Chapter 18.4, Problem 35SP

Interpretation Introduction

Interpretation: The concentration of sulfide ions in a 1.0 L solution of iron (II) sulfide to which 0.04 mol of iron(II) nitrate 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 35SP

The concentration of sulfide ions in a 1.0 L solution of iron (II) sulfide to which 0.04 mol of iron(II) nitrate is $2\times {\text{10}}^{-17}\text{M}$ .

Explanation of Solution

Given; ${\text{K}}_{\text{sp}}\text{of}\text{FeS}=8\times {10}^{-19}$

The dissociation equation for iron(II) sulfide solution is represented as;

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

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

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

The moles of iron(II) nitrate is 0.04 mol which means one mole of iron(II) nitrate gives 0.04 moles of iron (II) and 0.04 moles of nitrate ions. Thus, the moles of iron (II) ions will be 0.04 moles.

Assuming the concentration of both iron (II) and sulfide ions be s. Now, calculate the concentration of iron (II) ions as;

  $\begin{array}{c}\text{8}\times {\text{10}}^{-19}=\left[\frac{0.04\text{mol}}{1.0\text{L}}\right]\left[\text{s}\right]\\ \text{8}\times {\text{10}}^{-19}=0.04\text{M}\times \left[\text{s}\right]\\ \left[\text{s}\right]=\frac{\text{8}\times {\text{10}}^{-19}}{0.04}\text{M}\\ \left[{\text{S}}^{2-}\right]=2\times {\text{10}}^{-17}\text{M}\end{array}$