Chapter 22, Problem 22.62QP

Interpretation Introduction

Interpretation: The formula of fluorapatite is ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$. The $K_{\text{sp}}$ of ${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\text{F}$ is given to be $3.2\times {10}^{-60}$. The $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$ 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 $K_{\text{sp}}$.

To determine: The $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$.

Answer to Problem 22.62QP

Solution

The $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$ is $\underset{\_}{10.24\times 10^{-120}}$.

Explanation of Solution

Explanation

Given

The $K_{\text{sp}}$ value of ${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\text{F}$ is $3.2\times {10}^{-60}$.

The dissolution of ${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\text{F}$ in water is,

${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\text{F}\rightleftharpoons 5{\text{Ca}}^{2+}+3{\left({\text{PO}}_4\right)}^{3-}+{\text{F}}^{-}$

The $K_{\text{sp}}$ for ${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\text{F}$ is calculated by the formula,

$K_{\text{sp}}={\left[{\text{Ca}}^{2+}\right]}^5{\left[{\left({\text{PO}}_4\right)}^{3-}\right]}^3{\left[{\text{F}}^{-}\right]}^1$

Substitute the given value of $K_{\text{sp}}$ for ${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\text{F}$ in water in the above expression.

$3.2\times {10}^{-60}={\left[{\text{Ca}}^{2+}\right]}^5{\left[{\left({\text{PO}}_4\right)}^{3-}\right]}^3{\left[{\text{F}}^{-}\right]}^1$ (1)

The dissolution of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$ in water is,

${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2\rightleftharpoons 10{\text{Ca}}^{2+}+6{\left({\text{PO}}_4\right)}^{3-}+2{\text{F}}^{-}$

The $K_{\text{sp}}$ for ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$ is calculated by the formula,

$\begin{array}{c}{K}_{\text{sp}}\text{of}{\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2={\left[{\text{Ca}}^{2+}\right]}^{10}{\left[{\left({\text{PO}}_4\right)}^{3-}\right]}^6{\left[{\text{F}}^{-}\right]}^2\\ ={\left\{{\left[{\text{Ca}}^{2+}\right]}^5{\left[{\left({\text{PO}}_4\right)}^{3-}\right]}^3{\left[{\text{F}}^{-}\right]}^1\right\}}^2\end{array}$

Substitute the value of equation $\left(1\right)$ in above expression.

$\begin{array}{c}{K}_{\text{sp}}={\left(3.2\times {10}^{-60}\right)}^2\\ =\underset{\_}{10.24\times 10^{-120}}\end{array}$

Therefore, $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$ is $\underset{\_}{10.24\times 10^{-120}}$.

Conclusion

The $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\text{F}}_2$ is $\underset{\_}{10.24\times 10^{-120}}$.