Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{\mathrm{s}}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}_{\text {k }}\) whose \(K_{p}=1.0 \times 10^{-40}\), (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Short Answer

Expert verified
(a) The solubility-constant expressions are: 1. Hydroxyapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{OH}^-]\) 2. Fluoroapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{F}^-]\) (b) The molar solubility of each compound is: 1. Hydroxyapatite: \(1.08\times10^{-6}\,\text{mol/L}\) 2. Fluoroapatite: \(2.05\times10^{-8}\,\text{mol/L}\)

Step by step solution

01

(a) Write the expression for the solubility-constant for hydroxyapatite and fluoroapatite

: Let's first write the balanced chemical equations for the dissolution of each compound in water: 1. Hydroxyapatite: \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH}\rightleftharpoons5\mathrm{Ca}^{2+}+3\mathrm{PO}_4^{3-}+\mathrm{OH}^-\) 2. Fluoroapatite: \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F}\rightleftharpoons5\mathrm{Ca}^{2+}+3\mathrm{PO}_4^{3-}+\mathrm{F}^-\) Now let's write the solubility constant expressions for each compound using their respective balanced chemical equations: 1. Hydroxyapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{OH}^-]\) 2. Fluoroapatite: \(K_{sp}=[\mathrm{Ca}^{2+}]^5[\mathrm{PO}_4^{3-}]^3[\mathrm{F}^-]\) We have now written the expressions for the solubility constants of hydroxyapatite and fluoroapatite.
02

(b) Calculate the molar solubility of each compound

: 1. Hydroxyapatite: Let the common molar solubility of hydroxyapatite be \(s\) mol/L. Then, the concentrations of each ion are: \([\mathrm{Ca}^{2+}]=5s\) \([\mathrm{PO}_4^{3-}]=3s\) \([\mathrm{OH}^-]=s\) Substituting these values into the solubility constant expression: \(K_{sp} = [5s]^5[3s]^3[s]\) We are given that \(K_{sp}=6.8\times10^{-27}\), so we can solve for \(s\): \(6.8\times10^{-27}=(5s)^5(3s)^3(s)\) Solving for s, we get: \(s\approx1.08\times10^{-6}\,\text{mol/L}\) The molar solubility of hydroxyapatite is approximately \(1.08\times10^{-6}\) mol/L. 2. Fluoroapatite: Let the common molar solubility of fluoroapatite be \(s'\) mol/L. Then, the concentrations of each ion are: \([\mathrm{Ca}^{2+}]=5s'\) \([\mathrm{PO}_4^{3-}]=3s'\) \([\mathrm{F}^-]=s'\) Substituting these values into the solubility constant expression: \(K_{sp}=[5s']^5[3s']^3[s']\) We are given that \(K_{sp}=1.0\times10^{-40}\), so we can solve for \(s'\): \(1.0\times10^{-40}=(5s')^5(3s')^3(s')\) Solving for s', we get: \(s'\approx2.05\times10^{-8}\,\text{mol/L}\) The molar solubility of fluoroapatite is approximately \(2.05\times10^{-8}\) mol/L.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydroxyapatite
Hydroxyapatite is a naturally occurring mineral form of calcium apatite with the formula \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH}\). It is the main component of tooth enamel and bone, providing both rigidity and structure. The solubility product constant (\(K_{sp}\)) for hydroxyapatite is incredibly low, \(6.8 \times 10^{-27}\), indicating its lack of solubility in water.
This low solubility is crucial for structural integrity in biological materials, such as our bones and teeth. In water, hydroxyapatite dissolves into calcium ions, phosphate ions, and hydroxide ions:
  • \([\mathrm{Ca}^{2+}] = 5s\)
  • \([\mathrm{PO}_4^{3-}] = 3s\)
  • \([\mathrm{OH}^-] = s\)
The equilibrium expression for its solubility product is given by:
\[ K_{sp} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_4^{3-}]^3 [\mathrm{OH}^-] \]
This expression helps in determining the molar solubility, \(s\), of hydroxyapatite in various conditions.
Fluoroapatite
Fluoroapatite is another form of calcium apatite but with fluoride replacing the hydroxide group. Its chemical formula is \(\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F}\). Unlike hydroxyapatite, fluoroapatite is even less soluble, with a solubility product constant (\(K_{sp}\)) of \(1.0 \times 10^{-40}\).
This extremely low solubility means fluoroapatite is highly resistant to dissolution, which makes it a preferred mineral for strengthening tooth enamel. This mineral forms when fluoride reacts with hydroxyapatite, contributing to enhanced dental health by making teeth more resistant to acidic conditions and decay.
Like hydroxyapatite, fluoroapatite dissolves into calcium ions, phosphate ions, and fluoride ions. These concentrations in solution are:
  • \([\mathrm{Ca}^{2+}] = 5s'\)
  • \([\mathrm{PO}_4^{3-}] = 3s'\)
  • \([\mathrm{F}^-] = s'\)
The solubility constant expression for fluoroapatite is:
\[ K_{sp} = [\mathrm{Ca}^{2+}]^5 [\mathrm{PO}_4^{3-}]^3 [\mathrm{F}^-] \]
This helps in calculating the molar solubility, \(s'\), and understanding the conditions affecting fluoride incorporation in dental prophylaxis.
Molar Solubility
Molar solubility refers to the number of moles of a compound that can dissolve per liter of solution to reach saturation. For slightly soluble salts like hydroxyapatite and fluoroapatite, the concept of molar solubility is pivotal in predicting how much of these minerals can dissolve in biological systems.
The molar solubility \(s\) for hydroxyapatite, calculated from its solubility product \(K_{sp} = 6.8 \times 10^{-27}\), is approximately \(1.08 \times 10^{-6}\) mol/L. This reflects the minimal dissolution required for maintaining the integrity of tooth enamel and bone.
In contrast, fluoroapatite has a much lower molar solubility of \(2.05 \times 10^{-8}\) mol/L, derived from its \(K_{sp} = 1.0 \times 10^{-40}\). A lower molar solubility suggests a stronger and more resistant structure, explaining why fluoride treatments are beneficial for dental health.
These values highlight how a small change in solubility can significantly affect biological properties and the chemical stability of these minerals.
Dissolution Reactions
Dissolution reactions describe the process where a solid dissolves in a solvent, dissociating into its constituent ions. For hydroxyapatite and fluoroapatite, these reactions are crucial in understanding their roles in biological systems.
The dissolution of hydroxyapatite in water can be expressed as:
\[ \mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{OH}^- \]
This equilibrium shows how hydroxyapatite dissolves into calcium, phosphate, and hydroxide ions, albeit minimally due to its low solubility.
Comparatively, the dissolution of fluoroapatite is:
\[ \mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{F} \rightleftharpoons 5\mathrm{Ca}^{2+} + 3\mathrm{PO}_4^{3-} + \mathrm{F}^- \]
Given the very low \(K_{sp}\) of fluoroapatite, this reaction proceeds to an even lesser extent, resulting in very few ions in solution.
Understanding these reactions helps in predicting the stability and effects of these minerals in applications like toothpaste, where the dissolution of fluoride ions can enhance enamel strength and resistance to decay.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{Cr}(\mathrm{OH})_{2}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{4}\), (d) \(\mathrm{Ag}^{*}\) and \(\mathrm{Mn}^{2+}\), Suggest how each mixture might be separated.

The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C}\) ?

You are asked to prepare a pH \(=4.00\) buffer starting from \(1.50 \mathrm{~L}\) of \(0.0200 \mathrm{M}\) solution of benzoic acid \(\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{COOH}\right)\) and any amount you need of sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa}\right)\). (a) What is the \(\mathrm{pH}\) of the benzoic acid solution prior to adding sodium benzoate? (b) How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

A person suffering from anxiety begins breathing rapidly and as a result suffers alkalosis, an increase in blood \(\mathrm{pH}\). (a) Using Equation 17.10, explain how rapid breathing can cause the \(\mathrm{pH}\) of blood to increase. (b) One cure for this problem is breathing in a paper bag. Why does this procedure lower blood \(\mathrm{pH}\) ?

How many milliliters of \(0.0850 \mathrm{M} \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{MHNO}_{3}\), (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH}\), (c) \(50.0 \mathrm{~mL}\). of a solution that contains \(1.85 \mathrm{~g}\) of \(\mathrm{HCl}\) per liter?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free