Chapter 17.4, Problem 8PPB

(a)

Interpretation Introduction

Interpretation:

The solubility product constant of given compounds have to be calculated.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution.
• The solubility product constant ( ${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.

To calculate: The solubility product constant of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}$.

Answer to Problem 8PPB

(a)

The solubility product of the given compound is ${\text{K}}_{\text{sp}}\text{ =}1.\text{52}\text{×}{\text{10}}^{\text{-14}}$

(b)

The solubility product of the given compound is ${\text{K}}_{\text{sp}}\text{=}4.80\text{×}{\text{10}}^{\text{-29}}$

Explanation of Solution

The solubility product constant of ( ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}$) is calculated below.

The equilibrium of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}$ is

$\begin{array}{l}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }{\text{ 2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{ +}{\text{SO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\\ \text{Initial}\text{concentration }\left(\text{M}\right)\text{: }\text{0 }\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }\text{+2s }\text{+s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }2\text{s }\text{s}\end{array}$

$\begin{array}{l}\text{ Molar solubility}\text{of}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}\text{=}\frac{\text{4}\text{.6 ×}{\text{10}}^{\text{-3}}\text{g}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}}{\text{1}\text{L}}\text{×}\frac{\text{1mol}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{3}}}{\text{295}{\text{.8g Ag}}_{\text{2}}{\text{SO}}_{\text{3}}}\\ \text{S}\text{=}1.56\text{×}{\text{10}}^{\text{-5}}\text{mol/L}\\ {\text{K}}_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{]}}^{\text{2}}{\text{[SO}}_{\text{3}}{}^{2-}\text{]}\\ {\text{K}}_{\text{sp}}\text{= }{\left(\text{2s}\right)}^2\left(\text{s}\right)\\ {\text{K}}_{\text{sp}}\text{= }{\text{4s}}^{\text{3}}\\ {\text{K}}_{\text{sp}}\text{=}4\times {\text{(}}^{1.56}\\ {\text{K}}_{\text{sp}}\text{=}1.\text{52}\text{×}{\text{10}}^{\text{-14}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The solubility product constant of given compounds have to be calculated.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution.
• The solubility product constant ( ${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.

Answer to Problem 8PPB

(b)

The solubility product of the given compound is ${\text{K}}_{\text{sp}}\text{=}4.80\text{×}{\text{10}}^{\text{-29}}$

Explanation of Solution

To calculate: The solubility product constant of ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$.

The solubility product constant of ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$ is calculated below.

The equilibrium of ${\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}$ is

$\begin{array}{l}{\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }{\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}\text{+}{\text{2I}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial}\text{concentration }\left(\text{M}\right)\text{: }\text{0 }\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }\text{+s }\text{+2s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{s }2\text{s}\end{array}$

$\begin{array}{l}\text{ Molar solubility}\text{of}{\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}\text{=}\frac{\text{1}\text{.5 ×}{\text{10}}^{\text{-7}}\text{g}{\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}}{\text{1}\text{L}}\text{×}\frac{\text{1mol}{\text{Hg}}_{\text{2}}{\text{I}}_{\text{2}}}{\text{654}{\text{.98g Hg}}_{\text{2}}{\text{I}}_{\text{2}}}\\ \text{S}\text{=}2.29\text{×}{\text{10}}^{\text{-10}}\text{mol/L}\\ {\text{K}}_{\text{sp}}{\text{= [Hg}}_{\text{2}}{}^{\text{2+}}{\text{][I}}^{\text{-}}{\text{]}}^{\text{2}}\\ {\text{K}}_{\text{sp}}\text{= }\left(\text{s}\right){\left(\text{2s}\right)}^2\\ {\text{K}}_{\text{sp}}\text{= }{\text{4s}}^{\text{3}}\\ {\text{K}}_{\text{sp}}\text{=}4\times {\text{(}}^{2.29}\\ {\text{K}}_{\text{sp}}\text{=}4.80\text{×}{\text{10}}^{\text{-29}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The solubility product constant of given compounds have to be calculated.

Concept introduction:

• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• Solubility of a compound is expressed as concentration of its ions in saturated solution.
• The solubility product constant ( ${\text{K}}_{\text{sp}}$) is defined as the equilibrium between compound and its ions in an aqueous solution.
• Solubility product is the multiplication of concentration of dissolved ion,  raised to the power of coefficients.
• Ionic compound ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.

To calculate: The solubility product constant of ${\text{Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$.

Answer to Problem 8PPB

(c)

The solubility product of the given compound is ${\text{K}}_{\text{sp}}\text{=}\text{9}\text{×}{\text{10}}^{\text{-33}}$

Explanation of Solution

The solubility product constant of ${\text{Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is calculated below.

The equilibrium of ${\text{Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ is

$\begin{array}{l}{\text{Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftarrows }{\text{ 3Zn}}^{\text{2+}}{}_{\left(\text{aq}\right)}\text{ +}{\text{2PO}}_{\text{4}}{{}^{\text{3-}}}_{\left(\text{aq}\right)}\\ \text{Initial}\text{concentration }\left(\text{M}\right)\text{: }\text{0 }\text{0}\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: }\text{+3s }\text{+2s}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: }\text{3s }\text{2s}\end{array}$

$\begin{array}{l}\text{ Molar solubility}\text{of}{\text{Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\text{=}\frac{\text{5}\text{.9}\times {\text{10}}^{\text{-5}}\text{g}{\text{Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}{\text{1}\text{L}}\text{×}\frac{\text{1mol}{\text{Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}{\text{386}{\text{.11g Zn}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}}\\ \text{S}\text{=}1.52\text{×}{\text{10}}^{\text{-7}}\text{mol/L}\\ \\ {\text{K}}_{\text{sp}}{\text{= [Zn}}^{\text{2+}}{\text{]}}^{\text{3}}{{\text{[PO}}_4}^{{}^{\text{3-}}}\\ {\text{K}}_{\text{sp}}\text{= }{\left(\text{3s}\right)}^3{\left(\text{2s}\right)}^2\\ \\ {\text{K}}_{\text{sp}}\text{=}108{\text{(}}^{1.52}\\ {\text{K}}_{\text{sp}}\text{=}\text{9}\text{×}{\text{10}}^{\text{-33}}\end{array}$