# What is the structural formula for lysine with pH of 4?

Aug 3, 2018 #### Explanation:

Lysine has the following pKas:

• $p {K}_{\text{a1}} \left(\alpha - C O O H\right) = 2.2$

• $p {K}_{\text{a2}} \left(\alpha - N {H}_{3}^{+}\right) = 9.0$

• $p {K}_{\text{a3}} \left(R g r o u p\right) = 10.5$

Here are the structures of lysine as it transitions from fully protonated (dicationic form) to fully deprotonated (anionic form): The first proton to be lost as the pH is raised (when base is added) is the proton of the alpha-carboxyl group ($p {K}_{\text{a1}} = 2.2$). The next proton to be removed is in the alpha-aminium group ($p {K}_{\text{a2}} = 9.0$). The final proton to be removed is at the side chain's aminium group ($p {K}_{\text{a3}} = 10.5$).

The state of the amino acid at any given pH is determined by a combination of two equilibria.

Referring to the Henderson–Hasselbalch equation,

$p {K}_{a} = p H + \log \left(\frac{\left[H A\right]}{\left[{A}^{-}\right]}\right)$

When the $p {K}_{a}$ is the same as $p H$, it means,

$\log \left(\frac{\left[H A\right]}{\left[{A}^{-}\right]}\right) = 0$

which means, $\left[H A\right] = \left[{A}^{-}\right]$, meaning we have an equal amount of the 2 forms of the amino acid.

Starting at very low pH, the predominant structure of lysine is the dicationic form (refer to image above). As more base is added (pH increases), some of the monocationic form appears. At $p H = p {K}_{\text{a1}} = 2.2$, we have an equal amount of dicationic and monocationic form .

As the pH increases (past pH 2.2), there will be more monocationic form as compared to the dicationic form .

At $p H = p {K}_{\text{a2}} = 9.0$, there will be an equal amount of the monocationic form and dipolar ion .

Since the question is asking for the structure at pH 4 , that's before reaching $p {K}_{\text{a2}}$. That means, the predominant structure of lysine will be the monocationic form : 