Chapter 15, Problem 22E

Use the following data to calculate the K_sp value for each solid.

a. The solubility of Pb₃(PO₄) is 6.2 × 10⁻¹² mol/L.

b. The solubility of Li₂CO₃ is 7.4 × 10⁻² mol/L.

Interpretation Introduction

Interpretation: The solubility of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given. The solubility product of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is to be calculated.

Concept introduction: 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,

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

Answer to Problem 22E

Answer

The solubility product of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\underset{\_}{9.9\times {10}^{-55}}$.

Explanation of Solution

Explanation

To determine: The solubility product of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$.

The concentration of ${\text{Pb}}^{\text{2}+}$ is $\underset{\_}{18.6\times 10^{-12}mol/L}$.

Given

Solubility of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol/L}$.

Since, solid ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is,

${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)\stackrel{}{\to }{\text{3Pb}}^{\text{2}+}\left(aq\right)+{\text{2PO}}_{\text{4}}{}^{\text{3}-}\left(aq\right)$

Since, ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ does not dissolved initially, hence,

${\left[{\text{Pb}}^{\text{2}+}\right]}_{\text{initial}}={\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}_{\text{initial}}=\text{0}$

The concentration at equilibrium can be calculated from the measured solubility of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$. If $\text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol}$ of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is dissolved in $\text{1}\text{.0}\text{L}$ of solution, the change in solubility will be equal to $\text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol/L}$. The reaction is,

${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)\stackrel{}{\to }{\text{3Pb}}^{\text{2}+}\left(aq\right)+{\text{2PO}}_{\text{4}}{}^{\text{3}-}\left(aq\right)$

Therefore,

$\text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol/L}{\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\stackrel{}{\to }\text{3}\left(\text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol/L}\right){\text{Pb}}^{\text{2}+}+\text{2}\left(\text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol/L}\right){\text{PO}}_{\text{4}}$

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

$\left[{\text{Pb}}^{\text{2}+}\right]={\left[{\text{Pb}}^{\text{2}+}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}$

Substitute the value of ${\left[{\text{Pb}}^{\text{2}+}\right]}_{\text{initial}}$ and change to reach equilibrium in the above equation.

$\begin{array}{c}\left[{\text{Pb}}^{\text{2}+}\right]={\left[{\text{Pb}}^{\text{2}+}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\text{3}\left(\text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol/L}\right)\\ =\underset{\_}{18.6\times 10^{-12}mol/L}\end{array}$

The concentration of ${\text{PO}}_{\text{4}}{}^{\text{3}-}$ is $\underset{\_}{12.4\times 10^{-12}mol/L}$.

Given

Solubility of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

The equilibrium concentration of ${\text{PO}}_{\text{4}}{}^{\text{3}-}$ is written as,

$\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]={\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}$

Substitute the value of ${\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}_{\text{initial}}$ and change to reach equilibrium in the above equation.

$\begin{array}{c}\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]={\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\left(\text{2}\times \text{6}\text{.2}\times {\text{10}}^{-\text{12}}\text{mol/L}\right)\\ =\underset{\_}{12.4\times 10^{-12}mol/L}\end{array}$

The solubility product of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is $\underset{\_}{9.9\times {10}^{-55}}$.

The concentration of ${\text{Pb}}^{\text{2}+}$ is $\text{18}\text{.6}\times {\text{10}}^{-\text{12}}\text{mol/L}$.

The concentration of ${\text{PO}}_{\text{4}}{}^{\text{3}-}$ is $\text{12}\text{.4}\times {\text{10}}^{-\text{12}}\text{mol/L}$.

Formula

The solubility product of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is calculated as,

$K_{\text{sp}}={\left[{\text{Pb}}^{\text{2}+}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}^{\text{2}}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Pb}}^{\text{2}+}\right]$ is concentration of ${\text{Pb}}^{\text{2}+}$.
• $\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]$ is concentration of ${\text{PO}}_{\text{4}}{}^{\text{3}-}$

Substitute the values of $\left[{\text{Pb}}^{\text{2}+}\right]$ and $\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}={\left[{\text{Pb}}^{\text{2}+}\right]}^{\text{3}}{\left[{\text{PO}}_{\text{4}}{}^{\text{3}-}\right]}^{\text{2}}\\ ={\left(\text{18}\text{.6}\times {\text{10}}^{-\text{12}}\right)}^{\text{3}}{\left(\text{12}\text{.4}\times {\text{10}}^{-\text{12}}\right)}^{\text{2}}\\ =\underset{\_}{9.9\times {10}^{-55}}\end{array}$

Interpretation Introduction

Interpretation: The solubility of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is given. The solubility product of ${\text{Pb}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is to be calculated.

Concept introduction: 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,

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

Answer to Problem 22E

Answer

The solubility product of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\underset{\_}{1.6\times {10}^{-3}}$.

Explanation of Solution

Explanation

To determine: The solubility product of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$.

The concentration of ${\text{Li}}^{+}$ is $\underset{\_}{14.8\times 10^{-2}mol/L}$.

Given

Solubility of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}$.

Since, solid ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is,

${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)\stackrel{}{\to }{\text{2Li}}^{+}\left(aq\right)+{\text{CO}}_{\text{3}}{}^{\text{2}-}\left(aq\right)$

Since, ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ does not dissolved initially, hence,

${\left[{\text{Li}}^{+}\right]}_{\text{initial}}={\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]}_{\text{initial}}=\text{0}$

The concentration at equilibrium can be calculated from the measured solubility of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$. If $\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}$ of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is dissolved in $\text{1}\text{.0}\text{L}$ of solution, the change in solubility will be equal to $\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}$. The reaction is,

${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(s\right)\stackrel{}{\to }{\text{2Li}}^{+}\left(aq\right)+{\text{CO}}_{\text{3}}{}^{\text{2}-}\left(aq\right)$

Therefore,

$\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}{\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }\text{2}\left(\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}\right){\text{Li}}^{+}+\left(\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}\right){\text{CO}}_{\text{3}}$

The equilibrium concentration of ${\text{Li}}^{+}$ is written as,

$\left[{\text{Li}}^{+}\right]={\left[{\text{Li}}^{+}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}$

Substitute the value of ${\left[{\text{Li}}^{+}\right]}_{\text{initial}}$ and change to reach equilibrium in the above equation.

$\begin{array}{c}\left[{\text{Li}}^{+}\right]={\left[{\text{Li}}^{+}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\text{2}\left(\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}\right)\\ =\underset{\_}{14.8\times 10^{-2}mol/L}\end{array}$

The concentration of ${\text{CO}}_{\text{3}}{}^{\text{2}-}$ is $\underset{\_}{7.4\times 10^{-2}mol/L}$.

Given

Solubility of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

The equilibrium concentration of ${\text{CO}}_{\text{3}}{}^{\text{2}-}$ is written as,

$\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]={\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}$

Substitute the value of ${\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]}_{\text{initial}}$ and change to reach equilibrium in the above equation.

$\begin{array}{c}\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]={\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\left(\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}\right)\\ =\underset{\_}{7.4\times 10^{-2}mol/L}\end{array}$

The solubility product of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $\underset{\_}{1.6\times {10}^{-3}}$.

The concentration of ${\text{Li}}^{+}$ is $\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\text{mol/L}$.

The concentration of ${\text{CO}}_{\text{3}}{}^{\text{2}-}$ is $\text{14}\text{.8}\times {\text{10}}^{-\text{2}}\text{mol/L}$.

Formula

The solubility product of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as,

$K_{\text{sp}}={\left[{\text{Li}}^{+}\right]}^{\text{2}}\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Li}}^{+}\right]$ is concentration of ${\text{Li}}^{+}$.
• $\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]$ is concentration of ${\text{CO}}_{\text{3}}{}^{\text{2}-}$

Substitute the values of $\left[{\text{Bi}}^{\text{3}+}\right]$ and $\left[\text{I}\right]$ in the above expression.

$\begin{array}{c}{K}_{\text{sp}}={\left[{\text{Li}}^{+}\right]}^{\text{2}}\left[{\text{CO}}_{\text{3}}{}^{\text{2}-}\right]\\ ={\left(\text{14}\text{.8}\times {\text{10}}^{-\text{2}}\right)}^{\text{2}}\left(\text{7}\text{.4}\times {\text{10}}^{-\text{2}}\right)\\ =\underset{\_}{1.6\times {10}^{-3}}\end{array}$