Chapter 8, Problem 108E

a.

Interpretation Introduction

Interpretation: The solubility of lead iodide in water needs to be calculated.

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

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

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 108E

  $\text{1}\text{.5}\times {\text{10}}^{-3}\text{ M}$

Explanation of Solution

Given:

  ${\text{K}}_{\text{sp}}{\text{(PbI}}_2\text{) = 1}{\text{.4×10}}^{\text{-8}}$

The dissociation reaction for lead iodide is:

  ${\text{PbI}}_2(\text{s)}\rightleftharpoons {\text{Pb}}^{2+}({\text{aq) + 2I}}^{-}(\text{aq)}$

The expression for the solubility constant is:

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

Let the equilibrium concentrations be (as the water does not provide any common ion):

  $\begin{array}{l}{\text{Concentration (mol/L) PbI}}_2\rightleftharpoons {\text{ Pb}}^{2+}{\text{ + 2I}}^{-}\\ \text{ Initial x 0 0}\\ \text{ Change -x +x +2x}\\ \text{ Equilibrium 0 x 2x}\end{array}$

Substituting the values as:

  $\begin{array}{l}{\text{K}}_{\text{sp}}{\text{ = [Pb}}^{2+}]{[}^{{\text{I}}^{-}}\\ \text{1}\text{.4}\times {\text{10}}^{-8}{\text{ = [x][2x]}}^2\\ 4{\text{x}}^3\text{ = 1}\text{.4}\times {\text{10}}^{-8}\\ {\text{x}}^3\text{ = }\frac{\text{1}\text{.4}\times {\text{10}}^{-8}}{4}\\ {\text{x}}^3\text{ = 3}\text{.5}\times {\text{10}}^{-9}\\ \text{x = }\sqrt[3]{\text{3}\text{.5}\times {\text{10}}^{-9}}\\ \text{x = 1}\text{.5}\times {\text{10}}^{-3}\text{ M}\end{array}$

b.

Interpretation Introduction

Interpretation: The solubility of lead iodide in $\text{0}{\text{.1 M Pb(NO}}_3{)}_2$ needs to be calculated.

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

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

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 108E

  $\text{1}{\text{.9 × 10}}^{\text{-4}}\text{ M}$ .

Explanation of Solution

The concentration of ${\text{Pb}}^{\text{2+}}$ in $\text{0}{\text{.1 M Pb(NO}}_3{)}_2$ is 0.1 M as 1 mole of ${\text{Pb}}^{\text{2+}}$ in present in 1 mole of ${\text{Pb(NO}}_3{)}_2$ . So, the initial concentration of ${\text{Pb}}^{\text{2+}}$ ions in the solution is 0.1 M.

The dissociation reaction for lead iodide is:

  ${\text{PbI}}_2(\text{s)}\rightleftharpoons {\text{Pb}}^{2+}({\text{aq) + 2I}}^{-}(\text{aq)}$

The expression for the solubility constant is:

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

Let the equilibrium concentrations be:

  $\begin{array}{l}{\text{Concentration (mol/L) PbI}}_2\rightleftharpoons {\text{ Pb}}^{2+}{\text{ + 2I}}^{-}\\ \text{ Initial x 0}\text{.10 0}\\ \text{ Change -x +x +2x}\\ \text{ Equilibrium 0 0}\text{.10+x 2x}\end{array}$

Substituting the values as:

  $\begin{array}{l}{\text{K}}_{\text{sp}}{\text{ = [Pb}}^{2+}]{[}^{{\text{I}}^{-}}\\ \text{1}\text{.4}\times {\text{10}}^{-8}\text{ = [0}{\text{.10 + x][2x]}}^2\end{array}$

Since, $\text{x <<< 0}\text{.1}$ so,

  $\begin{array}{l}\text{1}{\text{.4 × 10}}^{\text{-8}}\text{ = [0}{\text{.10 ][2x]}}^{\text{2}}\\ {\text{x}}^{\text{2}}\text{ = }\frac{\text{1}{\text{.4 × 10}}^{\text{-8}}}{\text{0}\text{.4}}\\ \text{x = }\sqrt{\text{3}{\text{.5× 10}}^{\text{-8}}}\\ \text{x = 1}{\text{.9 × 10}}^{\text{-4}}\text{ M}\end{array}$

c.

Interpretation Introduction

Interpretation: The solubility of lead iodide in $\text{0}\text{.010 M NaI}$ needs to be calculated.

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

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

Where K_sp is solubility product constant and square brackets represents concentration.

Answer to Problem 108E

  $\text{1}\text{.4}\times {\text{10}}^{-4}\text{ M}$ .

Explanation of Solution

The concentration of ${\text{I}}^{\text{-}}$ in $\text{0}\text{.010 M NaI}$ is 0.01 M as 1 mole of ${\text{I}}^{\text{-}}$ in present in 1 mole of $\text{NaI}$ . So, the initial concentration of ${\text{I}}^{\text{-}}$ ions in the solution is 0.01 M.

The dissociation reaction for lead iodide is:

  ${\text{PbI}}_2(\text{s)}\rightleftharpoons {\text{Pb}}^{2+}({\text{aq) + 2I}}^{-}(\text{aq)}$

The expression for the solubility constant is:

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

Let the equilibrium concentrations be:

  $\begin{array}{l}{\text{Concentration (mol/L) PbI}}_2\rightleftharpoons {\text{ Pb}}^{2+}{\text{ + 2I}}^{-}\\ \text{ Initial x 0 0}\text{.01}\\ \text{ Change -x +x +2x}\\ \text{ Equilibrium 0 x 0}\text{.010+2x}\end{array}$

Substituting the values:

  $\begin{array}{l}{\text{K}}_{\text{sp}}{\text{ = [Pb}}^{2+}]{[}^{{\text{I}}^{-}}\\ \text{1}\text{.4}\times {\text{10}}^{-8}\text{ = [x][0}{\text{.010+2x]}}^2\end{array}$

Since, $\text{x <<< 0}\text{.01}$ so,

  $\begin{array}{l}\text{1}\text{.4 }\times {\text{10}}^{-8}\text{ = [x][0}{\text{.010]}}^2\\ \text{x = }\frac{\text{1}\text{.4 }\times {\text{10}}^{-8}}{0.0001}\\ \text{x = 1}\text{.4}\times {\text{10}}^{-4}\text{ M}\end{array}$