# 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?

Dec 16, 2016

A) Given the reaction:

${\text{NaCl"(s) stackrel("H"_2"O"(l)" ")(->) "Na"^(+)(aq) + "Cl}}^{-} \left(a q\right)$

The $\Delta {G}_{\text{rxn}}^{\circ}$ is as usual, and just like $\Delta {H}_{\text{rxn}}^{\circ}$:

$\textcolor{b l u e}{\Delta {G}_{\text{rxn}}^{\circ}} = {\sum}_{P} {\nu}_{P} \Delta {G}_{f , P}^{\circ} - {\sum}_{R} {\nu}_{R} \Delta {G}_{f , R}^{\circ}$

$= \left[{\nu}_{\text{Na"^(+)(aq))DeltaG_(f,Na^(+)(aq))^@ + nu_("Cl"^(-)(aq))DeltaG_(f,Cl^(-)(aq))^@] - [nu_("NaCl} \left(s\right)} \Delta {G}_{f , N a C l \left(s\right)}^{\circ}\right]$

$= \left[\left(1\right) \left(- \text{261.9 kJ/mol") + (1)(-"131.2 kJ/mol")] - [(1)(-"384.0 kJ/mol}\right)\right]$

$= - \text{261.9 kJ/mol" - "131.2 kJ/mol" + "384.0 kJ/mol}$

$= \textcolor{b l u e}{- \text{9.1 kJ/mol}}$

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

B)

The larger the ${K}_{\text{sp}}$, the more the equilibrium is skewed towards the aqueous products, and thus the more soluble the compound is in water. Obviously, $\text{NaCl}$ is extremely soluble in water ($\text{359 g/L}$), so ${K}_{\text{sp}}$ should be very large.

Since we have $\Delta {G}_{\text{rxn}}^{\circ}$ for this process, we can realize that $\Delta {G}_{\text{rxn}} = 0$ at equilibrium, so that:

$\cancel{\Delta {G}_{\text{rxn")^(0) = DeltaG_"rxn"^@ + RTlncancel(Q)^(K_"sp}}}$

$\implies \Delta {G}_{\text{rxn"^@ = -RTlnK_"sp}}$

$\implies \textcolor{b l u e}{{K}_{\text{sp") = e^(-DeltaG_"rxn"^@"/}} R T}$

$= {e}^{- \left(- \text{9.1 kJ/mol")"/"[("0.008314472 kJ/mol"cdot"K")("298.15 K}\right)}$

$= \textcolor{b l u e}{39.29}$

And indeed, ${K}_{\text{sp}}$ is fairly large. For poorly-soluble compounds, the ${K}_{\text{sp}}$ is often less than ${10}^{- 5}$.