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Tooth enamel is composed of the mineral hydroxyapatite. The K sp of hydroxyapatite, Ca 5 (PO 4 ) 3 OH, is 6.8 × 10 −37 . Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, Ca 5 (PO 4 ) 3 F, forms. The K sp of this substance is 1 × 10 −60 . Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 79AE
Textbook Problem
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Tooth enamel is composed of the mineral hydroxyapatite. The Ksp of hydroxyapatite, Ca5(PO4)3OH, is 6.8 × 10−37. Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, Ca5(PO4)3F, forms. The Ksp of this substance is 1 × 10−60. Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

Interpretation Introduction

Interpretation: The composition of tooth enamel is given. The solubility of Hydroxyapatite in water; effect of addition of acid over the solubility of Hydroxyapatite; solubility of fluoroapatite in water and an explanation for the fluoridation of water by the above calculations is to be stated.

Concept introduction: The salts that do not dissolve completely in a solvent or that show partial dissociation, for them solubility product that is Ksp is used.

Explanation of Solution

Explanation

To determine: The solubility of Hydroxyapatite in water; effect of addition of acid over the solubility of Hydroxyapatite; solubility of fluoroapatite in water and rationale provided for the fluoridation of water by the above calculations.

The solubility of Hydroxyapatite in water is 2.5×10-5mole/L_ .

Given

The Ksp of Hydroxyapatite is 6.8×1037 .

The solubility of solid Hydroxyapatite in water is shown as,

Ca5(PO4)3OH(s)5Ca2+(aq)+3PO43(aq)+OH(aq)

The value of Ksp for the above salt is,

Ksp=[Ca2+]5[PO43]3[OH]

Let xmole/L of the solid is dissolved in water. Therefore the equilibrium concentration of each ion is [Ca2+]=5xmole/L , [PO43]=3xmole/L , [OH]=xmole/L .

Substitute these values in above equation.

Ksp=[Ca2+]5[PO43]3[OH]6.8×1037=[5x]5[3x]3[x]6.8×1037=84,375x9x9=8.1×1042

The equation is further simplified.

x9=8.1×10429logx=log8.1×1042logx=4.6x=2.5×10-5mole/L_

The solubility of Hydroxyapatite in water is 2.5×10-5mole/L_ .

There occurs an increase in solubility when an acid is added to the aqueous solution of Hydroxyapatite.

There occurs an increase in solubility when an acid is added to the aqueous solution of Hydroxyapatite. This occurs because Hydroxyapatite. Exists in equilibrium with its ions Ca2+ , PO43 and OH . The solubility of solid Hydroxyapatite in water is shown as,

Ca5(PO4)3OH(s)5Ca2+(aq)+3PO43(aq)+OH(aq)

When acid is added in the above solution, then H+ reacts with PO43 and OH , as these are the strong bases

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Chapter 15 Solutions

Chemistry: An Atoms First Approach
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