**Problem Statement:**Typical "hard" water contains about 80.2 mg of Ca²⁺ per liter. Calculate the maximum concentration of fluoride ion which could be present in hard water. The solubility product constant (Kₛₚ) of CaF₂ is 4.00 x 10⁻¹¹.**Multiple Choice Answers:**1. ○ 4.53 x 10⁻⁵ M fluoride ions2. ○ 1.41 x 10⁻⁴ M fluoride ions3. ○ None of these4. ○ 4.96 x 10⁻³ M fluoride ions5. ○ 2.00 x 10⁻⁸ M fluoride ions6. ○ 0.0291 M fluoride ions**Explanation:**To solve this problem, we need to assess the conditions where calcium fluoride (CaF₂) might precipitate or dissolve in the presence of calcium ions (Ca²⁺) and fluoride ions (F⁻) in the given water sample. This involves calculating the fluoride ion concentration that results in equilibrium based on the provided Kₛₚ value.

Answered: Image /qna-images/answer/7bce1cd4-ff1c-4fbe-88d0-8416da710137.jpg

Typical "hard" water contains about 80.2 mg of Ca2* per liter. Calculate the maximum concentration of fluoride ion which could be present in hard water. Ksp of CaF2 = 4.00 x 10-11 4.53 x 10-5 M fluoride ions 1.41 x 10-4 M fluoride ions None of these 4.96 x 10-3 M fluoride ions 2.00 x 10-8 M fluoride ions 0.0291 M fluoride ions

Typical "hard" water contains about 80.2 mg of Ca2* per liter. Calculate the maximum concentration of fluoride ion which could be present in hard water. Ksp of CaF2 = 4.00 x 10-11 4.53 x 10-5 M fluoride ions 1.41 x 10-4 M fluoride ions None of these 4.96 x 10-3 M fluoride ions 2.00 x 10-8 M fluoride ions 0.0291 M fluoride ions

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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