Chapter 18.4, Problem 42LC

Interpretation Introduction

Interpretation: The concentration of lead ions and sulfide ions in a saturated solution of lead sulfide having ${\text{K}}_{\text{sp}}=3.0\times {10}^{-28}$ is to be calculated.

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 42LC

The concentration of lead ions and sulfide ions in a saturated solution of lead sulfide having ${\text{K}}_{\text{sp}}=3.0\times {10}^{-28}$ is $1.73\times {10}^{-14}$ .

Explanation of Solution

The chemical equation for the dissociation of lead sulfide 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 expression for $\text{PbS}$ is given as;

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

Now, calculating the concentration of lead ions and sulfide ions is calculated as;

  $\begin{array}{c}{\text{K}}_{\text{sp}}=\left[{\text{Pb}}^{2+}\right]\left[{\text{S}}^{2-}\right]\\ {\text{K}}_{\text{sp}}=\left[s\right]\left[s\right]\\ 3.0\times {10}^{-28}={\left[s\right]}^2\\ \left[s\right]=\sqrt{3.0\times {10}^{-28}}\\ \left[s\right]=1.73\times {10}^{-14}\end{array}$

Therefore, $\left[{\text{Ni}}^{\text{2+}}\right]=\left[{\text{S}}^{\text{2-}}\right]=1.73\times {10}^{-14}$ .