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

Answer:

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

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:

$[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}$

$[F^-]_{max}=4x10{-3}\frac{molF^-}{L}$

Best regards.