Chapter 17, Problem 17.50QP

(a)

Interpretation Introduction

Interpretation:

The solubility product of given compounds has to be calculated.

Concept introduction:

• 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 ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.
• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.

To calculate: solubility product of ${\text{SrF}}_{\text{2}}$

Answer to Problem 17.50QP

${\text{K}}_{\text{sp}}$ of ${\text{SrF}}_{\text{2}}$ is $\text{7}\text{.8}\text{×}{\text{10}}^{\text{-10}}$

Explanation of Solution

${\text{SrF}}_{\text{2}}\text{=}\text{7}\text{.3 ×}{\text{10}}^{\text{-2}}\text{g/L}$

Calculate the molar solubility

$\begin{array}{l}\text{=}\frac{\text{7}\text{.3 ×}{\text{10}}^{\text{-2}}{\text{g SrF}}_{\text{2}}}{\text{1L}}\text{×}\frac{{\text{1 mol SrF}}_{\text{2}}}{\text{125}{\text{.6 g SrF}}_{\text{2}}}\\ \text{=}\text{5}\text{.8 ×}{\text{10}}^{\text{-4}}\text{mol /L =}\text{s}\end{array}$

The molar solubility can be calculated using given moles and molecular weight of ${\text{SrF}}_{\text{2}}$

Calculate the ${\text{K}}_{\text{sp}}$ of ${\text{SrF}}_{\text{2}}$

For calculation convenience consider ''s" as molar solubility of ${\text{SrF}}_{\text{2}}$

$\begin{array}{l}{\text{SrF}}_{\text{2}}{}_{\text{(s)}}\to {\text{ Sr}}^{\text{2+}}{}_{\text{(aq)}}{\text{ + 2F}}^{\text{-}}{}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-s }+\text{s}\text{+2s}\\ \text{equilibrium concentration (M): }\text{s}\text{2s}\\ \\ {\text{K}}_{\text{sp}}{\text{= [Sr}}^{\text{2+}}{\text{][F}}^{\text{-}}{\text{]}}^{\text{2}}\\ {\text{= (s)(2s)}}^{\text{2}}{\text{ = 4s}}^{\text{3}}\\ \text{=}\text{4(5}\text{.8 ×}{\text{10}}^{\text{-4}}{\text{)}}^{\text{3}}\\ \text{=}\text{7}\text{.8}\text{×}{\text{10}}^{\text{-10}}\end{array}$

The ${\text{K}}_{\text{sp}}$ value of ${\text{SrF}}_{\text{2}}$ can be calculated using the concentrations of ions.  By doing simple mathematical operations in solubility product expression, ${\text{K}}_{\text{sp}}$ of ${\text{SrF}}_{\text{2}}$ can be determined.

(b)

Interpretation Introduction

Interpretation:

The solubility product of given compounds has to be calculated.

Concept introduction:

• 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 ${\text{A}}_{\text{3}}\text{B}$ ${\text{K}}_{\text{sp}}\text{=}{\left[\text{A}\right]}^{\text{3}}\left[\text{B}\right]$.
• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.

To calculate: the ${\text{K}}_{\text{sp}}$ of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$

Answer to Problem 17.50QP

${\text{K}}_{\text{sp}}$ of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{1}\text{.8}\text{×}{\text{10}}^{\text{-18}}$

Explanation of Solution

${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\text{6}\text{.7 ×}{\text{10}}^{\text{-3}}\text{g/L}$

Calculate the molar solubility of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$

$\begin{array}{l}\text{=}\frac{\text{6}\text{.7 ×}{\text{10}}^{\text{-3}}{\text{g Ag}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{1L}}\text{×}\frac{{\text{1 mol Ag}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{418}{\text{.7 g Ag}}_{\text{3}}{\text{PO}}_{\text{4}}}\\ \text{=}\text{1}\text{.6 ×}{\text{10}}^{\text{-5}}\text{mol /L =}\text{s}\end{array}$

The molar solubility can be calculated using given moles and molecular weight of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$

Calculate the ${\text{K}}_{\text{sp}}$ of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$

For calculation convenience consider ''s" as molar solubility of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$

$\begin{array}{l}{\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}{}_{\text{(s)}}\to {\text{ 3Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{ + PO}}_{\text{4}}{{}^{\text{3-}}}_{\text{(aq)}}\\ \\ \text{Initial concentration(M): }\text{0}\text{0}\\ \text{Change in concentration (M):}\text{-s }\text{-3s}\text{+s}\\ \text{equilibrium concentration (M): }\text{3s}\text{s}\\ \\ {\text{K}}_{\text{sp}}{\text{= [Ag}}^{\text{+}}{\text{]}}^{\text{3}}{\text{[PO}}_{\text{4}}{}^{\text{3-}}\text{]}\\ {\text{= (3s)}}^{\text{3}}{\text{(s) = 27s}}^{\text{4}}\\ \text{=}\text{27(1}\text{.6 ×}{\text{10}}^{\text{-5}}{\text{)}}^{\text{4}}\\ \text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-18}}\end{array}$

The ${\text{K}}_{\text{sp}}$ value of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ can be calculated using the concentrations of ions.  By doing simple mathematical operations in solubility product expression, ${\text{K}}_{\text{sp}}$ of ${\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4}}$ can be determined.