Chapter 18, Problem 64E

Interpretation Introduction

(a)

Interpretation:

When the ${\text{K}}_{\text{sp}}=8.7\times {10}^{-9}$ for calcium carbonate, the solubility in moles per liter is to be calculated.

Concept introduction:

Solubility product is the product of concentrations of ions raised to the power of their stoichiometric coefficients at equilibrium. It is represented by $K_{\text{sp}}$. It is applicable for saturated solutions.

Answer to Problem 64E

When the ${\text{K}}_{\text{sp}}=8.7\times {10}^{-9}$ for calcium carbonate, the solubility in moles per liter is $9.327\times {10}^{-5}\text{mol/L}$.

Explanation of Solution

The equilibrium equation for calcium carbonate is shown below.

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

The solubility product constant for calcium carbonate is calculated by the formula shown below.

${\text{K}}_{\text{sp}}=\left[{\text{Ca}}^{2+}\right]\left[{\text{CO}}_3{}^{2-}\right]$…(1)

Where,

• $\left[{\text{Ca}}^{2+}\right]$ is the molar concentration of calcium ion.

• $\left[{\text{CO}}_3{}^{2-}\right]$ is the molar concentration of carbonate ion.

The value of ${\text{K}}_{\text{sp}}=8.7\times {10}^{-9}$.

Suppose the solubility of ${\text{CaCO}}_{\text{3}}$ is $\text{x}\text{mol/L}$.

In the above reaction, the number of moles of ${\text{Ca}}^{2+}$ and ${\text{CO}}_{\text{3}}{}^{2-}$ produced are equal to the number of moles of ${\text{CaCO}}_{\text{3}}$.

Therefore, the concentration of ${\text{CaCO}}_{\text{3}}$ is equal to the concentrations of ${\text{Ca}}^{2+}$ and ${\text{CO}}_{\text{3}}{}^{2-}$

Substitute the values of ${\text{K}}_{\text{sp}}$ and concentrations of ${\text{Ca}}^{2+}$ and ${\text{CO}}_{\text{3}}{}^{2-}$ in equation (1).

$\begin{array}{rll}\text{8}\text{.7}\times {\text{10}}^{-9}& =& \text{x}\times \text{x}\\ \text{8}\text{.7}\times {\text{10}}^{-9}& =& {\text{x}}^2\\ \sqrt{\text{8}\text{.7}\times {\text{10}}^{-9}}& =& \text{x}\\ \text{x}& =& 9.327\times {10}^{-5}\text{mol/L}\end{array}$

So, the molar solubility of ${\text{CaCO}}_{\text{3}}$ is $9.327\times {10}^{-5}\text{mol/L}$.

Conclusion

The solubility in moles per liter is $9.327\times {10}^{-5}\text{mol/L}$.

Interpretation Introduction

(b)

Interpretation:

When the ${\text{K}}_{\text{sp}}=8.7\times {10}^{-9}$ for calcium carbonate, the solubility in grams per $100\text{mL}$ is to be calculated

Concept introduction:

Solubility product is the product of concentrations of ions raised to the power of their stoichiometric coefficients at equilibrium. It is represented by $K_{\text{sp}}$. It is applicable for saturated solutions.

Answer to Problem 64E

When the ${\text{K}}_{\text{sp}}=8.7\times {10}^{-9}$ for calcium carbonate, the solubility in grams per $100\text{mL}$ is $9.327\times {10}^{-4}\text{g per 100 mL}$.

Explanation of Solution

The formula to convert mole into mass is shown below.

$\text{Mass}=\left(\text{Moles}\times \text{Molar}\text{mass}\right)$

Substitute the value of moles of ${\text{CaCO}}_{\text{3}}$ in the above expression.

$\begin{array}{rll}\text{Mass}& =& \left(9.327\times {10}^{-5}\text{mol/L}\times \text{100}\text{g}/\text{mol}\right)\\ & =& 9.327\times {10}^{-3}\text{g/L}\end{array}$

The conversion of $\text{L}$ into $\text{mL}$ is done as shown below.

$1\text{L}=1000\text{mL}$

Therefore, $9.327\times {10}^{-3}\text{g/L}$ is converted to $\text{mL}$ as shown below.

$\frac{9.327\times {10}^{-3}\text{g/L}}{1000\text{mL}}=9.327\times {10}^{-6}\text{g/mL}$

So, the solubility in grams per $100\text{mL}$ is shown below.

$\begin{array}{r}9.327\times {10}^{-6}\text{g/mL}\times \text{100}\text{mL}\\ =9.327\times {10}^{-4}\text{g per 100 mL}\end{array}$

The solubility is $9.327\times {10}^{-4}\text{g per 100 mL}$.

Conclusion

The solubility in grams per $100\text{mL}$ is $9.327\times {10}^{-4}\text{g per 100 mL}$.