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The solubility of a saturated solution of calcium fluoride is `2xx10^(-4)` mol/L. Its solubility product isA. `12 xx 10^(-2)`B. `14 xx 10^(-4)`C. `22 xx 10^(-11)`D. `32 xx 10^(-12)` |
Answer» Correct Answer - D `CaF_(2) hArr underset(2 xx 10^(-4))(Ca^(2+))+underset(2 xx 2 xx 10^(-4))(2F^(-))` Solubility Product `=[Ca^(2+)][F^(-)]^(2)` `=[2 xx 10^(-4)][2 xx 2 xx 10^(-4)]^(2)` `=32 xx 10^(-12)` |
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