# Gibbs Free Energy of Solutes and Solutions

Using the equation $$\ce{CaSO4 (s) <=> Ca^2+ (aq) + SO4^2- (aq)},$$ and $\Delta H/\Delta S$ values, calculate $\Delta G$ at $50~^\circ\mathrm{C}$ when the solution is saturated with $\ce{Ca^2+}$ and $\ce{SO4^2-}$.

My Attempt $$\Delta G = \Delta G^\circ + RT \ln Q$$ I calculated $\Delta G^\circ$ at $50\ \mathrm{^\circ C}$, But I am confused as to how we know what $Q$ is if there is no reactant then does $Q$ not become $\ce{[Ca^2+][SO4^2- ]/0}$, but this is impossible.

• without reactant how can we get product? Any example? – Freddy Apr 10 '15 at 9:57
• No I do not think so. – user15480 Apr 10 '15 at 11:20
• This problem might be much easier than you think. If the solution is "saturated", the dissolution process is at equilibrium. What do you know about $\Delta G$ for processes at equilibrium? – Curt F. May 11 '15 at 4:41

For the reaction $$\ce{CaSO4 (s) <=> Ca^2+ (aq) + SO4^{2-} (aq)},$$ the appropriate equilibrium constant using activities can be set up as $$K_\mathrm{eq}=\frac{a(\ce{Ca^2+})\,a(\ce{SO4^2-})}{a(\ce{CaSO4})}.$$

The activity $a$ of a pure substances in condensed phases is approximately one, since their behaviour can be often treated ideally, i.e. the chemical potential $\mu$ at the considered conditions is approximately the same as the chemical potential at standard conditions $\mu^\circ$. This derives directly from the definition of the relative activity: $$a = \exp\left\{\frac{\mu-\mu^\circ}{\mathcal{R}T}\right\}$$

For substances where the solubility is very low, concentrations of the ions are very low, so in a further approximation we can assume that in these solutions the activity coefficient is approximately one, therefore, we can write concentrations instead of activities and obtain for the equilibrium constant the expression, that is also known a solubility product. $$K_\mathrm{eq} \approx K_\mathrm{sp} = c(\ce{Ca^2+})\,c(\ce{SO4^2-})$$

However, this is completely irrelevant to the question itself. Since the solution is saturated, it also means, that the solid is in equilibrium with the solution, which means that the kinetics of dissolution and precipitation are equal of magnitude. The Gibbs energy has a very distinct value for these equations.

chemical equilibrium
Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy, $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:
$$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + \mathcal{R}T\,\ln\,K = 0\\\Delta G_\mathrm{r}^\circ = - \mathcal{R}T\,\ln\,K\\$$ The equilibrium constant, $K$, is given by the mass-law effect.
Source: IUPAC goldbook

Our solution is at equilibrium so $\ce {Q} = {K}$. That is, $\ce{K_{eq}} = {K_{sp}}$. $$\Delta G = \Delta G^\circ + RT \ln K$$ At equilibrium, $\Delta G = 0$.

So, $$\Delta G^\circ = - RT \ln K$$ Now, you can use the value of $\ce{K_{sp}}$ and the other known values given to determine $\Delta G^\circ$.

Your issue is that you are falsely considering the $\ce{CaSO4}$ to be part of the system. The system under consideration is the solution, which the solid is not a part of. This is the idea of the solubility product.

Your reaction quotient is therefore: $Q= \ce{[Ca^{2+}]·[{SO4}^{2-}]}$

• I believe this is wrong. In a saturated solution, there always will be/ has to be solid. This is of course part of the system, as it is in equilibrium with it. That is what the chemical equation is all about. – Martin - マーチン May 11 '15 at 4:39
• @Martin while I do agree that the solid is part of the overall equilibrium system, the solid is not part of the solution. You cannot measure the concentration of something that is not dissolved in solution. – ringo May 11 '15 at 5:24
• While it is certainly correct, that you cannot measure the concentration of something, that is not in solution, neglecting the conditions, that lead to the equilibrium is wrong. And the explanation of the solubility product therefore is wrong. – Martin - マーチン May 11 '15 at 13:14

You can leave this solid phase alone. Formally it's part of Nernst equation of equilibrium and for the formality's sake you can put it there, BUT its activity is ONE, so it really doesn't influence the description of the equilibrium.