Would C6H6(NH3)OH be more soluble at a pH of 2.0 or 8.0?
To determine the solubility of C6H6(NH3)OH at different pH values, we need to understand the ionization behavior of the compound.
C6H6(NH3)OH is an organic compound called phenylamine or aniline. At low pH values, it exists primarily as the protonated form with the NH3 group protonated, resulting in C6H6NH3+OH-. On the other hand, at higher pH values, it can exist in a deprotonated form as an anion, C6H6N-.
To determine if it would be more soluble at a pH of 2.0 or 8.0, we need to consider the charges on the compound and the charges on the solvent molecules.
At a low pH of 2.0, there is a high concentration of protons (H+) in the solution. These protons can interact with the negatively charged anion (C6H6N-) and form stronger electrostatic attractions, leading to higher solubility.
At a higher pH of 8.0, there is a lower concentration of protons (H+) in the solution. This reduces the electrostatic interaction between protons and the anion, leading to lesser solubility.
Therefore, C6H6(NH3)OH would be more soluble at a pH of 2.0 compared to a pH of 8.0.