Chapter 18.4, Problem 41LC

Interpretation Introduction

Interpretation: The solubility product of nickel (II) sulfide having equilibrium concentration of $2\times {10}^{-10}\text{M}$ each of ${\text{Ni}}^{\text{2+}}$ and ${\text{S}}^{\text{2-}}$ 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 41LC

The solubility product of nickel (II) sulfide has an equilibrium concentration of $2\times {10}^{-10}\text{M}$ each of ${\text{Ni}}^{\text{2+}}$ and ${\text{S}}^{\text{2-}}$ is $4\times {10}^{-20}$ .

Explanation of Solution

The dissociation of nickel (II) sulfide is represented as;

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

Therefore, the solubility product expression for $\text{NiS}$ is given as;

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

Now, calculate the solubility product constant as;

  $\begin{array}{c}{\text{K}}_{\text{sp}}=\left[2\times {10}^{-10}\right]\left[2\times {10}^{-10}\right]\\ =4\times {10}^{-20}\end{array}$