Chapter 18.4, Problem 43LC

Interpretation Introduction

Interpretation: The concentration of sulfate ions in a saturated solution of barium sulfate having ${\text{K}}_{\text{sp}}=1.1\times {10}^{-10}$ 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 43LC

The concentration of sulfate ions in a saturated solution of barium sulfate having ${\text{K}}_{\text{sp}}=1.1\times {10}^{-10}$ is $7.3\times {\text{10}}^{-9}$ .

Explanation of Solution

The chemical equation for the dissociation of barium sulfate is represented as;

  ${\text{BaSO}}_4\left(\text{s}\right)\rightleftharpoons {\text{Ba}}^{\text{2+}}\left(\text{aq}\right)+{\text{SO}}_4^{2-}\left(\text{aq}\right)$

Therefore, the solubility product expression for ${\text{BaSO}}_4$ is given as;

  ${\text{K}}_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{SO}}_4^{2-}\right]$

As the moles of barium nitrate are 0.015 that means the moles of barium ion are 0.015. Thus, calculating the concentration of sulfate ions is calculated as;

  $\begin{array}{c}{\text{K}}_{\text{sp}}=\left[{\text{Ba}}^{2+}\right]\left[{\text{SO}}_4^{2-}\right]\\ \text{1}\text{.1}\times {\text{10}}^{-10}=\left[\frac{0.015}{1}\right]\left[{\text{SO}}_4^{2-}\right]\\ \left[{\text{SO}}_4^{2-}\right]=\frac{\text{1}\text{.1}\times {\text{10}}^{-10}}{0.015}\\ \left[{\text{SO}}_4^{2-}\right]=7.3\times {\text{10}}^{-9}\end{array}$