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GrogVix [38]
3 years ago
15

-

Chemistry
1 answer:
mart [117]3 years ago
3 0

Answer:

reaction B is the best one I would choose

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Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c
malfutka [58]

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.

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