Chapter 22, Problem 22.65QA

Interpretation Introduction

To find:

Some sources give the formula of hydroxyapatite as ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$. If the $K_{sp}$ of ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ is, $2.3\times {10}^{-59}$ what is the $K_{sp}$ of ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$?

Answer to Problem 22.65QA

Solution:

The $K_{sp}$ of ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2=5.3\times {10}^{-118}$.

Explanation of Solution

1) Concept:

Solubility product constant $K_{sp}$ is the product of the concentration of the ions raised to the power of their stoichiometric coefficients present in a saturated solution of an ionic compound.

The dissociation equation for ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$ is

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

The dissociation equation for ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ is

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)\rightleftharpoons 5{Ca}^{2+}\left(aq\right)+3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$; $K_{sp}=2.3\times {10}^{-59}$

As the ions produced from both are the same, we can compare the $K_{sp}$ expression and calculate $K_{sp}$ of ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$.

2) Given:

$K_{sp {Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)}=2.3\times {10}^{-59}$

3) Calculation:

The $K_{sp}$ expression for ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$ is

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2}={\left[{Ca}^{2+}\right]}^{10}{\left[{PO}_4^{3-}\right]}^6{\left[{OH}^{-}\right]}^2$ equation (1)

The $K_{sp}$ expression for ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ is

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

Squaring of equation (2) gives,

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

${\left[K_{sp {Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)}\right]}^2={\left[{Ca}^{2+}\right]}^{10}{\left[{PO}_4^{3-}\right]}^6{\left[{OH}^{-}\right]}^2$    equation (3)

The right hand side of equations (1) and (3) is the same. Therefore, we can write,

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2}={\left[K_{sp {Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)}\right]}^2$

Therefore,

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2}={\left(2.3\times {10}^{-59}\right)}^2$

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2}=5.29\times {10}^{-118}$

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2}=5.3\times {10}^{-118}$

Hence, the $K_{sp}$ of ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$ is $5.3\times {10}^{-118}$.

Conclusion:

We compared the $K_{sp}$ expression of both ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$ and ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ and calculated the $K_{sp}$ of ${Ca}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2$.