`K_(sp)` of `SrF_(2) = 2.8 xx 10^(-9)` at `25^(@)C`. How much `NaF` should be added to `100mL` of solution having `0.016M` in `Sr^(2+)` ions to reduce its concentration to `2.5 xx 10^(-3)M`?

Explanation

Initial `[Sr^(2+)] = 16 xx 10^(-3)M`
Left `[Sr^(2+)] = 16 xx 10^(-3)M`
`[Sr^(2+)] ppted = (16 - 2.5) xx 10^(3) = 13.5 xx 10^(-3)M`
`[F^(Θ)]` needed for this precipitation
`rArr 2 xx 13.5 xx 10^(-3)M = 27.0 xx 10^(-3)M`
(because `Sr^(2+) + 2F^(Θ) rarr SrF_(2))`
Also `[Sr^(2+)] [F^(Θ)]^(2) = K_(sp_(SrF_(2))) = 2.8 xx 10^(-9)`
`[F^(Θ)]^(2) = (2.8 xx 10^(-9))/(2.5 xx 10^(-3))`
`:. [F^(Θ)] = 1.058 xx 10^(-3)M,` i.e., the concentration of `F^(Θ)` which will also apperar in solution state.
Thus, `[F^(Θ)]` needed `= [27.0 + 1.058] xx 10^(-3)M`
`=28.058 xx 10^(-3)M`
`:. NaF` needed for `1L = 28.058 xx 10^(-3) xx 42g`
`:. NaF` needed for `100mL`
`= (28.058xx10^(-3)xx42)/(10)g`
`= 0.1178g`