Chapter U6, Problem 7RE

Interpretation Introduction

Interpretation:The concentration of ${\text{Pb}}^{\text{2+}}$ ions needs to be determined.

Concept introduction: A constant that defines the solubility equilibrium that exists between a compound in its solid-state and its ions in the aqueous state is said to be solubility product constant. Its general expression is written as:

$\begin{array}{l}{\text{A}}_{\text{x}}{\text{B}}_{\text{y}}\text{(s)}\rightleftharpoons {\text{xA}}^{\text{+}}{\text{(aq) + yB}}^{\text{-}}\text{(aq)}\end{array}$

$\begin{array}{c}{\text{K}}_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{]}}^{\text{x}}{{\text{[B}}^{\text{-}}}^{\text{]}}\end{array}$

Where K_sp is the solubility product constant and square brackets represent concentration.

Answer to Problem 7RE

The concentration of ${\text{Pb}}^{\text{2+}}$ ions is $\text{0}\text{.016 mol/L}$ .

Explanation of Solution

Given:

${\text{PbCl}}_{\text{2}}\text{(s) }\rightleftharpoons {\text{ Pb}}^{\text{2+}}{\text{(aq) + 2Cl}}^{\text{-}}{\text{(aq) K}}_{\text{sp}}=\text{ 1}\text{.78}\times {\text{10}}^{\text{-5}}$

The dissociation reaction of ${\text{PbCl}}_{\text{2}}$ is:

${\text{PbCl}}_{\text{2}}\text{(s) }\rightleftharpoons {\text{ Pb}}^{\text{2+}}{\text{(aq) + 2Cl}}^{\text{-}}\text{(aq) }$

The expression for ${\text{K}}_{\text{sp}}$ of ${\text{PbCl}}_{\text{2}}$ l is:

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

Let the solubility of ${\text{Pb}}^{\text{2+}}$ be $\text{s mol/L}$ so, the solubility of each is represented as:

$\begin{array}{l}{\text{PbCl}}_{\text{2}}\text{(s) }\rightleftharpoons {\text{ Pb}}^{\text{2+}}{\text{(aq) + 2Cl}}^{\text{-}}\text{(aq) }\end{array}$

$\begin{array}{c}\text{ s 2s}\end{array}$

Substituting the values:

$\begin{array}{c}1.78\times {10}^{-5}=\left(s\right){\left(2s\right)}^2\end{array}$

$\begin{array}{c}1.78\times {10}^{-5}=4s^3\end{array}$

    $\begin{array}{c}{s}^3=\frac{1.78\times {10}^{-5}}{4}\end{array}$

    $\begin{array}{c}{s}^3=0.445\times {10}^{-5}=4.45\times {10}^{-6}\end{array}$

    $\begin{array}{c}s=\sqrt[3]{4.45\times {10}^{-6}}\end{array}$

    $\begin{array}{c}s=0.016\text{ mol/L}\end{array}$

Thus, the concentration of ${\text{Pb}}^{\text{2+}}$ ions is $\text{0}\text{.016 mol/L}$ .

Conclusion

$\text{0}\text{.016 mol/L}$ is the concentration of ${\text{Pb}}^{\text{2+}}$ ions.