In the process of salting-in, protein solubility in a dilute salt solution is increased by adding more salt. Because the protein solubility depends on the total ion concentration as well as the ion charge, salts yielding doubly charged ions are often more effective than those yielding singly charged ions. (a) How many grams of MgCl₂ must dissolve to equal the ion concentration of 12.4 g of NaCl? (b) How many grams of CaS must dissolve? (c) Which of the three salt solutions would dissolve the most protein?

Moles of NaCl are,

$\begin{array}{rcl}\mathrm{Moles}& =& \frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}\\ & =& \frac{12.4\mathrm{g}}{58.44{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}\\ & =& 0.212\mathrm{mol}\end{array}$

NaCl produces two ions Na⁺ and Cl^-.

Let us take the solution volume be 1.0 L.

The total concentration of NaCl solution is,

$$\begin{gathered} =0.212{\mathrm{molNa}}^{+}+0.212{\mathrm{molCl}}^{-} \\ =0.424\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

MgCl₂ dissociate as,

${\mathrm{MgCl}}_2\left(\mathrm{aq}\right)\to {\mathrm{Mg}}^{+2}\left(\mathrm{aq}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

Let the concentration of MgCl₂ be x mol/L.

The total ion concentration is 3x.

$\begin{array}{rcl}0.424\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.& =& 3\mathrm{x}\\ \mathrm{x}& =& 0.141\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.\end{array}$

Moles of MgCl₂ needed = 0.141 mol

Mass of MgCl₂ is,

$\begin{array}{rcl}\mathrm{Mass}& =& \mathrm{moles}\times \mathrm{molar}\mathrm{mass}\\ & =& 0.141\mathrm{mol}\times 95.21\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\\ & =& 13.4\mathrm{g}\end{array}$

Thus the mass of MgCl₂ must dissolve to equal the ion concentration of 12.4 g of NaCl is 13.4 g.

CaS dissociate as,

$\mathrm{CaS}\left(\mathrm{aq}\right)\to {\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right)$

Let the concertation of CaS be x mol/L.

The total ion concentration is 2x.

$\begin{array}{rcl}0.424\mathrm{mol}/\mathrm{L}& =& 2\mathrm{x}\\ \mathrm{x}& =& 0.212\mathrm{mol}/\mathrm{L}\end{array}$

Moles of CaS needed = 0.212 mol

Mass of CaS is,

$\begin{array}{rcl}\mathrm{Mass}& =& \mathrm{moles}\times \mathrm{molar}\mathrm{mass}\\ & =& 0.212\mathrm{mol}\times 75.15\mathrm{g}/\mathrm{mol}\\ & =& 15.3\mathrm{g}\end{array}$

Thus the mass of CaS must dissolve is 15.3 g.

The solubility of the salt depends on the total ion concentration and the ion charge.

The ion concentration of the three salts NaCl, MgCl₂, and CaS is same. So, the solubility will depend on the ion charge. Among the given salts, CaS produces two doubly charged ions.

Thus, CaS salt solution will dissolve the most protein.