Chapter 21, Problem 102GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Calcium fluoride can be used in the fluoridation of municipal water supplies. If you want to achieve a fluoride ion concentration of 2.0 × 10−5 M, what mass of CaF2 must you use for 1.0 × 106 L of water? (Ksp for CaF2 is 5.3 × 10−11.)

Interpretation Introduction

Interpretation:

To calculate the mass of CaF2 must use for 1×106L of water.

Concept introduction:

Solubility product constant (Ksp) is equilibrium constant is defined for equilibrium between a solid and its respective ions in a solution.

The ionization of CaF2 in water is written as,

CaF2(s)Ca+2(aq)+2F(aq)

The Solubility product constant (Ksp) for CaF2 is written as,

Ksp=[Ca+2][2(F)]2 (1)

Numberofmoles=MassMolecularmassMass=Numberofmoles×Molecularmass

Explanation

The mass of CaF2 must use for 1Ã—106â€‰L of water is calculated below.

Given:

The Solubility product constant (Ksp) for CaF2 is 5.3Ã—10âˆ’11.

The volume of water is 1Ã—106â€‰L.

The concentration of fluoride ion in solution is 2Ã—10âˆ’5â€‰M.

The molecular weight of CaF2 is 40â€‰gâ‹…molâˆ’1.

Substitute the value of Solubility product constant (Ksp) for CaF2 and concentration of fluoride ion in equation (1) to calculate the concentration of calcium ion in the solution,

5.3Ã—10âˆ’11=[Ca+2][2(2Ã—10âˆ’5â€‰M)]2[Ca+2]=0.033â€‰M

The concentration of calcium ion is equal to the concentration of CaF2. Therefore, the concentration of CaF2 is 0.033â€‰M.

The number of moles of CaF2(nCaF2) is equal to the product of the concentration of CaF2 and volume of water

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