Question

Slaked lime $\left[{\text{Ca(OH)}}_{\text{2}}\right]$is used to soften hard water by removing calcium ions from hard water through the reaction

${\mathbf{\text{Ca(OH)}}}_{\mathbf{\text{2}}}{\mathbf{\text{(aq) + Ca}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq) + 2HC}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}}\mathbf{\to }$${\mathbf{\text{2CaCO}}}_{\mathbf{\text{3}}}{\mathbf{\text{(s) + 2H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}$

Although ${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}\mathbf{\text{(s)}}$is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of ${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}\mathbf{\text{(s)}}$in water$\left(K_{sp}=8.7\times {10}^{-9}\right)$

Short answer

The molar solubility of $CaC{O}_3$ is $9.3\times {10}^{-5}mol/L$.

Step-by-step solution

Definition for Molar solubility

Molar solubility is directly related to the solubility product, and it is the number of moles of the solute that can be dissolved per liter of the solution before the solution becomes saturated.

To find the molar solubility of $CaC{O}_3$, we must find the equilibrium concentrations of $C{a}^{2+}$and$\text{C}{{\text{O}}_3}^{2-}$

The dissociation equation is shown below.
${\text{CaCO}}_{\text{3}}\text{(s)}\rightleftharpoons {\text{Ca}}^{\text{2 +}}{\text{(aq) + CO}}_{\text{3}}^{\text{2 -}}\text{(aq)}$

Calculation for molar solubility of CaCO3

We can write an equilibrium expression for this process according to the law of mass action:

${\text{K}}_{\text{sp}}\text{=}\left[{\text{Ca}}^{\text{2 +}}\right]\left[{\text{CO}}_{\text{3}}^{\text{2 -}}\right]$

Where $\left[{\text{Ca}}^{2+}\right]$ and $\left[{\text{CO}}_3^{2-}\right]$ are expressed in $mol/L$.

The constant $K_{sp}$ is called the solubility product constant, or simply the solubility product for the equilibrium expression.

$K_{sp}=8.7\times {10}^{-3}$

We can write $\left[{\text{Ca}}^{\text{2 +}}\right]\text{=}\left[{\text{CO}}_{\text{3}}^{\text{2 -}}\right]\text{= s}$, to simplify the equation shown above.

${\text{K}}_{\text{sp}}\text{=}\left[{\text{Ca}}^{\text{2 +}}\right]\left[{\text{CO}}_{\text{3}}^{\text{2 -}}\right]{\text{=s×s=s}}^2$

$s=9.3\times {10}^{-5}$

Therefore, the molar solubility of ${\mathrm{CaCO}}_3$is$s=9.3\times {10}^{-5}$.