The well water in an area is "hard" because it is in equilibrium with ${\mathbf{CaCO}}_{\mathbf{3}}$in the surrounding rocks. What is the concentration of ${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}$in the well water (assuming the water'sis such that the $\mathbf{C}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}$ion is not significantly protonated)? (See Appendix$\mathbf{C}$for ${\mathbf{K}}_{\mathbf{sp}}$of${\mathbf{CaCO}}_{\mathbf{3}}$.)

A solid's solubility (commonly referred to as molar solubility) is measured as the concentration of the "dissolved solute" in a saturated solution. The solubility of a sparingly soluble salt can be estimated from the solubility product. The solubility product is the product of the concentrations of the constituent ions present in the solution raised to their stoichiometric coefficients.

The chemical equilibrium between the ${\mathrm{CaCO}}_3$and its constituent ions is given below.

data-custom-editor="chemistry" ${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ca}}^{2+}{\left(\mathrm{aq}\right)+\mathrm{CO}}_3^{2-}\left(\mathrm{aq}\right)$

Let us prepare an ICE table to estimate the equilibrium concentration.

Using the ${\mathrm{K}}_{\mathrm{sp}}$value we can calculate the solubility of ${\mathrm{Ca}}^{+2}$ions.

$\begin{array}{c}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{CO}}_3^{2-}\right]\\ 1.4\times {10}^{-8}=\mathrm{S}\times \mathrm{S}\\ \mathrm{S}=\sqrt{1.4\times {10}^{-8}}\\ =1.18\times {10}^{-4}\mathrm{M}\end{array}$

In the well water, the concentration ${\mathrm{Ca}}^{2+}$of is $1.18\times {10}^{-4}\mathrm{M}$.

Therefore, the required value is $1.18\times {10}^{-4}\mathrm{M}$.