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Chapter 20: Problem 21

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is \(1 \mathrm{mg} \mathrm{F}^{-}\) 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? ( \(K_{\text {sp }}\) for \(\left.\mathrm{CaF}_{2}=4.0 \times 10^{-11} .\right)\)

Short Answer

Expert verified
The maximum molarity of calcium ions in hard water with the USPHS recommended fluoride concentration is 2.87 x 10鈦烩伒 mol/L.

Step by step solution

01

Convert the concentration of fluoride ions from mg/L to mol/L.

First, we need to convert the given concentration of fluoride ions from mg per liter to mol per liter to work with molarity. The molar mass of F- is 19 g/mol. Given Concentration of F- = 1 mg/L Since 1 g = 1000 mg, we have 1 mg = 0.001 g Molarity of F- = (mass of F-) / (molar mass of F- * volume of solution) = (0.001 g) / (19 g/mol * 1L) So, the molarity of F- = \( \frac{0.001}{19} \) mol/L.
02

Write the balanced equation and Ksp expression for CaF2.

The balanced equation for the dissolution of CaF2 in water is: CaF2 (s) 鈫� Ca虏鈦� (aq) + 2 F鈦� (aq) The Ksp expression is given by: Ksp = [Ca虏鈦篯 x [F鈦籡虏 Using the given value of Ksp: 4.0 x 10鈦宦孤�
03

Find the maximum molarity of calcium ions with the given fluoride concentration.

Let [Ca虏鈦篯 be the maximum molarity of Ca虏鈦� in the water and [F鈦籡 = \( \frac{0.001}{19} \) mol/L. Then from the balanced equation, we have the concentration of F- ions as twice the concentration of Ca虏鈦� ions. Using the Ksp expression: Ksp = [Ca虏鈦篯 x [F鈦籡虏 4.0 x 10鈦宦孤� = [Ca虏鈦篯 x \( (\frac{0.001}{19}) \)虏 Now, solve for [Ca虏鈦篯: [Ca虏鈦篯 = \(\frac{4.0 \times 10^{-11}}{(\frac{0.001}{19})^2}\)
04

Calculate the maximum molarity of calcium ions.

By evaluating the previous expression: [Ca虏鈦篯 = 2.87 x 10鈦烩伒 mol/L Therefore, the maximum molarity of calcium ions in hard water with the USPHS recommended fluoride concentration is 2.87 x 10鈦烩伒 mol/L.

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Key Concepts

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

Fluoride Ion Concentration
Fluoride ions play a vital role in dental health by helping to prevent tooth decay. The U.S. Public Health Service suggests a fluoride concentration of 1 mg per liter in drinking water. To convert this to molarity, which is more useful for chemical calculations, we need to use the molar mass of fluoride ions, which is 19 g/mol.
To do this conversion:
  • Convert milligrams to grams: 1 mg is 0.001 grams.
  • Molarity is determined by dividing mass by molar mass and the volume of the solution.
Thus, we can calculate: \[ \text{Molarity of } \mathrm{F}^- = \frac{0.001}{19} \text{ mol/L}\]This allows us to use the Ksp expression efficiently in subsequent calculations related to calcium precipitation.
Calcium Ion Precipitation
Calcium ions can react with fluoride ions to form calcium fluoride (CaF2), a process that is critical when considering water chemistry. The dissociation of CaF2 in water is represented by:
  • \[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \]
The solubility product (Ksp) helps predict whether a precipitate will form. For CaF2, Ksp is given by:\[ K_{sp} = [\text{Ca}^{2+}] \times [\text{F}^-]^2 \]Given the Ksp of CaF2 as 4.0 \times 10^{-11}, we can derive the maximum concentration of calcium ions that will not cause precipitation, under a specific fluoride ion concentration. Understanding the relationship between Ksp and ion concentrations is essential for controlling unwanted precipitation.
Molarity Calculations
Molarity is a key concept when dealing with solutions in chemistry, representing the number of moles of a solute per liter of solution. Calculating molarity in exercises like this one involves understanding unit conversions and using the Ksp formula.The Ksp equation forCaF2 is used to find the calcium ion concentration.
  • Given: \[ K_{sp} = 4.0 \times 10^{-11} \text{ and } \text{[F}^-\text{] = } \frac{0.001}{19} \text{ mol/L}\]
  • Solve for [Ca虏鈦篯:\[ [\text{Ca}^{2+}] = \frac{4.0 \times 10^{-11}}{(\frac{0.001}{19})^2} \]
The result, 2.87 \times 10^{-5} mol/L, shows the maximum calcium ion concentration before precipitation occurs. Mastering molarity calculations gives you a strong foundation in analytical chemistry.
Hard Water Chemistry
Hard water contains high mineral content, primarily calcium and magnesium ions. These minerals can interact with fluoride, making water treatment essential to prevent issues like pipe scaling and decreased effectiveness of fluoride in dental health. When fluoride is added to hard water, calcium fluoride ( CaF2 ) may precipitate, reducing fluoride's availability for dental protection.
  • This is governed by Ksp values and equilibrium expressions.
  • Understanding these reactions can help in effectively managing water hardness.
Addressing hard water issues often requires balancing chemical equations and integrating chemistry principles to maintain fluoride benefits. A solid grasp of these concepts helps in both environmental and health applications.

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