   Chapter 16, Problem 80AE

Chapter
Section
Textbook Problem

# The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 mg F− per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? (Ksp for CaF2 = 4.0 × 10−11)

Interpretation Introduction

Interpretation: The fluoridation of water as a means for preventing tooth decay is recommended by the U.S. Public Health Service (USPHS) . The given recommended concentration is 1mgF per liter. The maximum molarity of the calcium ions present in hard water if the fluoride concentration is at the USPHS recommended level is to be calculated.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. The solubility product, Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated by the formula,

Ksp=[A]x[B]y

Explanation

Explanation

To determine: The maximum molarity of the calcium ions present in hard water if the fluoride concentration is at the USPHS recommended level.

The concentration of fluoride is 5.26×10-5M_ .

Given

The Ksp for CaF2 is 4.0×1011 .

The concentration of fluoride is 1mg/L(1×103mg/L) .

The molar mass of fluorine is 19g/mol .

The concentration in terms of molarity is calculated by the formula,

Concentration(M)=Concentration(g/L)MolarMass

Substitute the value of the concentration (in mg/L ) and the molar mass of fluorine in the above expression.

Concentration(M)=1×103mg/L19g/mol=5.26×10-5M_

The [Ca2+] is 0.014M_

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