Chapter 22, Problem 22.66QA

Interpretation Introduction

To find:

The $K_{sp}$ of ${Ca}_{10}{(PO}_4{)}_6{F}_2$ if the $K_{sp}$ of ${Ca}_5{(PO}_4{)}_{3}F$ is $3.2 \times {10}^{-60}$

Answer to Problem 22.66QA

Solution:

The $K_{sp}$ of ${Ca}_{10}{(PO}_4{)}_6{F}_2$ is $1.0\times {10}^{-119}$.

Explanation of Solution

1) Concept:

We are given $K_{sp}$ of ${Ca}_5{(PO}_4{)}_{3}F$,$3.2 \times {10}^{-60}$ and asked to find $K_{sp}$ of ${Ca}_{10}{(PO}_4{)}_6{F}_2$. As we noticed that each number of atoms in ${Ca}_{10}{(PO}_4{)}_6{F}_2$ is double to that of ${Ca}_5{(PO}_4{)}_{3}F$. The $K_{sp}$ of ${Ca}_{10}{(PO}_4{)}_6{F}_2$ can be calculated by squaring the $K_{sp}$ of ${Ca}_5{(PO}_4{)}_{3}F$ and it must be less than $K_{sp}$ of ${Ca}_5{(PO}_4{)}_{3}F$ since the number of atoms increases in the interested compound. 

2) Given:

$K_{sp}$ of ${Ca}_5{(PO}_4{)}_{3}F$ is $3.2 \times {10}^{-60}$     

3) Calculations:

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

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

The $K_{sp}$ expression for ${Ca}_5{\left(P{O}_4\right)}_{3}F$ is

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

Squaring of equation (2) gives

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

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

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

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

Therefore,

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2}={\left(3.2\times {10}^{-60}\right)}^2$

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2}=1.02\times {10}^{-119}$

$K_{sp {Ca}_{10}{\left(P{O}_4\right)}_6{F}_2}=1.0\times {10}^{-119}$

Hence the $K_{sp}$ of ${Ca}_{10}{\left(P{O}_4\right)}_6{F}_2$ is $1.0\times {10}^{-119}$.

Conclusion:

We compared the $K_{sp}$ expression of both ${Ca}_{10}{\left(P{O}_4\right)}_6{F}_2$ and ${Ca}_5{\left(P{O}_4\right)}_{3}F$ and calculated the $K_{sp}$ of ${Ca}_{10}{\left(P{O}_4\right)}_6{F}_2$.