Chapter 16, Problem 25E

(a)

Interpretation Introduction

Interpretation: The solubility of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{BiI}}_{\text{3}}$ is given. The solubility product of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{BiI}}_{\text{3}}$ 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 25E

Answer

The solubility product of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is $\underset{\_}{2.3\times {10}^{-9}}$.

Explanation of Solution

Explanation

To determine: The solubility product of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$.

The concentration of ${\text{Ca}}^{\text{2}+}$ is $\underset{\_}{4.8\times 10^{-5}mol/L}$.

Given

Solubility of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is $\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

Since, solid ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is placed in contact with water. Therefore, compound present before the reaction is ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{H}}_{\text{2}}\text{O}$. The dissociation reaction of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is,

${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\left(s\right)\stackrel{}{\to }{\text{Ca}}^{\text{2}+}\left(aq\right)+{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\left(aq\right)$

Since, ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ does not dissolved initially, hence,

${\left[{\text{Ca}}^{\text{2}+}\right]}_{\text{initial}}={\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]}_{\text{initial}}=\text{0}$

The concentration at equilibrium can be calculated from the measured solubility of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$. If $\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol}$ of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is dissolved in $\text{1}\text{.0}\text{L}$ of solution, the change in solubility will be equal to $\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}$. The reaction is,

${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\left(s\right)\stackrel{}{\to }{\text{Ca}}^{\text{2}+}\left(aq\right)+{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\left(aq\right)$

Therefore,

$\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}{\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}\stackrel{}{\to }\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}{\text{Ca}}^{\text{2}+}+\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$

The equilibrium concentration of $\left[{\text{Ca}}^{\text{2}+}\right]$ is written as,

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

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

$\begin{array}{c}\left[{\text{Ca}}^{\text{2}+}\right]={\left[{\text{Ca}}^{\text{2}+}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}\\ =\underset{\_}{4.8\times 10^{-5}mol/L}\end{array}$

The concentration of ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}$ is $\underset{\_}{4.8\times 10^{-5}mol/L}$.

Given

Solubility of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is $\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

The equilibrium concentration of $\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]$ is written as,

$\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]={\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}$

Substitute the value of ${\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]}_{\text{initial}}$ and change to reach equilibrium in the above equation.

$\begin{array}{c}\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]={\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}\\ =\underset{\_}{4.8\times 10^{-5}mol/L}\end{array}$

The solubility product of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is $\underset{\_}{2.3\times {10}^{-9}}$.

The concentration of ${\text{Ca}}^{\text{2}+}$ is $\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

The concentration of ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}$ is $\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

Formula

The solubility product of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ is calculated as,

$K_{\text{sp}}=\left[{\text{Ca}}^{\text{2}+}\right]\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]$

Where,

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

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

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2}+}\right]\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2}-}\right]\\ =\left(\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\right)\left(\text{4}\text{.8}\times {\text{10}}^{-\text{5}}\right)\\ =\underset{\_}{2.3\times {10}^{-9}}\end{array}$

(b)

Interpretation Introduction

Interpretation: The solubility of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{BiI}}_{\text{3}}$ is given. The solubility product of ${\text{CaC}}_{\text{2}}{\text{O}}_{\text{4}}$ and ${\text{BiI}}_{\text{3}}$ 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 25E

Answer

The solubility product of ${\text{BiI}}_{\text{3}}$ is $\underset{\_}{8.20\times {10}^{-19}}$.

Explanation of Solution

Explanation

To determine: The solubility product of ${\text{BiI}}_{\text{3}}$.

The concentration of ${\text{Bi}}^{\text{3}+}$ is $\underset{\_}{1.32\times 10^{-5}mol/L}$.

Given

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

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

${\text{BiI}}_{\text{3}}\left(s\right)\stackrel{}{\to }{\text{Bi}}^{\text{3}+}\left(aq\right)+{\text{3I}}^{-}\left(aq\right)$

Since, ${\text{BiI}}_{\text{3}}$ does not dissolved initially, hence,

${\left[{\text{Bi}}^{\text{3}+}\right]}_{\text{initial}}={\left[{\text{I}}^{-}\right]}_{\text{initial}}=\text{0}$

The concentration at equilibrium can be calculated from the measured solubility of ${\text{BiI}}_{\text{3}}$. If $\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}$ of ${\text{BiI}}_{\text{3}}$ is dissolved in $\text{1}\text{.0}\text{L}$ of solution, the change in solubility will be equal to $\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}$. The reaction is,

${\text{BiI}}_{\text{3}}\left(s\right)\stackrel{}{\to }{\text{Bi}}^{\text{3}+}\left(aq\right)+{\text{3I}}^{-}\left(aq\right)$

Therefore,

$\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}{\text{BiI}}_{\text{3}}\stackrel{}{\to }\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}{\text{Bi}}^{\text{3}+}+\left(\text{3}\times \text{1}\text{.32}\times {\text{10}}^{-\text{5}}\right)\text{mol/L}\text{I}$

The equilibrium concentration of ${\text{Bi}}^{\text{3}+}$ is written as,

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

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

$\begin{array}{c}\left[{\text{Bi}}^{\text{3}+}\right]={\left[{\text{Bi}}^{\text{3}+}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}\\ =\underset{\_}{1.32\times 10^{-5}mol/L}\end{array}$

The concentration of $\text{I}$ is $\underset{\_}{3.96\times 10^{-5}mol/L}$.

Given

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

The equilibrium concentration of $\text{I}$ is written as,

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

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

$\begin{array}{c}\left[\text{I}\right]={\left[\text{I}\right]}_{\text{initial}}+\text{change}\text{to}\text{reach}\text{equilibrium}\\ =\text{0}+\left(\text{3}\times \text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}\right)\\ =\underset{\_}{3.96\times 10^{-5}mol/L}\end{array}$

The solubility product of ${\text{BiI}}_{\text{3}}$ is $\underset{\_}{8.20\times {10}^{-19}}$.

The concentration of ${\text{Bi}}^{\text{3}+}$ is $\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

The concentration of $\text{I}$ is $\text{3}\text{.96}\times {\text{10}}^{-\text{5}}\text{mol/L}$.

Formula

The solubility product of ${\text{BiI}}_{\text{3}}$ is calculated as,

$K_{\text{sp}}=\left[{\text{Bi}}^{\text{3}+}\right]{\left[\text{I}\right]}^{\text{3}}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Bi}}^{\text{3}+}\right]$ is concentration of ${\text{Bi}}^{\text{3}+}$.
• $\left[\text{I}\right]$ is concentration of $\text{I}$

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

$\begin{array}{c}{K}_{\text{sp}}=\left[{\text{Bi}}^{\text{3}+}\right]{\left[\text{I}\right]}^{\text{3}}\\ =\left(\text{1}\text{.32}\times {\text{10}}^{-\text{5}}\right){\left(\text{3}\text{.96}\times {\text{10}}^{-\text{5}}\right)}^{\text{3}}\\ =\underset{\_}{8.20\times {10}^{-19}}\end{array}$