Chapter 18, Problem 67E

Interpretation Introduction

Interpretation:

The amount of calcium oxalate that dissolves in $2.5\times {10}^2\text{mL}$ of $0.22\text{M}{\text{Na}}_2{\text{C}}_2{\text{O}}_4$ if the ${\text{K}}_{\text{sp}}$ for ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is $2.4\times {10}^{-9}$, 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 ${\text{K}}_{\text{sp}}$. It is applicable for saturated solutions.

Answer to Problem 67E

The amount of calcium oxalate that will dissolve in $2.5\times {10}^2\text{mL}$ of $0.22\text{M}{\text{Na}}_2{\text{C}}_2{\text{O}}_4$, is $3.5\times {10}^{-7}\text{g}$.

Explanation of Solution

The equilibrium equation for calcium oxalate is shown below.

${\text{CaC}}_{\text{2}}{\text{O}}_4\left(\text{s}\right)\rightleftharpoons {\text{Ca}}^{2+}\left(\text{aq}\right)+{\text{C}}_{\text{2}}{\text{O}}_4^{2-}\left(\text{aq}\right)$

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

${\text{K}}_{\text{sp}}=\left[{\text{Ca}}^{2+}\right]\left[{\text{C}}_{\text{2}}{\text{O}}_4^{2-}\right]$…(1)

Where,

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

• $\left[{\text{C}}_{\text{2}}{\text{O}}_4^{2-}\right]$ is the molar concentration of oxalate ion.

The given value of ${\text{K}}_{\text{sp}}=2.4\times {10}^{-9}$.

The equilibrium equation for sodium oxalate is shown below.

${\text{Na}}_2{\text{C}}_{\text{2}}{\text{O}}_4\left(\text{s}\right)\rightleftharpoons 2{\text{Na}}^{+}\left(\text{aq}\right)+{\text{C}}_{\text{2}}{\text{O}}_4^{2-}\left(\text{aq}\right)$

In the above reaction, the concentration of ${\text{Na}}_2{\text{C}}_2{\text{O}}_4$ is equal to the concentration of ${\text{Na}}^{+}$ and ${\text{C}}_{\text{2}}{\text{O}}_4^{2-}$. Since the concentration of ${\text{Na}}_2{\text{C}}_2{\text{O}}_4$ is $0.22\text{M}$, so, the concentration of ${\text{C}}_{\text{2}}{\text{O}}_4^{2-}$ is also $0.22\text{M}$.

Substitute the values of ${\text{K}}_{\text{sp}}$ and concentration of ${\text{C}}_{\text{2}}{\text{O}}_4^{2-}$ in equation (1).

$\begin{array}{rll}2.4\times {10}^{-9}& =& \left[{\text{Ca}}^{2+}\right]\times 0.22\text{M}\\ \left[{\text{Ca}}^{2+}\right]& =& \frac{2.4\times {10}^{-9}}{0.22\text{M}}\\ & =& 1.09\times {10}^{-8}\text{M}\end{array}$

The concentration of $\left[{\text{Ca}}^{2+}\right]$ is $1.09\times {10}^{-8}\text{M}$.

According to the equilibrium equation for calcium oxalate shown above, the concentration of ${\text{CaC}}_{\text{2}}{\text{O}}_4$ must be equal to the concentration of $\left[{\text{Ca}}^{2+}\right]$. So, the concentration of ${\text{CaC}}_{\text{2}}{\text{O}}_4$ is $1.09\times {10}^{-8}\text{M}$.

The molar mass of ${\text{CaC}}_{\text{2}}{\text{O}}_4$ is $128.10\text{g}/\text{mol}$.

The volume of the solution is $2.5\times {10}^2\text{mL}$.

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

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

So, the conversion of $2.5\times {10}^2\text{mL}$ into $\text{L}$ is shown below.

$\frac{1\text{L}}{1000\text{mL}}\times 2.5\times {10}^2\text{mL}=\text{0}\text{.25}\text{L}$

The volume of the solution is $\text{0}\text{.25}\text{L}$.

The number of moles of solute is calculated as shown below.

$\text{Moles}=\frac{\text{Mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$…(2)

The molarity of the solution is calculated as shown below.

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{L}}$ …(3)

Substitute equation (2) in equation (3) as shown below.

$\text{Molarity}=\frac{\frac{\text{Mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}}{\text{Volume}\text{of}\text{solution}\text{in}\text{L}}$

Substitute the values of molarity, molar mass and volume of solution in above equation as shown below.

$\begin{array}{rll}\text{Molarity}& =& \frac{\frac{\text{Mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}}{\text{Volume}\text{of}\text{solution}\text{in}\text{L}}\\ 1.09\times {10}^{-8}\text{M}& =& \frac{\frac{\text{Mass}\text{of}\text{solute}}{\text{128}\text{.10}\text{g}/\text{mol}}}{\text{0}\text{.25}\text{L}}\end{array}$

Therefore, the mass of the solute is calculated as shown below.

$\begin{array}{rll}\text{Mass}\text{of}\text{solute}& =& 1.09\times {10}^{-8}\text{M}\times \text{128}\text{.10}\text{g}/\text{mol}\times \text{0}\text{.25}\text{L}\\ & =& 3.49\times {10}^{-7}\text{g}\\ & \approx & \text{3}\text{.5}\times {\text{10}}^{-7}\text{g}\end{array}$

Conclusion

The amount of calcium oxalate required is $\text{3}\text{.5}\times {\text{10}}^{-7}\text{g}$.