Chapter 17, Problem 70A

Interpretation Introduction

Interpretation:

The solubility of lead chromate is to be determined when the solubility product is given at a particular temperature.

Concept Introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

The solubility product is given as

  $\text{Ksp = }\left[{\text{Pb}}^{\text{+2}}\right]\text{×}\left[{\text{CrO}}_{\text{4}}{}^{\text{-2}}\right]$

Answer to Problem 70A

The solubility of lead chromate is $\text{4}{\text{.795 × 10}}^{\text{-7 }}\text{mol/L}$ .

Explanation of Solution

Given information: The solubility product of lead chromate is $\text{2}{\text{.3×10}}^{\text{-13}}$ .

PbCrO₄ Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

  $\begin{array}{l}{\text{ PbCrO}}_{\text{4}}\rightleftharpoons {\text{Pb}}^{\text{2+}}{\text{+CrO}}_{\text{4}}{}^{\text{-2}}\\ I\text{ I 0 0}\\ C-x\text{ +}x\text{ +}x\\ EI-xxx\end{array}$

The solubility product is given as

  ${\text{K}}_{\text{sp}}\text{= }\left[{\text{Pb}}^{\text{2+}}\right]\text{×}\left[{\text{CrO}}_{\text{4}}{}^{\text{2-}}\right]$

  $\text{= x×x}$

  ${\text{= x}}^{\text{2}}$

  $\sqrt{\text{Ksp}}{\text{ = x}}^{}$

Or,

  $\text{x = }\sqrt{\text{2}{\text{.3 × 10}}^{\text{-13}}}$

  $\text{= 4}{\text{.795 × 10}}^{\text{-7}}\text{mol/L}$

Conclusion

The solubility of lead chromate is $\text{4}{\text{.795 × 10}}^{\text{-7}}\text{mol/L}$ .