Question

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

A) What is the free energy for the dissolution of solid sodium
chloride in water at 25C?
NaCl(s) <-> Na+(aq) + Cl-(aq)

B) What is the solubility product constant for sodium chloride
in water at 25C?

Answer 1

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)`.