Chapter 21, Problem 21.63QP

(a)

Interpretation Introduction

Interpretation: The questions based on tooth enamel are to be answered.

Concept introduction: The product of concentration of salt ions raised to the 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: If this calcium mineral is more or less soluble than hydroxyapatite.

Answer to Problem 21.63QP

Solution

Yes, calcium mineral is more soluble than hydroxyapatite.

Explanation of Solution

Explanation

Given

The $K_{\text{sp}}$ value of hydroxyapatite is $2.3\times {10}^{-59}$.

The $K_{\text{sp}}$ value of calcium phosphate is $1.1\times {10}^{-47}$.

The formula of hydroxyapatite is ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ and the formula of calcium mineral is ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$.

The solubility of the salt depends upon the value of solubility product. If the value of the solubility product is more, then the salt is more soluble and if the value of the solubility product is less, then the salt is less soluble.

The $K_{\text{sp}}$ value of hydroxyapatite is $2.3\times {10}^{-59}$ and the $K_{\text{sp}}$ value of calcium phosphate is $1.1\times {10}^{-47}$.

So, $K_{\text{sp}}$ value of calcium phosphate is more than $K_{\text{sp}}$ value of hydroxyapatite.

Therefore, calcium mineral is more soluble than hydroxyapatite.

(b)

Interpretation Introduction

To determine: The solubility in moles per liter of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ in water at $\text{25}\text{°C}$ and at $\text{pH}$ value is $7.00$.

Answer to Problem 21.63QP

Solution

The solubility in moles per liter of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ in water at $\text{25}\text{°C}$ and at $\text{pH}$ value of $7.00$ is $\underset{\_}{1.203\times 10^{-490}mol/L}$.

Explanation of Solution

Explanation

Given

The $K_{\text{sp}}$ value of hydroxyapatite is $2.3\times {10}^{-59}$.

The value of $\text{pH}$ is $7.00$.

The $\text{pH}$ is calculated by the formula,

$\text{pH=}-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\text{pH}$ in the above expression.

$\begin{array}{c}7=-\log \left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]={10}^{-7}\end{array}$

The concentration of ${\text{OH}}^{-}$ is calculated by the formula,

$\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]={10}^{-14}$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{{10}^{-7}}\\ ={10}^{-7}\end{array}$

The dissociation of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ 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 $K_{\text{sp}}$ of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ is calculated by the formula,

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

The solubility of above ions is assumed to be $s$.

The concentration of ${\text{OH}}^{-}$ is small, that is ${10}^{-7}$, so it does not make much difference.

Substitute the values in the above expression.

$\begin{array}{c}2.3\times {10}^{-59}={\left(5s\right)}^5{\left(3s\right)}^3\left({10}^{-7}+s\right)\\ 2.3\times {10}^{-59}=84375s^9\\ s=\underset{\_}{1.203\times 10^{-490}mol/L}\end{array}$

Therefore, the solubility in moles per liter of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ in water is $\underset{\_}{1.203\times 10^{-490}mol/L}$.

(c)

Interpretation Introduction

To determine: The solubility of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ at $\text{pH}$ value of $5.00$.

Answer to Problem 21.63QP

Solution

The solubility of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ at $\text{pH}$ value of $5.00$ is $\underset{\_}{1.203\times 10^{-490}mol/L}$.

Explanation of Solution

Explanation

Given

The $K_{\text{sp}}$ value of hydroxyapatite is $2.3\times {10}^{-59}$.

The value of $\text{pH}$ is $5.00$.

The $\text{pH}$ is calculated by the formula,

$\text{pH=}-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\text{pH}$ in the above expression.

$\begin{array}{c}5=-\log \left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]={10}^{-5}\end{array}$

The concentration of ${\text{OH}}^{-}$ is calculated by the formula,

$\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]={10}^{-14}$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{{10}^{-5}}\\ ={10}^{-9}\end{array}$

The dissociation of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ 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 $K_{\text{sp}}$ of ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ is calculated by the formula,

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

The solubility of above ions is assumed to be $s$.

The concentration of ${\text{OH}}^{-}$ is small, that is ${10}^{-9}$, so, it does not make much difference.

Substitute the values in the above expression.

$\begin{array}{c}2.3\times {10}^{-59}={\left(5s\right)}^5{\left(3s\right)}^3\left({10}^{-9}+s\right)\\ 2.3\times {10}^{-59}=84375s^9\\ s=\underset{\_}{1.203\times 10^{-490}mol/L}\end{array}$

Therefore, the solubility of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ at $\text{pH}$ value of $5.00$ is $\underset{\_}{1.203\times 10^{-490}mol/L}$.

Conclusion

1. a) Yes, calcium mineral is more soluble than hydroxyapatite.
2. b) The solubility in moles per liter of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ in water at $\text{25}\text{°C}$ and at $\text{pH}$ value of $7.00$ is $\underset{\_}{1.203\times 10^{-490}mol/L}$.
3. c) The solubility of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ at $\text{pH}$ value of $5.00$ is $\underset{\_}{1.203\times 10^{-490}mol/L}$.