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

Options

A. 0.3210 g
B. 0.1178 g
C. 0.2529 g
D. 0.4213 g

Answer

B

Explanation

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