Q19.132CP - Chemistry: Molecular Nature Of Matter And Change

Question

Calcium ion present in water supplies is easily precipitated as calcite $\left({\text{CaCO}}_{\text{3}}\right)$:

 ${\mathbf{\text{Ca}}}^{\mathbf{\text{2 + }}}\mathbf{\text{(aq) + C}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{2 - }}}\mathbf{\text{(aq)}}\mathbf{\rightleftharpoons }{\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}\mathbf{\text{(s)}}$. Because the K_sp  decreases with temperature, heating hard water forms a calcite "scale," which clogs pipes and water heaters. Find the solubility of calcite in water

(a) At $\mathbf{10}\mathbf{°}\mathbf{C}\left(K_{\mathrm{sp}}=4.4\times {10}^{-9}\right)$  and 

(b) At  $\mathbf{30}\mathbf{°}\mathbf{C}\left(K_{\mathrm{sp}}=3.1\times {10}^{-9}\right)$.

Step-by-Step Solution

Answer

The solubility of calcite in water for given conditions are:

a) The required solubility is $6.6\times {10}^{-5} \mathrm{M}$.

b) The required solubility is $5.6\times {10}^{-5} \mathrm{M}$.

Solubility product

For a sparingly soluble salt, the solubility product of the salt is equal to the product of the constituent ions raised to their stoichiometric coefficients.

Find the solubility of calcite in water 10 ° C ( K sp = 4 . 4 × 10 - 9 )

a) The dissolution equation for calcium carbonate is given below.

 ${\text{CaCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{\text{ + 2}}\left(\text{aq}\right)\text{ + C}{{\text{O}}_{\text{3}}}^{\text{ - 2}}\left(\text{aq}\right)$

In order to find the solubility in water at $\mathbf{10}\mathbf{°}\mathbf{C}$ we use the K_sp expression. Let the solubility of ${\text{Ca}}^{\text{ + 2}}$ and ${\text{Cl}}^{\text{ - }}$  ions be S.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\text{Ca}}^{2+}\right]\left[\mathrm{C}{\text{O}}_3^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{S}}^2 \end{gathered}$$

$S^2=4.4\times {10}^{-9}$

$S=6.6\times {10}^{-5}\mathrm{M}$

Therefore, the solubility is $6.6\times {10}^{-5}\mathrm{M}$.

Find the solubility of calcite in water 30 ° C ( K sp = 3 . 1 × 10 - 9 )

b) 

Firstly, we can write the precipitation equation for ${\text{CaCO}}_3$ :

 ${\text{Ca}}^{2+}\left(aq\right)+{\text{CO}}_3^{2-}\left(aq\right)\leftrightharpoons {\text{CaCO}}_3\left(s\right)$

We are going to use the expression for  K_sp to find the solubility at $30°\mathrm{C}$,

$3.1\times {10}^{-9}=S^2$

 $\mathrm{S}=5.6\times {10}^{-5}\mathrm{M}$

Therefore, the required solubility is $5.6\times {10}^{-5}\mathrm{M}$.