What is the free energy for the dissolution of solid sodium chloride in water at 25C?

A) Given the reaction:

#"NaCl"(s) stackrel("H"_2"O"(l)" ")(->) "Na"^(+)(aq) + "Cl"^(-)(aq)#

The #DeltaG_"rxn"^@# is as usual, and just like #DeltaH_"rxn"^@#:

#color(blue)(DeltaG_"rxn"^@) = sum_P nu_PDeltaG_(f,P)^@ - sum_R nu_RDeltaG_(f,R)^@#

#= [nu_("Na"^(+)(aq))DeltaG_(f,Na^(+)(aq))^@ + nu_("Cl"^(-)(aq))DeltaG_(f,Cl^(-)(aq))^@] - [nu_("NaCl"(s))DeltaG_(f,NaCl(s))^@]#

#= [(1)(-"261.9 kJ/mol") + (1)(-"131.2 kJ/mol")] - [(1)(-"384.0 kJ/mol")]#

#= -"261.9 kJ/mol" - "131.2 kJ/mol" + "384.0 kJ/mol"#

#= color(blue)(-"9.1 kJ/mol")#

It should be this small; that's fine, for dissolving solutes.

B)

The larger the #K_"sp"#, the more the equilibrium is skewed towards the aqueous products, and thus the more soluble the compound is in water. Obviously, #"NaCl"# is extremely soluble in water (#"359 g/L"#), so #K_"sp"# should be very large.

Since we have #DeltaG_"rxn"^@# for this process, we can realize that #DeltaG_"rxn" = 0# at equilibrium, so that:

#cancel(DeltaG_"rxn")^(0) = DeltaG_"rxn"^@ + RTlncancel(Q)^(K_"sp")#

#=> DeltaG_"rxn"^@ = -RTlnK_"sp"#

#=> color(blue)(K_"sp") = e^(-DeltaG_"rxn"^@"/"RT)#

#= e^(-(-"9.1 kJ/mol")"/"[("0.008314472 kJ/mol"cdot"K")("298.15 K")]#

#= color(blue)(39.29)#

And indeed, #K_"sp"# is fairly large. For poorly-soluble compounds, the #K_"sp"# is often less than #10^(-5)#.