How does entropy of mixing relate to solubility?

As solubility increases, entropy associated with mixing a solute into a solvent increases.

Let's compare #"NaCl"# with #"RbCl"# (solubilities are #359# and #91 "g/L"#, respectively), but ignore Debye-Huckel theory on this for simplicity.

ENTROPY OF MIXING

There is an entropy associated with mixing two compounds together, called #DeltaS_"mix"#. When assuming a simple ideal binary mixture, we have the following equation:

#\mathbf(DeltaS_"mix" = -nR(chi_1lnchi_1 + chi_2lnchi_2))#

where #R# is the universal gas constant, #n# is the total number of #"mol"#s of both the solute and solvent combined, and #chi_i# is the mole fraction of component #i#.

ENTROPY OF MIXING VS MOLE FRACTION OF DISSOLVED SOLUTE

Now, let us consider mixing each salt into water, where water is component #1# and the salt is component #2#. Let's say we have #"50 mL"# of water and we add a mass of the salt from #0.01# up until and including #"1 g"# in #"0.01 g"#-increments (#100# separate solvation trials). You can also do this in Excel if you want to.

For each one, because we have normalized this to a fraction or percentage, the graph's curvature looks the same for each solute when assuming that both are ideal. The graph for entropy of mixing #DeltaS_"mix"# vs. mole fraction #chi# of the ideal solute, following the experiment above, looks like this:

With this, we know that entropy increases as more solute is dissolved.

MOLE FRACTION OF WATER AND THE SALT

As a specific calculation example, let's see what the entropy is for dissolving each #"1 g"# of each salt in water.

For #"1 g NaCl"#, let's determine the moles of each component at #25^o"C"#. You can do a similar calculation using #"RbCl"# on your own. Again, ignore Debye-Huckel theory for simplicity. We are only looking at entropy of mixing.

#color(green)(n_("H"_2"O")) = "50" cancel("mL H"_2"O") xx (cancel"0.9970749 g")/cancel("mL H"_2"O", 25^o"C") xx ("1 mol H"_2"O")/("18.015" cancel("g H"_2"O"))#

#=# #color(green)("2.7673 mol H"_2"O")#

#color(green)(n_"NaCl") = cancel"1 g NaCl" xx "1 mol NaCl"/("58.44" cancel"g NaCl")#

#=# #color(green)("0.017112 mol NaCl")#

Now, we can determine the mole fraction of each.

#\mathbf(chi_i = (n_i)/(n_i + n_j))# and #\mathbf(chi_j = (n_j)/(n_i + n_j))#

where #i ne j# because #"NaCl" ne "H"_2"O"#. Thus:

#color(green)(chi_1) = (n_("H"_2"O"))/(n_("H"_2"O") + n_"NaCl")#

#= ("2.7673 mol")/("2.7673 mol"+"0.017112 mol")#

#=# #color(green)(0.9939) = "99.39% H"_2"O"#

#color(green)(chi_2) = (n_"NaCl")/(n_("H"_2"O") + n_"NaCl")#

#= ("0.017112 mol")/("2.7673 mol"+"0.017112 mol")#

#=# #color(green)(0.006145) = "0.6145% NaCl"#

TOTAL NUMBER OF MOLS

The last variable we don't know is the total number of #"mol"#s:

#color(green)(n) = n_("H"_2"O") + n_"NaCl"#

#=# #"2.7673 mol" + "0.017112 mol" = color(green)("2.7845 mol")#

Lastly, we should know that since entropy has units of #"J/K"#, we are using #R = "8.314472 J/mol"cdot"K"#.

FINAL RESULT

Now that we have everything, we can determine #DeltaS_"mix"# for mixing #"1 g NaCl"# into water:

#color(blue)(DeltaS_"mix") = (-2.7845)(8.314472)(0.9939ln0.9939 + 0.006145ln0.006145)#

#color(blue)("= 0.8663 J/K")#

When you do the same calculation for #"1 g RbCl"#, which is less soluble, you should get #color(blue)"0.4686 J/K"#.

Thus, as solubility increases, entropy associated with mixing increases.