Chapter 18.4, Problem 36SP

Interpretation Introduction

Interpretation: The equilibrium concentration of sulfate ions in a 1.0 L solution of strontium sulfate to which 0.10 mol of $\text{Sr}{\left({\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\right)}_{\text{2}}$ is added 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 36SP

The equilibrium concentration of sulfate ions in a 1.0 L solution of strontium sulfate to which 0.10 mol of $\text{Sr}{\left({\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}\right)}_{\text{2}}$ is $3.2\times {\text{10}}^{-6}\text{M}$ .

Explanation of Solution

Given; ${\text{K}}_{\text{sp}}\text{of}{\text{SrSO}}_4=3.2\times {10}^{-7}$

The dissociation equation for strontium sulfate solution is represented as;

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

Therefore, the solubility product $\left({\text{K}}_{\text{sp}}\right)$ expression for ${\text{SrSO}}_4$ is given as;

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

One mole of strontium acetate would give one mole of strontium ion and two moles of acetate ion. Thus, 0.10 moles of strontium acetate would give 0.10 moles of strontium ions.

Now, assuming the concentration of both strontium (II) and sulfate ions be s. Now, calculate the concentration of sulfate ions as;

  $\begin{array}{c}\text{3}\text{.2}\times {\text{10}}^{-7}=\left[\frac{0.10\text{mol}}{1.0\text{L}}\right]\left[\text{s}\right]\\ \text{3}\text{.2}\times {\text{10}}^{-7}=0.10\text{M}\times \left[\text{s}\right]\\ \left[\text{s}\right]=\frac{\text{3}\text{.2}\times {\text{10}}^{-7}}{0.10}\text{M}\\ \left[{\text{SO}}_4^{2-}\right]=3.2\times {\text{10}}^{-6}\text{M}\end{array}$