Chapter 21, Problem 21.64QP

(a)

Interpretation Introduction

Interpretation: The questions based on $K_{\text{sp}}$ value of the given tooth enamel are to be answered.

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: If the tooth enamel is more soluble than pure hydroxyapatite.

Answer to Problem 21.64QP

Solution

The tooth enamel is more soluble than pure hydroxyapatite.

Explanation of Solution

Explanation

Given

The $K_{\text{sp}}$ value of actual tooth enamel is $1\times {10}^{-58}$.

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

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

The $K_{\text{sp}}$ value of tooth enamel is $1\times {10}^{-58}$ and the $K_{\text{sp}}$ value of pure hydroxyapatite is $2.3\times {10}^{-59}$.

So, $K_{\text{sp}}$ value of actual tooth enamel is more than $K_{\text{sp}}$ value of pure hydroxyapatite.

Therefore, the tooth enamel is more soluble than pure hydroxyapatite.

(b)

Interpretation Introduction

To determine: If the measured value of $K_{\text{sp}}$ for tooth enamel support the idea that tooth enamel is a mixture of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ and a calcium phosphate, ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$.

Answer to Problem 21.64QP

Solution

The $K_{\text{sp}}$ for tooth enamel supports the idea that tooth enamel is a mixture of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ and a calcium phosphate, ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

Explanation

Given

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

The $K_{\text{sp}}$ value of actual tooth enamel is $1\times {10}^{-58}$.

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

The values of the solubility product of actual tooth enamel and the hydroxyapatite are very close. It supports the idea that tooth enamel could be a mixture of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ and a calcium phosphate ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$.

The values of both the solubility products are equal.

Therefore, the tooth enamel could be a mixture of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ and a calcium phosphate ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$.

(c)

Interpretation Introduction

To determine: The solubility in moles per liter of ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$ in water at $\text{25}\text{°C}$ and at $\text{pH}$ value of $7.00$.

Answer to Problem 21.64QP

Solution

The solubility in moles per liter of ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$ in water at $\text{25}\text{°C}$ and at $\text{pH}$ value of $7.00$ is $\underset{\_}{5.137\times 10^{-338}mol/L}$.

Explanation of Solution

Explanation

Given

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

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

The formula of calcium phosphate 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 $\text{pH}$ 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 ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$ is,

${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}\left(s\right)\rightleftharpoons {\text{8Ca}}^{2+}\left(aq\right)+2{\text{HPO}}_4^{2-}\left(aq\right)+4{\text{PO}}_4^{3-}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The $K_{\text{sp}}$ of ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$ is calculated by the formula,

$K_{\text{sp}}={\left[{\text{Ca}}^{2+}\right]}^8{\left[{\text{HPO}}_4^{2-}\right]}^2{\left[{\text{PO}}_4^{3-}\right]}^4{\left[{\text{H}}_{\text{2}}\text{O}\right]}^6$

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

The concentration of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ is small that is ${10}^{-7}$ so concentration of ${\text{H}}_{\text{2}}\text{O}$ is neglected.

Substitute the values in the above expression.

$\begin{array}{c}1.1\times {10}^{-47}={\left(8s\right)}^2{\left(2s\right)}^2{\left(4s\right)}^4\\ 1.1\times {10}^{-47}=17179869184s^{14}\\ s=\underset{\_}{5.137\times 10^{-338}mol/L}\end{array}$

The solubility in moles per liter of ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$ in water is $\underset{\_}{5.137\times 10^{-338}mol/L}$.

Conclusion

1. a) The tooth enamel is more soluble than pure hydroxyapatite.
2. b) The $K_{\text{sp}}$ for tooth enamel support the idea that tooth enamel is a mixture of hydroxyapatite, ${\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\left(\text{OH}\right)$ and calcium phosphate ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$.
3. c) The solubility in moles per liter of ${\text{Ca}}_8{\left({\text{HPO}}_{\text{4}}\right)}_2{\left({\text{PO}}_4\right)}_4\cdot 6{\text{H}}_{\text{2}}\text{O}$ in water at $\text{25}\text{°C}$ and at $\text{pH}$ value of $7.00$ is $\underset{\_}{5.137\times 10^{-338}mol/L}$.