Chapter 17, Problem 48A

Interpretation Introduction

Interpretation:

The solubility product is to be calculated.

Concept Introduction:

The solubility product is the product of the concentration of its constituent ions with each ion concentration term raised to the power equal to the number of ions produced or dissociated.

Answer to Problem 48A

The solubility product of ${\text{ZnCO}}_{\text{3}}$ is $\text{7}{\text{.978×10}}^{\text{-13}}$ .

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

  $\begin{array}{l}{\text{ ZnCO}}_{\text{3}}\left(\text{s}\right)\rightleftharpoons {\text{Zn}}^{\text{+2}}\left(\text{aq}\right)\text{+}{\text{CO}}_{\text{3}}{}^{\text{-2}}\left(\text{aq}\right)\\ I\text{ 0 0}\\ C+x+x\\ Exx\end{array}$

Given information:

Solubility $\text{=1}{\text{.12×10}}^{\text{-4}}\text{g/L}$

Thus,

  $\begin{array}{c}S=\frac{\text{1}{\text{.12×10}}^{\text{-4}}\text{g/L}}{\text{125}\text{.38}\text{mol/L}}\\ =\text{0}{\text{.0089×10}}^{\text{-4}}\text{mol/L}\end{array}$

The Molar mass of zinc carbonate = $\text{125}\text{.38}\text{g/mol}$

So, the solubility of zinc ion and carbonate ion is equal $\text{0}{\text{.0089×10}}^{\text{-4}}\text{mol/L}$ .

The solubility product is calculated as:

  $\begin{array}{c}{\text{K}}_{\text{sp}}\text{=}\left[{\text{Zn}}^{\text{+2}}\right]\text{×}\left[{\text{CO}}_{\text{3}}{}^{\text{-2}}\right]\\ {\text{K}}_{\text{sp}}\text{=}\left(\text{0}{\text{.0089×10}}^{\text{-4}}\right)\text{×}\left(\text{0}{\text{.0089×10}}^{\text{-4}}\right)\\ {\text{K}}_{\text{sp}}\text{=0}{\text{.00007978×10}}^{\text{-8}}\\ {\text{K}}_{\text{sp}}\text{=}\text{7}{\text{.978×10}}^{\text{-13}}\end{array}$

The solubility product of ${\text{ZnCO}}_{\text{3}}$ is $\text{7}{\text{.978×10}}^{\text{-13}}$ .

Conclusion

The solubility product of ${\text{ZnCO}}_{\text{3}}$ is $\text{7}{\text{.978×10}}^{\text{-13}}$ .