Chapter 15, Problem 39E

The K_sp for silver sulfate (Ag₂SO₄) is 1.2 × 10⁻⁵. Calculate the solubility of silver sulfate in each of the following.

a. water

b. 0.10 M AgNO₃

c. 0.20 M K₂SO₄.

Interpretation Introduction

Interpretation: The solubility product of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is given. By using this value, the solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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 39E

Answer

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in water is $\underset{\_}{0.0144mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in water.

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in water is $\underset{\_}{0.0144mol/L}$.

Given

Solubility product of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{1}\text{.2}\times {\text{10}}^{-\text{5}}$.

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

${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)\stackrel{}{\to }{\text{2Ag}}^{+}\left(aq\right)+{\text{SO}}_{\text{4}}{}^{\text{2}-}\left(aq\right)$

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

${\left[{\text{Ag}}^{+}\right]}_{\text{initial}}={\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]}_{\text{initial}}=\text{0}$

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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 $\text{2}:\text{1}$ stoichiometry of salt is,

$s\text{mol/L}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }\text{2}s\text{mol/L}{\text{Ag}}^{+}+s\text{mol/L}{\text{SO}}_{\text{4}}{}^{\text{2}-}$

Make the ICE table for the dissociation reaction of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$.

$\begin{array}{ccccc}& {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{2Ag}}^{+}\left(aq\right)& {\text{+ SO}}_{\text{4}}{}^{\text{2}-}\left(aq\right)\\ \text{Initial}\text{(M)}:& & & 0& 0\\ \text{Chang}\left(\text{M}\right):& & & 2s& s\\ \text{Equilibrium}\left(\text{M}\right):& & & 2s& s\end{array}$

Formula

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

$\begin{array}{c}{K}_{\text{sp}}={\left[{\text{Ag}}^{+}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]\\ ={\left(\text{2}s\right)}^{\text{2}}\left(s\right)\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is concentration of ${\text{Ag}}^{+}$.
• $\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]$ is concentration of ${\text{SO}}_{\text{4}}{}^{\text{2}-}$.
• $s$ is the solubility.

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

$\begin{array}{c}{K}_{\text{sp}}={\left(\text{2}s\right)}^{\text{2}}\left(s\right)\\ \text{1}\text{.2}\times {\text{10}}^{-\text{5}}=\text{4}{s}^{\text{3}}\\ s=\underset{\_}{0.0144mol/L}\end{array}$

Interpretation Introduction

Interpretation: The solubility product of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is given. By using this value, the solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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 39E

Answer

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{0}\text{.10}\text{M}$ ${\text{AgNO}}_{\text{3}}$ is $\underset{\_}{0.0012mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{0}\text{.10}\text{M}$ ${\text{AgNO}}_{\text{3}}$.

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{0}\text{.10}\text{M}$ ${\text{AgNO}}_{\text{3}}$ is $\underset{\_}{0.0012mol/L}$.

Given

Solubility product of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{1}\text{.2}\times {\text{10}}^{-\text{5}}$.

Concentration of ${\text{AgNO}}_{\text{3}}$ is $\text{0}\text{.10}\text{M}$.

Since, ${\text{AgNO}}_{\text{3}}$ is soluble in aqueous solution. The dissociation reaction of ${\text{AgNO}}_{\text{3}}$ is,

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

Since, the ratio of moles existed between dissolved ions and solid is $\text{1}:\text{1}$, $\text{0}\text{.10}\text{M}$ of ${\text{AgNO}}_{\text{3}}$ contains,

$\left[{\text{Ag}}^{+}\right]=\left[{\text{NO}}_{\text{3}}{}^{-}\right]=\text{0}\text{.10}\text{M}$

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be calculated from the concentration of ions at equilibrium.

Make the ICE table for the dissociation reaction of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$.

$\begin{array}{ccccc}& {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{2Ag}}^{+}\left(aq\right)& {\text{+ SO}}_{\text{4}}{}^{\text{2}-}\left(aq\right)\\ \text{Initial}\text{(M)}:& & & 0.10& 0\\ \text{Chang}\left(\text{M}\right):& & & 2s& s\\ \text{Equilibrium}\left(\text{M}\right):& & & 0.10+2s& s\end{array}$

Formula

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

$\begin{array}{c}{K}_{\text{sp}}={\left[{\text{Ag}}^{+}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]\\ ={\left(\text{2}s+\text{0}\text{.10}\right)}^{\text{2}}\left(s\right)\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is concentration of ${\text{Ag}}^{+}$.
• $\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]$ is concentration of ${\text{SO}}_{\text{4}}{}^{\text{2}-}$.
• $s$ is the solubility.

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

$\text{2}s+\text{0}\text{.10}\approx \text{0}\text{.10}$

Substitute this value in the solubility product expression.

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

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

$\begin{array}{c}{K}_{\text{sp}}={\left(\text{0}\text{.10}\right)}^{\text{2}}\left(s\right)\\ \text{1}\text{.2}\times {\text{10}}^{-\text{5}}={\left(\text{0}\text{.10}\right)}^{\text{2}}\left(s\right)\\ s=\underset{\_}{0.0012mol/L}\end{array}$

Interpretation Introduction

Interpretation: The solubility product of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is given. By using this value, the solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ 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 39E

Answer

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{0}\text{.20}\text{M}$ ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\underset{\_}{0.0039mol/L}$.

Explanation of Solution

Explanation

To determine: The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{0}\text{.20}\text{M}$ ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{0}\text{.20}\text{M}$ ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\underset{\_}{0.0039mol/L}$.

Given

Solubility product of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{1}\text{.2}\times {\text{10}}^{-\text{5}}$.

Concentration of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{0}\text{.20}\text{M}$.

Since, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is soluble in aqueous solution. The dissociation reaction of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is,

${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\stackrel{}{\to }{\text{2K}}^{+}\left(aq\right)+{\text{SO}}_{\text{4}}{}^{-}\left(aq\right)$

The concentration of ${\text{SO}}_{\text{4}}{}^{-}$ is,

$\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]=\text{0}\text{.20}\text{M}$

The solubility of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be calculated from the concentration of ions at equilibrium.

Make the ICE table for the dissociation reaction of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$.

$\begin{array}{ccccc}& {\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}\left(s\right)& \stackrel{}{\rightleftharpoons }& {\text{2Ag}}^{+}\left(aq\right)& {\text{+ SO}}_{\text{4}}{}^{\text{2}-}\left(aq\right)\\ \text{Initial}\text{(M)}:& & & 0& 0.20\\ \text{Chang}\left(\text{M}\right):& & & 2s& s\\ \text{Equilibrium}\left(\text{M}\right):& & & 2s& 0.20+s\end{array}$

Formula

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

$\begin{array}{c}{K}_{\text{sp}}={\left[{\text{Ag}}^{+}\right]}^{\text{2}}\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]\\ ={\left(\text{2}s\right)}^{\text{2}}\left(\text{0}\text{.20}+s\right)\end{array}$

Where,

• $K_{\text{sp}}$ is solubility product.
• $\left[{\text{Ag}}^{+}\right]$ is concentration of ${\text{Ag}}^{+}$.
• $\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]$ is concentration of ${\text{SO}}_{\text{4}}{}^{\text{2}-}$.
• $s$ is the solubility.

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

$\text{0}\text{.20}+s\approx \text{0}\text{.20}$

Substitute this value in the solubility product expression.

$\begin{array}{c}{K}_{\text{sp}}={\left(\text{2}s\right)}^{\text{2}}\left(\text{0}\text{.20}+s\right)\\ ={\left(\text{2}s\right)}^{\text{2}}\left(\text{0}\text{.20}\right)\end{array}$

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

$\begin{array}{c}{K}_{\text{sp}}={\left(\text{2}s\right)}^{\text{2}}\left(\text{0}\text{.20}\right)\\ \text{1}\text{.2}\times {\text{10}}^{-\text{5}}=\text{0}\text{.8}{s}^{\text{2}}\\ s=\underset{\_}{0.0039mol/L}\end{array}$