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Chapter 20: Problem 30

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

Short Answer

Expert verified
The maximum molarity of calcium ions in hard water that can support the USPHS recommended fluoride concentration of 1 mg/L without causing precipitation is \(1.44 \times 10^{-2} \: M\).

Step by step solution

01

Write the balanced chemical equation for calcium fluoride

The balanced chemical equation for calcium fluoride (CaF2) dissolving in water is: \[CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}\]
02

Write the expression for the solubility product constant (Ksp)

The expression for the solubility product constant (Ksp) of calcium fluoride (CaF2) is: \[K_{sp} = [Ca^{2+}][F^{-}]^2\] The given Ksp value for CaF2 is \(4.0 \times 10^{-11}\).
03

Identify the concentration of fluoride ions

The concentration of fluoride ions is given at the USPHS recommended level, which is: \(1 \: mg \: F^- / L\) To convert this concentration into molarity, we need to divide by the molar mass of fluoride: \(\frac{1 \: mg \: F^- / L}{18.998 \: g \: F^- / mol} \times \frac{10^{-3} \: g}{1 \: mg} = 5.26 \times 10^{-5} \: M\) This is the concentration of F- ions in the water.
04

Calculate the concentration of calcium ions using the Ksp expression

Since we know the Ksp value for CaF2 and the concentration of fluoride ions, we can determine the maximum molarity of calcium ions without causing precipitation. Plugging the concentration of fluoride ions into the Ksp expression: \[4.0 \times 10^{-11} = [Ca^{2+}](5.26 \times 10^{-5})^2\] Now, solve for the concentration of calcium ions: \([Ca^{2+}] = \frac{4.0 \times 10^{-11}}{(5.26 \times 10^{-5})^2} = 1.44 \times 10^{-2} \: M\)
05

State the final answer

The maximum molarity of calcium ions in hard water that can support the USPHS recommended fluoride concentration of 1 mg/L without causing precipitation is \(1.44 \times 10^{-2} \: M\).

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