Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the \(\mathrm{F}^{-}\) ion concentration is adjusted to about \(1 \mathrm{ppm}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is \(8 \mathrm{ppm}\). Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

Short Answer

Expert verified
In the given solution with F- ion concentration of \(1 \, ppm\) and Ca2+ ion concentration of \(8 \, ppm\), the ion product (Q) is found to be \(5.503 \times 10^{-13}\), which is less than the solubility product constant (Ksp) of CaF2, \(3.9 \times 10^{-11}\). As \(Q < Ksp\), the solution is not saturated, and a precipitate of CaF2 will not form under these conditions.

Step by step solution

01

Convert ppm to molar concentration

Ppm (parts per million) is a unit of concentration that expresses the ratio of the solute to the solvent. To convert ppm to molarity, we need to know the molar mass of the ion and the density of water. The molar mass of F- ion: \(m_{F^-} = 19.00 \, g/mol\) The molar mass of Ca2+ ion: \(m_{Ca^{2+}} = 40.08 \, g/mol\) Density of water: \(\rho_{water} = 1 \, g/cm^3\) For F- ion: \[C_{F^-} = \dfrac{1 \, ppm \cdot \rho_{water}}{m_{F^-}}\] For Ca2+ ion: \[C_{Ca^{2+}} = \dfrac{8 \, ppm \cdot \rho_{water}}{m_{Ca^{2+}}}\] Keep in mind that \(1 \, ppm = 1 \frac{mg}{L}\), so we have:
02

Calculate the molar concentration of each ion

For F- ion: \[C_{F^-} = \dfrac{1 \frac{mg}{L} \cdot 1 \frac{g}{1000 \, mg}}{19.00 \frac{g}{mol}} = 5.26 \times 10^{-5} \, M\] For Ca2+ ion: \[C_{Ca^{2+}} = \dfrac{8 \frac{mg}{L} \cdot 1 \frac{g}{1000 \, mg}}{40.08 \frac{g}{mol}} = 1.997 \times 10^{-4} \, M\]
03

Calculate the ion product (Q)

Next, we need to calculate the ion product (Q) by multiplying the molar concentrations of the ions: \[Q = [Ca^{2+}][F^-]^2]\] Plugging in the values: \[Q = (1.997 \times 10^{-4} M)(5.26 \times 10^{-5} M)^2 = 5.503 \times 10^{-13}\]
04

Compare Q with Ksp

The solubility product constant (Ksp) of CaF2 is \(3.9 \times 10^{-11}\). Now we compare the ion product (Q) with the Ksp: \(Q = 5.503 \times 10^{-13}\) \(Ksp = 3.9 \times 10^{-11}\) Since \(Q < Ksp\), the solution is not saturated, and a precipitate of CaF2 will not form under these conditions.

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Most popular questions from this chapter

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid HA and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is 9.2 . (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{b}\) for \(B\) ?

A concentration of \(10-100\) parts per billion (by mass) of \(\mathrm{Ag}^{+}\) is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the \(\mathrm{Ag}^{+}\) can cause adverse health effects. One way to maintain an appropriate concentration of \(\mathrm{Ag}^{+}\) is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix \(D\), calculate the equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that would exist in equilibrium with (a) AgCl, (b) AgBr, (c) AgI.

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