`K_(sp)` of `SrF_(2)` (`s`) in water is `3.2xx10^(-11)`. The solubility `SrF_(2)` (`s`) in `0.1 (M) NaCl` solution is

Options

A. `3.2xx10^(-9) (M)`
B. `2xx10^(-4) (M)`
C. `4xx10^(-4) (M)`
D. slightly higher than `2xx10^(-4) (M)`

Answer

D

Explanation

`SrF_(2)(s)hArrunderset(s)(Sr^(2+))+underset(2s)(2F^(-))`
(where `s` is the solubility)
`:. 4s^(2)=32xx10^(-12)` or `s=2xx10^(-4)(M)`
But practically the solubility of `SrF_(2) (s)` in `NaCl` solution is slightly greater than `2xx10^(-4)` because `NaCl` increases ionic strength of the solution.