Question

Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that could be present in hard water. Assume fluoride is the only anion present that will precipitate calcium ion.

Answer 1

`[F^-]_(max)=4x10{-3}(molF^-)/(L)`

Step-by-step explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

`Ca^(+2)(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)`

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

`[F^-]_(max)=2.0x10^(-3)(molCa^(2+))/(L)*(2molF^-)/(1molCa^(2+)) \\`

`[F^-]_(max)=4x10{-3}(molF^-)/(L)`

Best regards.