Chapter 17, Problem 51A

Interpretation Introduction

Interpretation : The solubility of lead chromate in grams/L at 25°C needs to be determined.

Concept Introduction : The solubility product can be defined as the product of the equilibrium concentrations of the ions in a saturated solution of a salt.  In the expression of solubility product each concentration is raised to the power of their coefficient of ion as given in the balanced equation. For example; $\text{BaIO3}$ will show the equilibrium equation as:

  $\text{BaIO3(s)}\leftrightarrow {\text{Ba}}^{\text{+2}}{\text{(aq)+ IO3}}^{\text{-2}}\text{(aq)}$

The solubility expression will be:

  ${\text{Ksp= [Ba}}^{\text{+2}}{\text{] [ IO3}}^{\text{-2}}\text{]}$

The product of the concentrations of the ions at any moment in time for the given salt is called as ion product that is represented as $\text{Qsp}$ . The expression for it is also same as $\text{Ksp}$ but this is not at equilibrium condition.

Answer to Problem 51A

  $\text{1}{\text{.7×10}}^{\text{-4}}\text{ g/L}$

Explanation of Solution

Given information:Chromate ion is used as a qualitative test for lead (II) ions as it forms bright yellow precipitate of lead chromate.

${\text{K}}_{\text{sp}}$ for ${\text{PbCrO}}_{\text{4}}$ = $2.8\times {10}^{-13}$

  ${\text{PbCrO}}_{\text{4}}\to {\text{Pb}}^{\text{2+ }}{\text{ + CrO}}_{{}_{\text{4}}}^{-2}$

Thus, the ${\text{K}}_{\text{sp}}$ expression should be:

  $\begin{array}{c}{\text{PbCrO}}_{\text{4}}\to {\text{Pb}}^{\text{2+ }}{\text{ + CrO}}_4^{2-}\\ {\text{K}}_{\text{sp}}={\text{[Pb}}^{\text{2+ }}{\text{] [CrO}}_4^{2-}\text{]}\\ {\text{K}}_{\text{sp}}=[s][s]=s^2\\ \text{2}{\text{.8×10}}^{\text{-13}}=s^2\\ s\text{ = 5}{\text{.3×10}}^{\text{-7}}\text{M}\end{array}$

Molar mass of ${\text{PbCrO}}_{\text{4}}$ = 323.19 g/mol

Calculate the solubility in g/L:

  $\frac{\text{5}{\text{.3×10}}^{\text{-7}}\text{mole}}{\text{1 L}}\text{×}\frac{\text{323}\text{.19 g}}{\text{1 mole}}\text{ = 1}{\text{.7×10}}^{\text{-4}}\text{ g/L}$

Conclusion

Thus, the solubility is $\text{1}{\text{.7×10}}^{\text{-4}}\text{ g/L}$ .