# Why does the dissolution of calcium hydroxide have a negative entropy?

We had an experiment on the dissolution of calcium hydroxide and we obtained the experimental entropy value $$\pu{-203 J mol^-1 K^-1}$$ with a percent error of $$26.7\%.$$ Doesn't dissolving a solid in a solvent increase entropy?

• It usually does increase the entropy, but compound solubility usually does increase with temperature. It is not the case of $\ce{Ca(OH)2}$. May 21 at 7:33

When ionic compounds dissolve in water, water molecules form a complex ion around the calcium cation. The presence of the hydroxide anion has a similar effect, though the polarities of the water ligands is reversed. Thus, dissolution of Ca(OH)2 in water is actually a chemical reaction. In some cases, complex ions may be more structured than the crystals of undissolved ions. In such a case, the entropy change of solution will be negative.

Entropy can be measured and aproached differently, (i) from an equation that relates entropy and Enthalpy to measure the heat of reaction, the remainder of the work done per degree of Temperature (°) is

or if the change in enthalpy is measured then simply.

In this case the Vector of entropy has a negative direction hence an endothermic process and the magnitude is 203.

(ii) Considering the amount of microstates, Boltzmann constant would relate the energy and the temperature, between solid and liquid dissolution; As a function of Omega (W).

For Omega (W) evaluate the microstates in your solid sample and the liquid volume using Boltzmann Relation.

The change in Entropy between solid and liquid state is measure by Math approach, and is proven by Covention I and II (Physical Chemistry; Ira N. Levine Pg.295 6th Ed) for the Activity Coefficients on Gibbs free energy of standard solid state (S* solid 83.36 J/mol*K) and ideal model for gas state (S°gas,1 bar 285.62).

An increase in Entropy would be, then proven, as a solid State entropy is of approx. 83 J/molK and raise to 203 J/molK, through an endothermic process.