How much `NaF` should be added to `100mL` of solution having `0.016M in Sr^(2+)` ions to reduces its concentration to `2.5 xx 10^(-3)M? K_(sp) SrF_(2) = 2.8 xx 10^(-9)at 298K`.

Explanation

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