Chapter 18, Problem 98A

Interpretation Introduction

Interpretation: If preparing a solution by combining $0.025\text{ mol}$ ${\text{CaCl}}_2$ and $0.015\text{ mol}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2$ in $1.0\text{ L}$ water, will it precipitate out ${\text{PbCl}}_{\text{2}}$ is to be explained.

Concept introduction: The solubility product $\left(K_{\text{sp}}\right)$ is the product of the concentrations of ions in the salt, with each concentration multiplied by the dissociation reaction coefficient for that ion. The lower value of the solubility product constant decreases the compound's solubility.

Answer to Problem 98A

The ionic product is larger than the solubility product constant, hence, the solution will form ${\text{PbCl}}_{\text{2}}$ precipitate.

Explanation of Solution

A solution is prepared by combining $0.025\text{ mol}$ ${\text{CaCl}}_2$ and $0.015\text{ mol}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2$ in $1.0\text{ L}$ water.

  $0.015\text{ mol}$ of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2$ gives $0.015\text{ mol}$ of ${\text{Pb}}^{2+}$ ions.

  $0.025\text{ mol}$ of ${\text{CaCl}}_2$ gives $0.05\text{ mol}$ of ${\text{Cl}}^{-}$ ions.

The formula to calculate the ionic product of ${\text{PbCl}}_{\text{2}}$ is as follows:

  $IP=\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^2$

Where,

  $IP$ is the ionic product.

  $\left[{\text{Pb}}^{2+}\right]$ is the concentration of ${\text{Pb}}^{2+}$ ion.

  $\left[{\text{Cl}}^{-}\right]$ is the concentration of ${\text{Cl}}^{-}$ ion.

Substitute the values of $\left[{\text{Pb}}^{2+}\right]$ and $\left[{\text{Cl}}^{-}\right]$ in the above formula.

  $\begin{array}{c}IP=\left[{\text{Pb}}^{2+}\right]{\left[{\text{Cl}}^{-}\right]}^2\\ =\left(0.015\right){\left(0.05\right)}^2\\ =0.015\times 0.0025\\ =3.75\times {10}^{-7}\end{array}$

Ionic product for ${\text{PbCl}}_{\text{2}}$ is $3.75\times {10}^{-7}$ .

The solubility product constant for ${\text{PbCl}}_{\text{2}}$ is $1.6\times {10}^{-5}$ .

The ionic product of ${\text{PbCl}}_{\text{2}}$ is larger than the solubility product constant. Therefore, ${\text{PbCl}}_{\text{2}}$ form a precipitate.

Conclusion

The ionic product is larger than the solubility product constant, hence, the solution will form ${\text{PbCl}}_{\text{2}}$ precipitate.