9.2: Background

Background

In this lab students will calculate the solubility product (K_sp) for an slightly soluble hydroxide salt (Ca(OH)₂) at two temperatures and use that information to calculate the standard state Gibbs Free Energy, entropy and enthalpy for the dissolution reaction. The ion concentrations will be determined by measuring the pH.

Solubility Product

For the slightly soluble salt calcium hydroxide the equilibrium equation can be written as:

\[CaOH_{2}(s) \leftrightharpoons Ca^{+2}(aq) + 2OH^{-}(aq) \label{tag1}\]

The equilibrium constant expression (sec. 15.2 & 17.4.2) is:

\[K_{sp}=[Ca^{+2m}][OH^{-n}]^2 \label{eqtag2}\]

 

We can measure [OH^-] by measuring pH, noting

pH + pOH = 14, so pOH = 14-pH:

so

\[ [OH^-]=10^{-pOH}=10^{-(14-pH)}=10^{pH-14} \]

From  \ref{tag1}:

\[ [Ca^{+2}]=\frac{[OH^-]}{2} \\ so \\ [Ca^{+2}]= \frac{10^{pH-14}}{2}  \]

Substituting into K_sp (\ref{eqtag2} gives:

\[K_{sp}=[Ca^{+2}][OH^-]^2= \frac{10^{pH-14}}{2}\left ( 10^{pH-14}  \right )^2= \frac{\left ( 10^{pH-14}  \right )^3}{2}\]

 

Free Energy

Relating K to \(\Delta G^o\) (sec. 18.5.4.1)

\[ \Delta G^o =-RTlnK \label{eqtag3} \]

We also know from section 18.5.2 that

\[\Delta G^o =\Delta H^o - T\Delta S^o \label{eqtag4}\]

Eq. \ref{eqtag3} calculates \( \Delta G^o\) as a function of T by measuring K at a given T. In \ref{eqtag4} we can calculate the temperature dependence of \(\Delta G^o\) on \(\Delta H^o\)  and \(\Delta S^o\), but it needs to be noted that in this treatment we are considering both \(\Delta H^o\)  or \(\Delta S^o\) to be constant with respect to temperature.  This is normally true for the enthalpy but is often not true for the entropy, especially in the gas phase. 

We can rearrange Gibbs Free Energy Expression so it is equal the constant Enthalpy value

\[\Delta H^o = \Delta G^o +T \Delta S^o \label{eqtag5}\]

If we measure K_sp at two temperature we can use eq. \ref{eqtag3} to calculate the free energy at those two temperatures, which when substituted into eq \ref{eqtag5} gives:

\[\Delta H^o = \Delta G_1^o +T_1 \Delta S^o \\ \; \\ \Delta H^o = \Delta G_2^o +T_2 \Delta S^o\]

Equating these gives:

\[ \Delta G_1^o +T_1 \Delta S^o = \Delta G_2^o +T _2\Delta S^o \] 

The only unknown is ∆S^o, which can now be calculated:

\[\Delta S^o=\frac{\Delta G_1^o - \Delta G_2^o}{T_2-T_1}\]

Now that we know one \(\Delta G^o\) and \(\Delta G^o\) we can solver for ∆H^o at any temperature. 

\[\Delta H^o = \Delta G^o +T \Delta S^o\]

 

Theoretical Values

In this experiment you are going to calculate the percent error for  ∆G^o_rxn,  ∆H^o_rxn and ∆S^o_rxn.  The theoretical value will be determined from the standard thermodynamic formation values data at 298K.  The reaction we are studying is:

\[Ca(OH)_2(s) \leftrightharpoons Ca(OH)_2(aq)\] 

Standard thermodynamic formation energies at 298 K

Species	∆H^of (kJ/mol)	S^of (J/mol-K)	∆G^of (kJ/mol)
Ca(OH)2(s)	-986.09	83.39	-898.49
Ca(OH)2(aq)	-1002.82	-74.5	-868.07

We will consider  ∆H^o_rxn and ∆S^o_rxn to be independent of temperature and so you will use the above values of  ∆H^o_f and S^o_f at 25^oC to calculate  ∆H^o_rxn and ∆S^o_rxn at both temperatures.  On the other hand ∆G^o_rxn is dependent on temperature and so you will need to calculate it at both of the temperature values you measured the pH at using Gibb's Free energy equation:

\[\Delta G_{rxn}^{o} =\Delta H_{rxn}^{o}-T\Delta S_{rxn}^{o}\]

 

Percent Error

In this lab you will be required to report the percent error, which is defined as:

\[\% Error=\left | \frac{theoretical-experimental}{theoretical} \right |100\]

Entropy of Solution

At first glance you would think that the entropy of dissolving solid calcium hydroxide would be positive because the ions start in the structured form of a crystal and end up dispersed by the solvent, but in fact the entropy of dissolution is negative and the product (dissolved) state is more structured than the reactant state. The reason is that you need to properly identify the system, which is the solute (Ca(OH)₂) and the solvent (H₂O), and it is the water that is driving the decrease in entropy.  Before dissolution you have a very structured solute and a fairly unstructured solvent.  As the solute dissolves both the cation and the anion interact with water. The cation (Ca⁺²) functions as a Lewis base and forms a complex ion (last lab) with the water which functions as a Lewis base and donates electrons to the vacant metal valence orbitals (forming a coordinate covalent bond) and thus the water becomes more structured.  Due to ion-dipole interactions the complex ion also orients water molecules around it in a manner that maximizes attractions and minimizes repulsions (review section 11.2) .  The anion (OH^-) is part of the dissociation of water and according to Le Chatlier's principle (review section 15.6) the added hydroxide will push the water equilibria to the undissociated side and thus further reduce entropy (OH^- + H⁺ --> H₂O).  So the solute is undergoing an increase in entropy while the solvent is undergoing a decrease, and the latter effect is the greater, resulting in an overall decrease in entropy for the system.