Answer:
[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]
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:
[tex]Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)[/tex]
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:
[tex][F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}} \\[/tex]
[tex][F^-]_{max}=4x10{-3}\frac{molF^-}{L}[/tex]
Best regards.
