# 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

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