Chapter 19, Problem 19.122P

Interpretation Introduction

Interpretation:

Solubility of $C{a}_5{\left(P{O}_4\right)}_{3}OH$ and $C{a}_5{\left(P{O}_4\right)}_{3}OH$ has to be calculated. 

Concept Introduction:

Solubility Product:

The solubility product constant, $K_{sp}$ is the equilibrium constant for a solid substance dissolving in an aqueous solution.  It represents the level at which a solute dissolves in solution. 

Consider the general dissolution reaction below (in aqueous solutions):

$a{A}_{\left(s\right)}\rightleftarrows c{C}_{\left(aq\right)}+d{D}_{\left(aq\right)}$

The solubility product for the above solid can be written as

$K_{sp}={\left[C\right]}^c{\left[D\right]}^d$

Where,

$\begin{array}{l}\left[C\right]\text{}=ConcentrationofC\\ \\ \left[D\right]=ConcentrationofD\end{array}$

Solubility:

The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.  It is expressed by the symbol S.  Unit of solubility is g/L. 

Answer to Problem 19.122P

The solubility of $C{a}_5{\left(P{O}_4\right)}_{3}OH$ and $C{a}_5{\left(P{O}_4\right)}_{3}OH$ is $0.014g/L$ and $3.110^{-5}g/L$ respectively. 

Explanation of Solution

Given Data:

$\begin{array}{l}{K}_{sp}ofC{a}_5{\left(P{O}_4\right)}_{3}OH=6.8\times {10}^{-37}\\ \\ K_{sp}ofC{a}_5{\left(P{O}_4\right)}_{3}F=1.0\times {10}^{-60}\end{array}$

• Calculating the solubility of $C{a}_5{\left(P{O}_4\right)}_{3}OH$:

Writing the equation and ion-product expression:

  $C{a}_5{\left(P{O}_4\right)}_{3}OH\left(s\right)\rightleftarrows 5C{a}^{2+}\left(aq\right)+3P{O}_4{}^{3-}\left(aq\right)+O{H}^{-}\left(aq\right)$

  $K_{sp}={\left[C{a}^{2+}\right]}^5{\left[P{O}_4{}^{3-}\right]}^3\left[O{H}^{-}\right]=6.8\times {10}^{-37}$

  $\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{Concentration\left(M\right)C{a}_5{\left(P{O}_4\right)}_{3}OH\left(s\right)\rightleftarrows 5C{a}^{2+}\left(aq\right)+3P{O}_4{}^{3-}\left(aq\right)+O{H}^{-}\left(aq\right)}\\ Initial-000\\ Change-+5S+3S+S\end{array}}\\ Equilibrium-5S3SS\end{array}$

Substituting into the ion-product expression and solving for S:

$\begin{array}{l}{K}_{sp}=6.8\times {10}^{-37}\\ \\ ={\left[C{a}^{2+}\right]}^5{\left[P{O}_4{}^{3-}\right]}^3\left[O{H}^{-}\right]\\ \\ ={\left(5S\right)}^5{\left(3S\right)}^3\left(S\right)\\ \\ =84375{\left(S\right)}^9\end{array}$

$\begin{array}{l}S=\sqrt[9]{\frac{6.8\times {10}^{-37}}{84375}}\\ \\ =2.7\times {10}^{-5}M\\ \\ =2.7\times {10}^{-5}\frac{mol}{L}\times 502.32\frac{g}{mol}\left[\because MolecularWeightofC{a}_5{\left(P{O}_4\right)}_{3}OH=502.32g/mol\right]\\ \\ =0.01356g/L=0.014g/L.\end{array}$

Therefore, the solubility value of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\text{OH}$ was $0.014g/L.$

• Calculating the solubility of$C{a}_5{\left(P{O}_4\right)}_{3}F$:

Write the equation and ion-product expression:

$C{a}_5{\left(P{O}_4\right)}_{3}F\left(s\right)\rightleftarrows 5C{a}^{2+}\left(aq\right)+3P{O}_4{}^{3-}\left(aq\right)+F^{-}\left(aq\right)$

$K_{sp}={\left[C{a}^{2+}\right]}^5{\left[P{O}_4{}^{3-}\right]}^3\left[F^{-}\right]=1.0\times {10}^{-60}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{Concentration\left(M\right)C{a}_5{\left(P{O}_4\right)}_{3}F\left(s\right)\rightleftarrows 5C{a}^{2+}\left(aq\right)+3P{O}_4{}^{3-}\left(aq\right)+F^{-}\left(aq\right)}\\ Initial-000\\ Change-+5S+3S+S\end{array}}\\ Equilibrium-5S3SS\end{array}$

Substituting into the ion-product expression and solving for S:

$\begin{array}{l}{K}_{sp}=6.8\times {10}^{-37}\\ \\ ={\left[C{a}^{2+}\right]}^5{\left[P{O}_4{}^{3-}\right]}^3\left[F^{-}\right]\\ \\ ={\left(5S\right)}^5{\left(3S\right)}^3\left(S\right)=84375{\left(S\right)}^9.\end{array}$

$\begin{array}{l}S=\sqrt[9]{\frac{1.0\times {10}^{-60}}{84375}}\\ \\ =6.1090861\times {10}^{-8}M=6.1\times {10}^{-8}M\\ \\ =6.1090861\times {10}^{-8}\frac{mol}{L}\times 504.32\frac{g}{mol}\\ \\ =3.0808732\times {10}^{-5}g/L\left[\because MolecularWeightofC{a}_5{\left(P{O}_4\right)}_{3}F=504.32g/mol\right]\\ \\ =3.1\times {10}^{-5}g/L.\end{array}$

Therefore, the solubility value of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}F$ was $3.1\times 10^{-5}g/L.$