Chapter 21, Problem 21.65QP

Interpretation Introduction

Interpretation: The formula of hydroxyapatite is stated to be ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\left(\text{OH}\right)}_2$. The value of $K_{\text{sp}}$ of ${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\left(\text{OH}\right)$ is given to be $2.3\times {10}^{-59}$. The $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\left(\text{OH}\right)}_2$ 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 ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\left(\text{OH}\right)}_2$.

Answer to Problem 21.65QP

Solution

The $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\left(\text{OH}\right)}_2$ is $\underset{\_}{5.29\times 10^{-118}}$.

Explanation of Solution

Explanation

Given

The $K_{\text{sp}}$ value of ${\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\left(\text{OH}\right)$ is $2.3\times {10}^{-59}$.

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

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

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

$K_{\text{sp}}\text{of}{\text{Ca}}_5{\left({\text{PO}}_4\right)}_3\left(\text{OH}\right)={\left[{\text{Ca}}^{2+}\right]}^5{\left[{\left({\text{PO}}_4\right)}^{3-}\right]}^3{\left[{\text{OH}}^{-}\right]}^1$

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

$2.3\times {10}^{-59}={\left[{\text{Ca}}^{2+}\right]}^5{\left[{\left({\text{PO}}_4\right)}^{3-}\right]}^3{\left[{\text{OH}}^{-}\right]}^1$ (1)

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

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

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

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

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

$\begin{array}{c}{K}_{\text{sp}}={\left(2.3\times {10}^{-59}\right)}^2\\ =\underset{\_}{5.29\times 10^{-118}}\end{array}$

Therefore, $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\left(\text{OH}\right)}_2$ is $\underset{\_}{5.29\times 10^{-118}}$.

Conclusion

The $K_{\text{sp}}$ value of ${\text{Ca}}_{10}{\left({\text{PO}}_4\right)}_6{\left(\text{OH}\right)}_2$ is $\underset{\_}{5.29\times 10^{-118}}$.