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The solubility product of `SrF_(2)` in water is `8xx10^(-10)`. Calculate its solubility in `0.1M` NaF aqueous solution.

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`K_(sp)=[Sr^(2+)][F^(-2)]`
`8xx10^(-10)=s[2s+0.1]^(2)=s[0.1]^(2)` (neglecting the F(-) coming from sparingly soluble solution
`S=(8xx10^(-10))/((0.1)^(2))=8xx10^(-8)M`

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