Chapter 21, Problem 21.62QP

Interpretation Introduction

Interpretation: The equilibrium constant expression for the given reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions which lead to the formation of calcium fluoride is to be written. The value of $K$ for the given reaction is to be calculated.

Concept introduction: The product of concentration of salt ions raised to power of the coefficients of ions is known as the solubility product of a salt. The solubility product constant is denoted by $K_{\text{sp}}$.

The equilibrium constant expression for the given reaction is written by ratio of concentration of products to concentrations of reactants.

To determine: The equilibrium constant expression for the given reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions which lead to the formation of calcium fluoride and the value of $K$ for the given reaction.

Answer to Problem 21.62QP

Solution

The equilibrium constant expression for the given reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions which lead to the formation of calcium fluoride is,

$K=\frac{{\left[{\text{CaF}}_2\right]}^5{\left[{\text{PO}}_4^{3-}\right]}^3\left[{\text{OH}}^{-}\right]}{\left[\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)\right]{\left[{\text{F}}^{{}^{-}}\right]}^{10}}$

The value of $K$ for the given reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions which lead to the formation of calcium fluoride is $\underset{\_}{2.5\times 10^{-7}}$.

Explanation of Solution

Explanation

Given

The solubility product constant for equation $1$ is $2.3\times {10}^{-59}$.

The solubility product constant for equation $2$ is $3.9\times {10}^{-11}$.

The given reaction is,

$\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)\left(s\right)+10{\text{F}}^{-}\to 5{\text{CaF}}_2\left(s\right)+3{\text{PO}}_4^{3-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

In the above reaction, too much fluoride ions lead to the formation of calcium fluoride ions.

The equilibrium constant expression for the given reaction is written by ratio of concentration of products to concentrations of reactants.

$K=\frac{{\left[{\text{CaF}}_2\right]}^5{\left[{\text{PO}}_4^{3-}\right]}^3\left[{\text{OH}}^{-}\right]}{\left[\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)\right]{\left[{\text{F}}^{{}^{-}}\right]}^{10}}$

The dissociation of $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ into its ions is,

$\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)\left(s\right)\iff 5{\text{Ca}}^{2+}\left(aq\right)+3{\text{PO}}_4^{3-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$ (1)

The dissociation of calcium fluoride into ions is,

${\text{CaF}}_2\left(s\right)\iff {\text{Ca}}^{2+}\left(aq\right)+2{\text{F}}^{-}\left(aq\right)$ (2)

The reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions gives ${\text{CaF}}_2$. So, the formation of ${\text{CaF}}_2$ indicates the reverse of the reaction $\left(2\right)$ takes place.

Then,

${\text{Ca}}^{2+}\left(aq\right)+2{\text{F}}^{-}\left(aq\right)\iff {\text{CaF}}_2\left(s\right)$ (3)

$\begin{array}{c}{K}^{\text{'}}=1/K_{\text{sp}}\\ =1/3.9\times {10}^{-11}\end{array}$

The value of $K$ is calculated by adding equation $\left(1\right)$ and multiplying the equation $\left(3\right)$ with $5$.

$\begin{array}{l}\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)\left(s\right){\text{+5Ca}}^{2+}\left(aq\right)+10{\text{F}}^{-}\left(aq\right)\rightleftharpoons \\ 5{\text{CaF}}_2\left(s\right)+5{\text{Ca}}^{2+}\left(aq\right)+3{\text{PO}}_4^{3-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\end{array}$

The result of the above equation is,

$\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)\left(s\right)+10{\text{F}}^{-}\to 5{\text{CaF}}_2\left(s\right)+3{\text{PO}}_4^{3-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The value of $K^{\text{'}}$ for equation $\left(3\right)$ is calculated by the formula,

$\begin{array}{c}{K}^{\text{'}}=\frac{1}{{\left(3.9\times {10}^{-11}\right)}^5}\\ =902.2\times {10}^{-55}\end{array}$

The value of $K$ is calculated by the formula,

$K=\frac{K_{{\text{sp}}_{{}_1}}}{K^{\text{'}}}$

Where,

• $K$ is equilibrium constant.
• $K_{{\text{sp}}_{{}_1}}$ is solubility product constant for equation $1$.
• $K^{\text{'}}$ is inverse solubility product for equation $3$.

Substitute the value in the above expression.

$\begin{array}{c}K=\frac{2.3\times {10}^{-59}}{902.2\times {10}^{-55}}\\ =0.0025\times {10}^{-4}\\ =\underset{\_}{2.5\times 10^{-7}}\end{array}$

Therefore, the value of $K$ for the given reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions which lead to the formation of calcium fluoride is $\underset{\_}{2.5\times 10^{-7}}$.

Conclusion

The equilibrium constant expression for the given reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions which lead to the formation of calcium fluoride is,

$K=\frac{{\left[{\text{CaF}}_2\right]}^5{\left[{\text{PO}}_4^{3-}\right]}^3\left[{\text{OH}}^{-}\right]}{\left[\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)\right]{\left[{\text{F}}^{{}^{-}}\right]}^{10}}$

The value of $K$ for the given reaction between $\text{Ca}{\left({\text{PO}}_{\text{4}}\right)}_3\left(\text{OH}\right)$ and fluoride ions which lead to the formation of calcium fluoride is $\underset{\_}{2.5\times 10^{-7}}$.