Chapter 22, Problem 22.64QA

Interpretation Introduction

To find:

The $K_{sp}$ of actual tooth enamel is reported to be  $1\times {10}^{-58}$.

a) Does this mean that tooth enamel is more soluble than pure hydroxyapatite

$(K_{sp}=2.3\times {10}^{-59})$?

b) Does the measured value of $K_{sp}$ for tooth enamel support the idea that tooth enamel is a mixture of hydroxyapatite,${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ and a calcium phosphate,

${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$($K_{sp}=1.1\times {10}^{-47})$?

c) Calculate the solubility in moles per liter of a calcium phosphate,

${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$($K_{sp}=1.1\times {10}^{-47})$ in water at ${25}^{o}C$.

Answer to Problem 22.64QA

Solution:

a) The tooth enamel is more soluble because it has higher $K_{sp}$ than that of hydroxyapatite.

b) Yes, the measured value of $K_{sp}$ for the tooth enamel supports the idea that the tooth enamel is a mixture of hydroxyapatite and a calcium phosphate, but the amount of hydroxyapatite would be much more than that of calcium phosphate.

c) The solubility of calcium phosphate, ${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$ is $8.2\times {10}^{-5} M$.

Explanation of Solution

1) Concept:

$K_{sp}$(solubility product constant) is the equilibrium between the solid compound and its respective ions in the solution. The $K_{sp}$ expression is the product of concentration of ions raised to a power equal to the coefficient of that ion in the balanced equation.

$K_{sp}={\left[cation\right]}^x{\left[anion\right]}^y$

where $x$ and $y$ are the coefficient of cations and anions in the balanced dissociation equation.

To determine which compound is more or less soluble, we need to compare the $K_{sp}$ values, higher the $K_{sp}$ more soluble the compound is.

2) Formula:

$K_{sp}={\left[cation\right]}^x{\left[anion\right]}^y$

3) Given:

i) Tooth enamel, $K_{sp}=1\times {10}^{-58}$

ii) Calcium mineral, ${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$, $K_{sp}=1.1\times {10}^{-47}$

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

4) Calculation:

a) Is the tooth enamel more or less soluble than hydroxyapatite?

We are given the $K_{sp}$ for both as

Tooth enamel, $K_{sp}=1\times {10}^{-58}$

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

The compound with higher $K_{sp}$ value is more soluble.  Therefore, the tooth enamel is more soluble because it has higher $K_{sp}$ than that of hydroxyapatite.

b) The value of $K_{sp}$ for the tooth enamel is $1\times {10}^{-58}$

The measured value of $K_{sp}$ of the tooth enamel is in between $K_{sp}$ values of calcium phosphate and hydroxyapatite. Hence, we can say that the tooth enamel is a mixture of calcium phosphate and hydroxyapatite, but the $K_{sp}$  is not an average of $K_{sp}$ of both, it is more close to the $K_{sp}$ of hydroxyapatite, ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$,$K_{sp}=2.3\times {10}^{-59}$. This indicates that the mixture will be of greater amount of hydroxyapatite, ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ than calcium mineral, ${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$,$K_{sp}=1.1\times {10}^{-47}$.

c) The solubility in moles per liter of calcium phosphate,${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$:

The dissociation equation for calcium phosphate, ${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$ is,

${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O\mathrm{ }\left(s\right)\rightleftharpoons 8{Ca}^{2+}\left(aq\right)+4{PO}_4^{3-}\left(aq\right)+2H{PO}_4^{2-}\left(aq\right)+6H_{2}O\left(l\right)$

Therefore, the $K_{sp}$ expression for this equation is

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

According to the stoichiometry of the reaction,

$\left[H{PO}_4^{2-}\right]={\left(2x\right)}^{ } mol/L$ hence, $\left[{Ca}^{2+}\right]=(8{x)}^{ }$ and $\left[{PO}_4^{3-}\right]=(4{x)}^{ }$

Inserting these variables in the $K_{sp}$ expression,

$K_{sp}=\left(8{x)}^8\right(4{x)}^4{\left(2x\right)}^2$

Plug in the value of $K_{sp}$

$1.1\times {10}^{-47}=\left(8{x)}^8\right(4{x)}^4{\left(2x\right)}^2$

$1.1\times {10}^{-47}=17179869184 \times x^{14}$

$x^{14}=6.4028\times {10}^{-58}$

$x=8.218\times {10}^{-5} mol/L$

$x=8.2\times {10}^{-5} M$

Therefore, the solubility of calcium phosphate, ${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$ is $8.2\times {10}^{-5} M$.

Conclusion:

We compared the $K_{sp}$ values to determine if the compound is more or less soluble and used $K_{sp}$ expression to determine the solubility of calcium phosphate.