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The solubility of `CaF_(2) (K_(sp)=3.4xx10^(-11))` in `0.1 M` solution of `NaF` would beA. `3.4xx10^(-12) M`B. `3.4xx10^(-10) M`C. `3.4xx10^(-9) M`D. `3.4xx10^(-13) M` |
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Answer» Correct Answer - C `underset(0.1)(NaF) rarr underset(0.1)(Na^(+))+underset(0.1)(F^(-))` `CaF_(2) rarr Ca^(2+)+underset((2x+0.1)=0.1)(2F^(-))` `K_(sp)=x(0.1)^(2)=3.4xx10^(-11)x=3.4xx10^(-9)` |
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