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Solid iron (II) flouride is placed into an acidic solution of HF. If the concentration of...

Solid iron (II) flouride is placed into an acidic solution of HF. If the concentration of the hydrofluoric acid is 0.15 M, then what is the concentration of the metal and flouride ion?

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Answer #1

Solution.

The first approximation.

The dissociation equation of iron (II) fluoride is

The solubility product of iron (II) fluoride is

The dissociation constant of HF is 6.31*10-4, the concentration of ions from a dissociation of the acid are

The concentrations of ions from a solubility of the salt are

The total concentration of fluoride-ions is 0.017+0.0094 = 0.026 M;

Second approximation.

This concentration will supress a dissociation of HF:

It means that the HF gives

The total concentration of fluoride-ions is 0.017+0.0035 = 0.021 M;

Third approximation.

The total concentration of fluoride-ions is 0.017+0.0044 = 0.021 M.

From the solubility equation,

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