightarrow \mathrm{H}_2 \mathrm{NCH}_2 \mathrm{COOH} + \mathrm{H}_3 \mathrm{O}^{+} \) with \( K_{a1} = 4.47 \times 10^{-3} \) for this process.

Write the equilibrium equation for the second proton transfer

For the second proton transfer, the zwitterionic form of glycine, \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH} \) will lose another proton from the carboxyl group to form \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COO}^{-} \) and another hydronium ion \( \mathrm{H}_3 \mathrm{O}^{+} \). The equilibrium equation is \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH} \longleftrightarrow \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COO}^{-} + \mathrm{H}_3 \mathrm{O}^{+} \) with \( K_{a2} = 1.66 \times 10^{-10} \) for this process.

Determine the dominant form at pH 2

At \( \mathrm{pH} = 2 \), the \( \mathrm{pH} \) is lower than both \( pK_{a1} \) and \( pK_{a2} \) values (which can be calculated as the negative logarithm of \( K_{a} \) values). This means the solution is highly acidic, and the protonated form \( { }^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{COOH} \) will be dominant.

Determine the dominant form at pH 5

At \( \mathrm{pH} = 5 \), the \( \mathrm{pH} \) is greater than \( pK_{a1} \) but still lower than \( pK_{a2} \). The solution is less acidic. The dominant form of glycine will be the zwitterionic form \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH} \) because it has lost the proton from the ammonium group, but not yet from the carboxyl group.

Determine the dominant form at pH 12

At \( \mathrm{pH} = 12 \), the \( \mathrm{pH} \) is higher than both \( pK_{a1} \) and \( pK_{a2} \), which means the solution is very basic. The dominant form of glycine will be the negatively charged form \( \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COO}^{-} \) as both the ammonium and the carboxyl groups would have lost protons.

Key concepts

Proton Transfer Equilibria

In understanding proton transfer equilibria, we consider the behavior of compounds, like amino acids, that can donate and accept protons. Glycine, a simple amino acid, can exhibit different forms based on its protonation state. At a low pH, it tends to hold onto its protons, while at a high pH, it will lose them.

Glycine’s ability to gain or lose protons in water makes it a perfect candidate to illustrate proton transfer equilibria, which is the balance between the protonated (with extra hydrogen ions) and deprotonated (with fewer hydrogen ions) states of a molecule. The specific pH level of the solution dictates how these equilibria are established. In a low pH environment, the excess of hydrogen ions (protons) prompts glycine to take on its protonated form. As the pH rises, the equilibrium shifts, and the amino acid begins to lose protons. This shift is described by two key reactions, corresponding to the dissociation of protons from the amino and carboxyl groups of glycine.

Acid Dissociation Constants

The acid dissociation constants, denoted as \( K_a \) values, are critical in predicting the protonation state of molecules like glycine at different pH levels. These constants are a quantitative measure of a compound's tendency to lose protons, thus reflecting its strength as an acid.

With glycine, two \( K_a \) values are associated with its two ionizable groups: the ammonium group (\( K_{a1} \) of 4.47 x 10^-3) and the carboxyl group (\( K_{a2} \) of 1.66 x 10^-10). By comparing the pH of the solution to the \( pK_a \) values (the negative logarithm of \( K_a \) ), which are 2.35 and 9.78 respectively, we can predict the dominant form of glycine at that pH level. If the pH is lower than the \( pKa \) of the group, the group will likely be protonated. Conversely, if the pH is higher, deprotonation is favored. This relationship allows us to calculate the protonation state of glycine under various pH conditions.

pH-dependent Molecular Forms

The molecular form of substances like glycine can change notably with pH, a characteristic that is pivotal in many biological functions. The fundamental concept here is that each form predominates in a certain pH range.

At a very low pH (below \( pK_{a1} \) of the amino group), glycine is found in its fully protonated form; as the pH exceeds \( pK_{a1} \) but is still below \( pK_{a2} \) of the carboxylic acid group, it exists mainly as a zwitterion, which has both positive and negative charges but is overall neutral. If the pH is greater than \( pK_{a2} \) (a high pH environment), the carboxyl group also loses its proton, resulting in a negatively charged molecule. Therefore, the structure and charge of glycine are highly dependent on the pH of its surroundings. These characteristics have essential implications, particularly in biochemistry, affecting how molecules interact with each other and with their environment.