Can anyone help??
What are the Concentrations of the various ionic species of a 0.1 M solution of lysine at pH4?
http://cti.itc.virginia.edu/~cmg/Demo/analyzeAA/lysine/lysine.html
Certainly! To determine the concentrations of the various ionic species of lysine in a 0.1 M solution at pH 4, you need to consider the pKa values of lysine's ionizable groups and utilize the Henderson-Hasselbalch equation.
Lysine is an amino acid with three ionizable groups: the carboxyl group (pKa ~ 2.2), the amino group (pKa ~ 9.1), and the side chain amino group (pKa ~ 10.5). At pH 4, all three groups will be protonated.
To calculate the concentrations of each species, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the deprotonated (ionized) form of the species, while [HA] represents the concentration of the protonated (non-ionized) form.
At pH 4, the carboxyl group, the amino group, and the side chain amino group will all be protonated, meaning that [A-] will be 0 for each group. Therefore, we only need to consider the concentration of the protonated form, [HA].
Since the initial concentration of lysine is 0.1 M, the concentration of the protonated form of each group will also be 0.1 M.
In summary, at pH 4, the concentrations of the various ionic species of lysine in a 0.1 M solution will be:
- [CH3CH(NH3+)COOH] = 0.1 M (protonated carboxyl group)
- [CH3CH(NH3+)COO-] = 0 M (deprotonated carboxyl group)
- [CH3CH(NH3+)NH2] = 0.1 M (protonated amino group)
- [CH3CH(NH2)NH3+] = 0 M (deprotonated amino group)
- [CH3CH2CH(NH3+)COOH] = 0.1 M (protonated side chain amino group)
- [CH3CH2CH(NH3+)COO-] = 0 M (deprotonated side chain amino group)
Remember that the concentration of [A-] would only be significant if the pH were higher than the pKa, facilitating the deprotonation of the respective group.