# Problem: Fluoridation of drinking water is employed in many places to aid in the prevention of dental caries. Typically the F- ion concentration is adjusted to about 1 ppb. Some water supplies are also "hard"; that is, they contain certain cations such as Ca2+ that interfere with the action of soap. Consider a case where the concentration of Ca2+ is 8 ppb.Could a precipitate of CaF2 form under these conditions? (Make any necessary approximations.)

###### FREE Expert Solution

In this problem, we are asked if CaF2 precipitate would form under the conditions set in the problem above.

For this, we need to compare the reaction quotient (Q) vs. the solubility product constant (Ksp)

Recall that when:

Q > Ksp: the solution is supersaturated and a precipitate will form. Reactants are favored.

Q = Ksp: the solution is at equilibrium and no precipitate will form.

Q < Ksp: the solution is unsaturated and no precipitate will form. Products are favored.

To solve this problem:

Step 1. Convert the concentrations from ppb to M.

Step 2. Write and balance the disassociation of the compound into ions.

Step 3. Solve for Q using I.C.E. chart.

Step 4. Compare the Q value with the Ksp.

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###### Problem Details

Fluoridation of drinking water is employed in many places to aid in the prevention of dental caries. Typically the F- ion concentration is adjusted to about 1 ppb. Some water supplies are also "hard"; that is, they contain certain cations such as Ca2+ that interfere with the action of soap. Consider a case where the concentration of Ca2+ is 8 ppb.

Could a precipitate of CaF2 form under these conditions? (Make any necessary approximations.)