Chapter 19.3, Problem 19.8BFP

(a)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{CaF}}_{\text{2}}$ in water has to be calculated.

Concept Introduction:

Solubility product constant:

The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.

The equilibrium constant of more soluble ionic compound is given by ${\text{K}}_{\text{SP}}$ and it is expressed by product of each ion present in the compound which are raised to the power of number respective ion present in the compound to give a maximum solubility of the compound.

$\begin{array}{l}{\text{M}}_{\text{m}}{\text{X}}_{\text{x}}{}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ mM}}^{\text{n+}}{}_{\text{(aq)}}{\text{ + xX}}^{\text{y-}}{}_{\text{(aq)}}\\ \\ {\text{K}}_{\text{sp}}{\text{ = [M}}^{\text{n+}}{\text{]}}^{\text{m}}{{\text{[X}}^{\text{y-}}\text{]}}^{\text{x}}\end{array}$

Molar solubility:

Molar solubility (S) is given as solubility in moles per litre.

$\text{S =}\frac{\text{Solubility}\text{in}\text{gram}}{\text{1L}}\text{×}\frac{\text{1}\text{mol}}{\text{Molarmass}\text{(g)}}$

Explanation of Solution

Given information,

  The ${\text{K}}_{\text{SP}}$ value of ${\text{CaF}}_{\text{2}}$ is $\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}$.

Write the solubility equilibrium for ${\text{CaF}}_{\text{2}}$,

${\text{CaF}}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{Ca}}_{\left(\text{aq}\right)}^{\text{2+}}\text{+}\text{2}{\text{F}}_{}^{\text{-}}{}_{\left(\text{aq}\right)}$

Construct ICE table for the above equilibrium reaction,

$\begin{array}{l}{\text{CaF}}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{Ca}}_{\left(\text{aq}\right)}^{\text{2+}}\text{+}\text{2}{\text{F}}_{}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial}\text{(M):}\text{-}\text{0}\text{0}\\ \text{Change}\text{(M):}\text{-}\text{+S}\text{+2S}\\ \text{Equilibrium}\text{(M):}\text{-}\text{S}\text{2S}\end{array}$

Calculate the molar solubility of ${\text{CaF}}_{\text{2}}$ by using solubility product expression,

$\begin{array}{l}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}\text{=}\left(\text{S}\right){\left(\text{2S}\right)}^{\text{2}}\text{=}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\\ \\ \text{4}{\text{S}}^{\text{3}}\text{=}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\\ \\ \text{therefore,}\text{S}\text{=}\text{2}\text{.0}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{.}\end{array}$

Therefore, the molar solubility of ${\text{CaF}}_{\text{2}}$ in water is $\text{2}\text{.0}\text{×}{\text{10}}^{\text{-4}}\text{M}$.

(b)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{CaF}}_{\text{2}}$ in $\text{0}\text{.20}\text{M}{\text{CaCl}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Solubility product constant:

The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.

The equilibrium constant of more soluble ionic compound is given by ${\text{K}}_{\text{SP}}$ and it is expressed by product of each ion present in the compound which are raised to the power of number respective ion present in the compound to give a maximum solubility of the compound.

$\begin{array}{l}{\text{M}}_{\text{m}}{\text{X}}_{\text{x}}{}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ mM}}^{\text{n+}}{}_{\text{(aq)}}{\text{ + xX}}^{\text{y-}}{}_{\text{(aq)}}\\ \\ {\text{K}}_{\text{sp}}{\text{ = [M}}^{\text{n+}}{\text{]}}^{\text{m}}{{\text{[X}}^{\text{y-}}\text{]}}^{\text{x}}\end{array}$

Molar solubility:

Molar solubility (S) is given as solubility in moles per litre.

$\text{S =}\frac{\text{Solubility}\text{in}\text{gram}}{\text{1L}}\text{×}\frac{\text{1}\text{mol}}{\text{Molarmass}\text{(g)}}$

Explanation of Solution

Given information,

  The ${\text{K}}_{\text{SP}}$ value of ${\text{CaF}}_{\text{2}}$ is $\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}$.

Write the solubility equilibrium for ${\text{CaF}}_{\text{2}}$,

${\text{CaF}}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{Ca}}_{\left(\text{aq}\right)}^{\text{2+}}\text{+}\text{2}{\text{F}}_{}^{\text{-}}{}_{\left(\text{aq}\right)}$

Construct ICE table for the above equilibrium reaction,

$\begin{array}{l}{\text{CaF}}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{Ca}}_{\left(\text{aq}\right)}^{\text{2+}}\text{+}\text{2}{\text{F}}_{}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial}\text{(M):}\text{-}\text{0}\text{.20}\text{0}\\ \text{Change}\text{(M):}\text{-}\text{+S}\text{+2S}\\ \text{Equilibrium}\text{(M):}\text{-}\text{0}\text{.20}\text{+}\text{S}\text{2S}\end{array}$

Assume that $\text{0}\text{.20}\text{+}\text{S}$ is approximately equal to 0.20, which is a good approximation based on the fact that $\text{0}\text{.20}\text{>}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\text{,}{\text{K}}_{\text{SP}}$.

Calculate the molar solubility of ${\text{CaF}}_{\text{2}}$ by using solubility product expression,

$\begin{array}{l}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}\text{=}\left(\text{0}\text{.20}\right){\left(\text{2S}\right)}^{\text{2}}\text{=}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\\ \\ \left(\text{0}\text{.20}\right)\text{4}{\text{S}}^{\text{2}}\text{=}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\\ \\ \text{therefore,}\text{S}\text{=}\text{6}\text{.3}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{.}\end{array}$

Therefore, the molar solubility of ${\text{CaF}}_{\text{2}}$ in $\text{0}\text{.20}\text{M}{\text{CaCl}}_{\text{2}}$ is $\text{6}\text{.3}\text{×}{\text{10}}^{\text{-6}}\text{M}$.

(c)

Interpretation Introduction

Interpretation:

The molar solubility of ${\text{CaF}}_{\text{2}}$ in $\text{0}\text{.20}\text{M}{\text{NiF}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Solubility product constant:

The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.

The equilibrium constant of more soluble ionic compound is given by ${\text{K}}_{\text{SP}}$ and it is expressed by product of each ion present in the compound which are raised to the power of number respective ion present in the compound to give a maximum solubility of the compound.

$\begin{array}{l}{\text{M}}_{\text{m}}{\text{X}}_{\text{x}}{}_{\text{(s)}}\stackrel{}{\rightleftharpoons }{\text{ mM}}^{\text{n+}}{}_{\text{(aq)}}{\text{ + xX}}^{\text{y-}}{}_{\text{(aq)}}\\ \\ {\text{K}}_{\text{sp}}{\text{ = [M}}^{\text{n+}}{\text{]}}^{\text{m}}{{\text{[X}}^{\text{y-}}\text{]}}^{\text{x}}\end{array}$

Molar solubility:

Molar solubility (S) is given as solubility in moles per litre.

$\text{S =}\frac{\text{Solubility}\text{in}\text{gram}}{\text{1L}}\text{×}\frac{\text{1}\text{mol}}{\text{Molarmass}\text{(g)}}$

Explanation of Solution

Given information,

  The ${\text{K}}_{\text{SP}}$ value of ${\text{CaF}}_{\text{2}}$ is $\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}$.

Write the solubility equilibrium for ${\text{CaF}}_{\text{2}}$,

${\text{CaF}}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{Ca}}_{\left(\text{aq}\right)}^{\text{2+}}\text{+}\text{2}{\text{F}}_{}^{\text{-}}{}_{\left(\text{aq}\right)}$

There are two fluoride ions per ${\text{NiF}}_{\text{2}}$, so an ${\text{NiF}}_{\text{2}}$ concentration of $\text{0}\text{.20}\text{M}$ gives a fluoride ion concentration of $\text{0}\text{.40}\text{M}$.

Construct ICE table for the above equilibrium reaction,

$\begin{array}{l}{\text{CaF}}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\rightleftharpoons }{\text{Ca}}_{\left(\text{aq}\right)}^{\text{2+}}\text{+}\text{2}{\text{F}}_{}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial}\text{(M):}\text{-}\text{0}\text{0}\text{.40}\\ \text{Change}\text{(M):}\text{-}\text{+S}\text{+2S}\\ \text{Equilibrium}\text{(M):}\text{-}\text{S}\text{0}\text{.40}\text{+}\text{2S}\end{array}$

Assume that $\text{0}\text{.40}\text{+}\text{S}$ is approximately equal to 0.40, which is a good approximation based on the fact that $\text{0}\text{.40}\text{>}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\text{,}{\text{K}}_{\text{SP}}$.

Calculate the molar solubility of ${\text{CaF}}_{\text{2}}$ by using solubility product expression,

$\begin{array}{l}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}\text{=}\left(\text{S}\right){\left(\text{0}\text{.40}\text{+}\text{2S}\right)}^{\text{2}}\text{=}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\\ \\ \left(\text{S}\right){\left(\text{0}\text{.40}\right)}^{\text{2}}\text{=}\text{3}\text{.2}\text{×}{\text{10}}^{\text{-11}}\\ \\ \text{therefore,}\text{S}\text{=}\text{2}\text{.0}\text{×}{\text{10}}^{\text{-10}}\text{M}\text{.}\end{array}$

Therefore, the molar solubility of ${\text{CaF}}_{\text{2}}$ in $\text{0}\text{.20}\text{M}{\text{NiF}}_{\text{2}}$ is $\text{2}\text{.0}\text{×}{\text{10}}^{\text{-10}}\text{M}$.