Question #44065

A crystalline native protein, $P_{N(s)}$, is in equilibrium with its aqueous solution at $298\text{ K}$. At its isoelectric point ($\text{pH} = 6.0$), its intrinsic solubility is $S_0 = 1.0 \times 10^{-4}\text{ mol/L}$. In its native folded state, the dominant ionizable groups governing solubility have $pK_{a1} = 4.0$ (protonation of basic residues) and $pK_{a2} = 8.0$ (deprotonation of acidic residues).

When the solution is acidified to $\text{pH} = 2.0$ using $\ce{HCl}$, the protein undergoes complete and irreversible acid denaturation, precipitating as a new solid phase, $P_{D(s)}$. For this denatured state, the exposure of the hydrophobic core reduces its intrinsic solubility to $S_{0,D} = 1.0 \times 10^{-7}\text{ mol/L}$. Furthermore, the loss of the native tertiary structure and salt bridges shifts the $pK_{a1}$ of the basic residues to $5.0$.

Assuming the activity coefficients are unity, what is the actual equilibrium solubility of the protein at $\text{pH} = 2.0$?