Chapter 18.4, Problem 33SP

Interpretation Introduction

Interpretation: The concentration of lead(II) ions in a saturated solution of $\text{PbS}$ at ${\text{25}}^{\text{ο}}\text{C}$ .

Concept Introduction: The expression representing the product of the concentrations of the ions each raised to the power (equal to their stoichiometric coefficient) in the dissociation equation is termed as solubility product constant $\left({\text{K}}_{\text{sp}}\right)$ .

Answer to Problem 33SP

The concentration of lead(II) ions in a saturated solution of $\text{PbS}$ at ${\text{25}}^{\text{ο}}\text{C}$ is $1.7\times {\text{10}}^{-14}\text{M}$ .

Explanation of Solution

Given; ${\text{K}}_{\text{sp}}=3.0\times {10}^{-28}$

The dissociation equation for lead sulfide solution is represented as;

  $\text{PbS}\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\left(\text{aq}\right)+{\text{S}}^{2-}\left(\text{aq}\right)$

Therefore, the solubility product $\left({\text{K}}_{\text{sp}}\right)$ expression for $\text{PbS}$ is given as;

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

Assuming the concentration of both lead (II) and sulfide ions are s. now, calculate the concentration of lead (II) ions as;

  $\begin{array}{c}\text{3}\text{.0}\times {\text{10}}^{-28}=\left[\text{s}\right]\left[\text{s}\right]\\ \text{3}\text{.0}\times {\text{10}}^{-28}={\left[\text{s}\right]}^2\\ \left[\text{s}\right]=\sqrt{\text{3}\text{.0}\times {\text{10}}^{-28}}\\ \left[{\text{Pb}}^{\text{2+}}\right]=1.7\times {\text{10}}^{-14}\text{M}\end{array}$