Chapter 15, Problem 40E

(a)

Interpretation Introduction

Interpretation: The solubility product of ${\text{PbI}}_{\text{2}}$ is given. The solubility of ${\text{PbI}}_{\text{2}}$ is to be calculated for each given conditions.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, $K_{\text{sp}}$ is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of ${\text{A}}_x{\text{B}}_y$ is calculated as,

$K_{\text{sp}}={\left[\text{A}\right]}^x{\left[\text{B}\right]}^y$

Where,

• $x$ is coefficient of concentration of $\text{A}$.
• $y$ is coefficient of concentration of $\text{B}$.

Answer to Problem 40E

Answer

The solubility of ${\text{PbI}}_{\text{2}}$ in water is $\underset{\_}{1.5\times {10}^{-3}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{PbI}}_{\text{2}}$ in water.

The solubility of ${\text{PbI}}_{\text{2}}$ in water is $\underset{\_}{0.0015mol/L}$.

Given

Solubility product of ${\text{PbI}}_{\text{2}}$ is $\text{1}\text{.4}\times {\text{10}}^{-\text{8}}$.

Since, solid ${\text{PbI}}_{\text{2}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{PbI}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{PbI}}_{\text{2}}$ is,

${\text{PbI}}_{\text{2}}\left(s\right)\stackrel{}{\to }{\text{Pb}}^{\text{2}+}\left(aq\right)+{\text{2I}}^{-}\left(aq\right)$

Since, ${\text{Pb}}^{\text{2}+}$ does not dissolved initially, hence,

${\left[{\text{Pb}}^{\text{2}+}\right]}_{\text{initial}}={\left[{\text{I}}^{-}\right]}_{\text{initial}}=\text{0}$

The solubility of ${\text{PbI}}_{\text{2}}$ can be calculated from the concentration of ions at equilibrium.

It is assumed that $s\text{mol/L}$ of solid is dissolved to reach the equilibrium. The meaning of $\text{1}:\text{2}$ stoichiometry of salt is,

$s\text{mol/L}{\text{PbI}}_{\text{2}}\stackrel{}{\to }s\text{mol/L}{\text{Pb}}^{\text{2}+}+\text{2}s\text{mol/L}{\text{I}}^{-}$

Make the ICE table for the dissociation reaction of ${\text{PbI}}_{\text{2}}$.

$\begin{array}{ccccc}& {\text{PbI}}_{\text{2}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Pb}}^{2+}\left(aq\right)& \text{+}{\text{I}}^{-}\left(aq\right)\\ \text{Initial}\text{(M)}:& & & 0& 0\\ \text{Chang}\left(\text{M}\right):& & & s& 2s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& 2s\end{array}$

Formula

The solubility product of ${\text{PbI}}_{\text{2}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Pb}}^{\text{2}+}\right]{\left[{\text{I}}^{-}\right]}^{\text{2}}\\ =\left(s\right){\left(\text{2}s\right)}^{\text{2}}\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Pb}}^{\text{2}+}\right]$ is concentration of ${\text{Pb}}^{\text{2}+}$.
• $\left[{\text{I}}^{-}\right]$ is concentration of ${\text{I}}^{-}$.
• $s$ is the solubility.

Substitute the value of $K_{\text{sp}}$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{2}s\right)}^{\text{2}}\\ \text{1}\text{.4}\times {\text{10}}^{-\text{8}}=\text{4}{s}^{\text{3}}\\ s=\underset{\_}{0.0015mol/L}\end{array}$

(b)

Interpretation Introduction

Interpretation: The solubility product of ${\text{PbI}}_{\text{2}}$ is given. The solubility of ${\text{PbI}}_{\text{2}}$ is to be calculated for each given conditions.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, $K_{\text{sp}}$ is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of ${\text{A}}_x{\text{B}}_y$ is calculated as,

$K_{\text{sp}}={\left[\text{A}\right]}^x{\left[\text{B}\right]}^y$

Where,

• $x$ is coefficient of concentration of $\text{A}$.
• $y$ is coefficient of concentration of $\text{B}$.

Answer to Problem 40E

Answer

The solubility of ${\text{PbI}}_{\text{2}}$ in $\text{0}\text{.10}\text{M}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\underset{\_}{1.9\times {10}^{-4}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{PbI}}_{\text{2}}$ in $\text{0}\text{.10}\text{M}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$.

The solubility of ${\text{PbI}}_{\text{2}}$ in $\text{0}\text{.10}\text{M}$ $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\underset{\_}{1.9\times {10}^{-4}mol/L}$.

Given

Solubility product of ${\text{PbI}}_{\text{2}}$ is $\text{1}\text{.4}\times {\text{10}}^{-\text{8}}$.

Concentration of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\text{0}\text{.10}\text{M}$.

The dissociation reaction of ${\text{PbI}}_{\text{2}}$ is,

${\text{PbI}}_{\text{2}}(s)\stackrel{}{\to }{\text{Pb}}^{\text{2}+}\left(aq\right)+{\text{2I}}^{-}\left(aq\right)$

The ratio of moles between ions is $\text{1}:\text{2}$.

Since, $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is soluble in aqueous solution. The dissociation reaction of $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is,

$\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(aq\right)\stackrel{}{\to }{\text{Pb}}^{\text{2}+}\left(aq\right)+{\text{2NO}}_{\text{3}}{}^{-}\left(aq\right)$

The concentration of ${\text{Pb}}^{\text{2}+}$ is,

$\left[{\text{Pb}}^{\text{2}+}\right]=\text{0}\text{.10}\text{M}$

The solubility of ${\text{PbI}}_{\text{2}}$ can be calculated from the concentration of ions at equilibrium.

Make the ICE table for the dissociation reaction of ${\text{PbI}}_{\text{2}}$.

$\begin{array}{ccccc}& {\text{PbI}}_{\text{2}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Pb}}^{2+}\left(aq\right)& \text{+}{\text{2I}}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right):& & & 0.10& 0\\ \text{Chang}\left(\text{M}\right):& & & s& 2s\\ \text{Equilibrium}\left(\text{M}\right):& & & 0.10+s& 2s\end{array}$

Formula

The solubility product of ${\text{PbI}}_{\text{2}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Pb}}^{\text{2}+}\right]{\left[{\text{I}}^{-}\right]}^{\text{2}}\\ =\left(s+\text{0}\text{.10}\right){\left(\text{2}s\right)}^{\text{2}}\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Pb}}^{\text{2}+}\right]$ is concentration of ${\text{Pb}}^{\text{2}+}$.
• $\left[{\text{I}}^{-}\right]$ is concentration of ${\text{I}}^{-}$.
• $s$ is the solubility.

Since, value of $s$ is expected to be small value hence, it is assumed that,

$s+\text{0}\text{.10}\approx \text{0}\text{.10}$

Substitute this value in the solubility product expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s+\text{0}\text{.10}\right){\left(\text{2}s\right)}^{\text{2}}\\ =\left(\text{0}\text{.10}\right){\left(\text{2}s\right)}^{\text{2}}\end{array}$

Substitute the value of $K_{\text{sp}}$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(\text{0}\text{.10}\right){\left(\text{2}s\right)}^{\text{2}}\\ \text{1}\text{.4}\times {\text{10}}^{-\text{8}}=\left(\text{0}\text{.10}\right){\left(\text{2}s\right)}^{\text{2}}\\ s=\underset{\_}{1.9\times {10}^{-4}mol/L}\end{array}$

(c)

Interpretation Introduction

Interpretation: The solubility product of ${\text{PbI}}_{\text{2}}$ is given. The solubility of ${\text{PbI}}_{\text{2}}$ is to be calculated for each given conditions.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, $K_{\text{sp}}$ is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of ${\text{A}}_x{\text{B}}_y$ is calculated as,

$K_{\text{sp}}={\left[\text{A}\right]}^x{\left[\text{B}\right]}^y$

Where,

• $x$ is coefficient of concentration of $\text{A}$.
• $y$ is coefficient of concentration of $\text{B}$.

Answer to Problem 40E

Answer

The solubility of ${\text{PbI}}_{\text{2}}$ in $\text{0}\text{.010}\text{M}$ $\text{NaI}$ is $\underset{\_}{1.4\times {10}^{-4}mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{PbI}}_{\text{2}}$ in $\text{0}\text{.010}\text{M}$ $\text{NaI}$.

The solubility of ${\text{PbI}}_{\text{2}}$ in $\text{0}\text{.010}\text{M}$ $\text{NaI}$ is $\underset{\_}{1.4\times {10}^{-4}mol/L}$.

Given

Solubility product of ${\text{PbI}}_{\text{2}}$ is $\text{1}\text{.2}\times {\text{10}}^{-\text{5}}$.

Concentration of $\text{NaI}$ is $\text{0}\text{.010}\text{M}$.

Since, $\text{NaI}$ is soluble in aqueous solution. The dissociation reaction of $\text{NaI}$ is,

$\text{NaI}\left(aq\right)\stackrel{}{\to }{\text{Na}}^{+}\left(aq\right)+{\text{I}}^{-}\left(aq\right)$

Since, the ratio of moles existed between dissolved ions and solid is $\text{1}:\text{1}$, $\text{0}\text{.010}\text{M}$ of $\text{NaI}$ contains,

$\left[{\text{Na}}^{+}\right]=\left[{\text{I}}^{-}\right]=\text{0}\text{.010}\text{M}$

The solubility of ${\text{PbI}}_{\text{2}}$ can be calculated from the concentration of ions at equilibrium.

Make the ICE table for the dissociation reaction of ${\text{PbI}}_{\text{2}}$.

$\begin{array}{ccccc}& {\text{PbI}}_{\text{2}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{Pb}}^{2+}\left(aq\right)& {\text{+ 2I}}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right):& & & 0& 0.010\\ \text{Chang}\left(\text{M}\right):& & & s& 2s\\ \text{Equilibrium}\left(\text{M}\right):& & & s& 0.010+2s\end{array}$

Formula

The solubility product of ${\text{PbI}}_{\text{2}}$ is calculated as,

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Pb}}^{\text{2}+}\right]{\left[{\text{I}}^{-}\right]}^{\text{2}}\\ =\left(s\right){\left(\text{0}\text{.010}+\text{2}s\right)}^{\text{2}}\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Pb}}^{\text{2}+}\right]$ is concentration of ${\text{Pb}}^{\text{2}+}$.
• $\left[{\text{I}}^{-}\right]$ is concentration of ${\text{I}}^{-}$.
• $s$ is the solubility.

Since, value of $s$ is expected to be small value, hence, it is assumed that,

$\left(\text{0}\text{.010}+\text{2}s\right)\approx \text{0}\text{.010}$

Substitute this value in the solubility product expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{0}\text{.010}+\text{2}s\right)}^{\text{2}}\\ =\left(s\right){\left(\text{0}\text{.010}\right)}^{\text{2}}\end{array}$

Substitute the value of $K_{\text{sp}}$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}=\left(s\right){\left(\text{0}\text{.010}\right)}^{\text{2}}\\ \text{1}\text{.4}\times {\text{10}}^{-\text{8}}=\left(s\right){\left(\text{0}\text{.010}\right)}^{\text{2}}\\ s=\underset{\_}{1.4\times {10}^{-4}mol/L}\end{array}$