# Why doesn't the expermintal molar solubility match the one calculated by the Ksp value?

For $$\ce{HgCl2}$$ ($$M = \pu{271.52 g mol-1}$$), the reported $$K_\mathrm{sp}$$ value is $$10^{-15.69}$$.

This means that the calculated molar solubility of $$\ce{HgCl2}$$ in water is $$(10^{-15.69}/4)^{1/3}\cdot\pu{271.52 g mol-1} \approx \pu{0.0010 g L-1}$$, or equivalently, $$\pu{0.010 g cm-3}$$.

However, in Pubchem, the reported molar solubility is 6.9 g/100 cc. What is the source of the large deviation between the experimental and the theoretically calculated molar solubility of $$\ce{HgCl2}$$?

• It does match, you're confusing HgCl2 - soluble and Hg2Cl2 - almost insoluble. – Mithoron Aug 4 '19 at 0:03
• What do you mean. Both of my data values are for $\ce{HgCl2}$. – DrPepper Aug 4 '19 at 3:59
• Just use, at very least, Wikipedia instead of some obscure and dubious sources. – Mithoron Aug 4 '19 at 19:39
• @Mithoron bruh Wikipedia doesn't have Ksp data for it smh – DrPepper Aug 5 '19 at 0:37

The $$\mathrm{p}K_\mathrm{sp}$$ value of $$15.69$$ for $$\ce{HgCl2}$$ given by your reference must have an error. First of all, $$\ce{HgCl2}$$ is fairly soluble in water at $$\pu{20 ^\circ C}$$. Wikipedia lists it as $$\pu{65.7 g L-1}$$, yet lists solubility of $$\ce{Hg2Cl2}$$ as $$\pu{3.25e-4 g L-1}$$. Meantime, I found an University of Arkansas, Little Rock website, which has listed $$K_\mathrm{sp}$$ value of $$\ce{Hg2Cl2}$$ as $$\pu{1.43e-18}$$ ($$\mathrm{p}K_\mathrm{sp} = 17.84$$).
The molar mass of $$\ce{Hg2Cl2}$$ is $$\pu{472.09 g mol-1}$$. Thus, theoretically, the solubility ($$s$$) of $$\ce{Hg2Cl2}$$ can be calculated similar way you did:
$$s = \left({\frac{\pu{1.43e-18 mol3 L-3}}{4}}\right)^{\frac13}\times \pu{472.09 g mol-1} = \pu{3.35e-4 g L-1}$$