Chapter 18, Problem 12SAQ

Interpretation Introduction

Interpretation:

The molar solubility of lead (II) bromide $\left(PbB{r}_2\right)$ for lead (II) bromine ${\text{K}}_{\text{sp}}\text{=4}{\text{.67×10}}^{-6}$ should be calculated.

Concept introduction:

Molar solubility is defined as the number of moles of given substance dissolve per liter of the solution without getting saturated or before getting saturated. It is determined by the solubility product ( ${\text{K}}_{\text{sp}}$ ) and stoichiometry.

${\text{K}}_{\text{sp}}$ is the solubility product constant which is equal to the product of the concentration of dissolved ions raised to their stoichiometry coefficients.