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The solubility product of `SrF_(2)` in water is `8xx10^(-10)`. Calculate its solubility in `0.1M` NaF aqueous solution. |
Answer» `K_(SP)=[Sr^(2+)][F^(-)]^(2)` `8xx10^(-10)=S[2S+0.1]^(2)` `:. S= (8xx10^(-10))/((0.1)^(2))=8xx10^(-8)mol//litre` |
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