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Chapter 18: Problem 7

Fluoridated drinking water contains about 1 part per million (ppm) of \(\mathrm{F}^{-}\). Is \(\mathrm{CaF}_{2}\) sufficiently soluble in water to be used as the source of fluoride ion for the fluoridation of drinking water? Explain. [Hint: Think of 1 ppm as signifying \(1 \mathrm{g} \mathrm{F}^{-}\) per \(10^{6} \mathrm{g}\) solution.

Short Answer

Expert verified
No, \( \mathrm{CaF}_{2} \) is not sufficiently soluble to provide the required fluoride ion concentration for fluoridation of drinking water.

Step by step solution

01

Calculate Molar Mass of \( \mathrm{CaF}_{2} \) and \( \mathrm{F}^{-} \)

First, calculate the molar mass of \( \mathrm{CaF}_{2} \) which is \(40.08 g/mol + 2*18.9984 g/mol = 78.0768 g/mol\). Molar mass of fluoride ion \( \mathrm{F}^{-} \) is \(18.9984 g/mol\).
02

Understanding \( \mathrm{CaF}_{2} \) Dissociation

When 1 mole of \( \mathrm{CaF}_{2} \) dissolves in water, it dissociates into 1 mole of \( \mathrm{Ca}^{2+} \) and 2 moles of \( \mathrm{F}^{-} \). Therefore, for 1 mole of \( \mathrm{CaF}_{2} \), we get 2 moles of \( \mathrm{F}^{-} \).
03

Calculate \( \mathrm{F}^{-} \) Concentration

So if we dissolve 1 ppm (1g) of \( \mathrm{CaF}_{2} \) in \(10^{6} \mathrm{g}\) of water, it would provide \( \frac{1}{78.0768} = 0.0128 mol \) of \( \mathrm{CaF}_{2} \), and thus \( 2*0.0128 = 0.0256 mol \) of \( \mathrm{F}^{-} \) or 25.6 millimoles of \( \mathrm{F}^{-} \).
04

Compare with Required Concentration

The required concentration for fluoridation of water is 1 ppm, which signifies \( \frac{1 g}{18.9984 g/mol} = 0.0526 mol \) or 52.6 millimoles of \( \mathrm{F}^{-} \) per million grams of water.
05

Concluding Remarks

Comparing the obtained and required concentrations, \( \mathrm{CaF}_{2} \) is not sufficiently soluble in water to be used as the source of fluoride ion. As the obtained concentration is less than half of the required.

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