Chapter 22, Problem 22.63QA

Interpretation Introduction

To find:

The tooth enamel is actually a composite material containing both hydroxyapatite and a calcium phosphate, ${Ca}_8{\left(HP{O}_4\right)}_2{\left(P{O}_4\right)}_4·6H_{2}O$($K_{sp}=1.1\times {10}^{-47})$

a) Is this calcium mineral more or less soluble than hydroxyapatite ($K_{sp}=2.3\times {10}^{-59})$?

b) Calculate the solubility in moles per liter of hydroxyapatite,

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$, ($K_{sp}=2.3\times {10}^{-59})$ in water at ${25}^{o}C$.

c) Explain why the production of weak acids by bacteria on teeth and gums increases the solubility of hydroxyapatite?

Answer to Problem 22.63QA

Solution:

a) Calcium mineral is more soluble because it has higher $K_{sp}$ than that of hydroxyapatite

b) The solubility of hydroxyapatite, ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ is $8.6\times {10}^{-8} M$.

c) Weak acid produced on teeth and gums reacts with ${OH}^{-}$ ions of hydroxyapatite, ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ which results in a shift of equilibrium to the right and hence, increases the solubility of hydroxyapatite,${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$.

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

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

4) Calculation:

a) Is calcium mineral more or less soluble than hydroxyapatite?

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

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

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, calcium mineral is more soluble because it has higher $K_{sp}$ than that of hydroxyapatite.

b) The solubility in moles per liter of hydroxyapatite,${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$:

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

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)\rightleftharpoons 5{Ca}^{2+}\left(aq\right)+3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$

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

$K_{sp}={\left[{Ca}^{2+}\right]}^5{\left[{PO}_4^{3-}\right]}^3\left[{OH}^{-}\right]$

According to the stoichiometry of the reaction,

$\left[{OH}^{-}\right]=x mol/L$ hence, $\left[{Ca}^{2+}\right]=(5{x)}^{ }$ and $\left[{PO}_4^{3-}\right]=(3{x)}^{ }$

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

$K_{sp}=\left(5{x)}^5\right(3{x)}^{3}x$

Plug in the value of $K_{sp}$.

$2.3\times {10}^{-59}=\left(5{x)}^5\right(3{x)}^{3}x$

$2.3\times {10}^{-59}=84375x^9$

$x^9=2.7259\times {10}^{-64}$

$x=8.655\times {10}^{-8} mol/L$

$x=8.6\times {10}^{-8} M$

Therefore, the solubility of hydroxyapatite, ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ is  $8.6\times {10}^{-8} M$.

c) When the weak acid is produced by bacteria on teeth and gums, the acid ($H^{+}$) reacts with the hydroxide ions produced by hydroxyapatite.

 ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)+H^{+} \to {Ca}_5{\left(P{O}_4\right)}_3^{-}+H_{2}O$

Therefore, the equilibrium shifts to the right in the dissociation of ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ which means the solubility of hydroxyapatite increases. Hence, the production of weak acids by bacteria on teeth and gums increases the solubility of hydroxyapatite ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$.

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 hydroxyapatite.