# Calculating Standard Gibbs Energy of Dissolution

The solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$ in water at $25\ \mathrm{^\circ C}$. What is the standard Gibbs energy of dissolution of the salt? The reaction is $$\ce{Hg2I2(s) -> Hg2^2+(aq) + 2I- (aq)}$$

Here is what I think I need in order to solve the question:

$$\Delta G=-RT\ln(K)$$ where $K$ will be the solubility constant. When the question says “solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$” I don’t exactly understand what this means.
The units make it seem like a concentration to me.

• The solubility is given in form of the concentration of $\ce{Hg_2^2+}$ that would result when maximum of mercurous iodide has dissolved. It helps you calculate the $K_{sp}$ which is the equilibrium constant of the reaction, and thereby calculate $\Delta G$. – stochastic13 Dec 18 '13 at 11:52
• @SatwikPasani is the solubility the concentration of just the $Hg_2^{2+}$, or is for both the $Hg_2^{2+}$ and $I^-$ as a total? – user66807 Dec 19 '13 at 1:25

## 1 Answer

You are correct that you need $\Delta G^\circ =-RT\ln{K}$ and that the equilibrium constant in this case is $K_{sp}$ the solubility product constant, which is defined for a process like this one to be:

$$\ce{A_{m} B_{n} (s) <=> mA^{n+}(aq) + nB^{m-}(aq)}$$ $$K_{sp} = \ce{[A^{n+}]^{m} [B^{m-}]^{n}}$$

In your case $K_{sp}=[\ce{Hg2^{2+}}][\ce{I-}]^2$

You are given the solubility (maximum obtainable concentration) of $\ce{Hg2I2}$ in water at $25 \ ^\circ \text{C}$. How can you use that information to find $[\ce{Hg2^{2+}}]$ and $[\ce{I-}]$ when the solution is saturated?

• Using what you've said this is what I've understood so far: If I let the concentration of $Hg_2^{2+}=S$, then the concentration of $I^-=2S$. Thus the total concentration is $3S=5.5X10^{-15}$ and $S=1.83X10^{-15}$. So to calculate the $K_{sp}$ I do $[1.83X10^{-15}][2*1.83X10^{-15}]^2=2.46X10^{-44}$. Am I on the right track, or completely off? Thanks! – user66807 Dec 18 '13 at 17:51
• You are given the concentration of $C=\ce{[Hg2I2]}=5.5\times10^{-15}\text{ M}$, since there is $1\ \ce{Hg2^{2+}}$ for every unit of $\ce{Hg2I2}$, then $[\ce{Hg2^{2+}}]=[\ce{Hg2^{2+}}]$ – Ben Norris Dec 19 '13 at 12:45
• @BenNorris Shouldn't we get a dimensionless quantity inside the natural logarithm function? – Mockingbird Feb 9 '17 at 10:04
• @Mockingbird - Yes. Equilibrium constants are unitless quantities because they should be calculated using activities instead of concentrations. Activities are unitless. Concentrations are easier, and for dilute solutions (like this one) concentrations are good approximations of activities. – Ben Norris Feb 9 '17 at 12:01