The solubility of `CaF_(2)` in a solution of `0.1M Ca(NO_(3))_(2)` is

Options

A. `[Ca^(2+)]`
B. `2[F^(Theta)]`
C. `([F^(Theta)])/(2)`
D. `2[NO_(3)^(Theta)]`

Answer

C

Explanation

`CaF_(2) hArr Ca^(2+) + 2F^(Theta)`
(Let `S` is the solubility of `CaF_(2)` in `H_(2)o)`?
` :. [Ca^(2+)] "in" H_(2)O = S, [F^(Theta)]_(H_(2)O) = 2S`
`[Ca^(2+)] "in" 0.1 M Ca(NO_(3))_(2) = S + 0.1`
`[F^(Theta)] "in" 0.1M Ca(NO_(3))_(2) = 2S`
` :. 2S = [F^(Theta)]_("in") H_(2)O)`
`S = ([F^(Theta)])/(2)`.