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If the solubility of `Pbbr_(2)` is S g-mole per litre, its solubility product, considering it to be 80% ionized, isA. `2.048 S^(2)`B. `20.48 S^(3)`C. `2.048 S^(3)`D. `2.048 S^(4)` |
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Answer» Correct Answer - C `PbBr_(2) hArr Pb^(2+) + 2Br^(-)` `S (S xx 80)/(100) (2S xx 80)/(100)` [`because PbBr_(2)` ionises to 80%] `K_(sp) = [(S xx 80)/(100)][(2S xx 80)/(100)]^(2)` `= 0.8 S xx 2.56 S^(2) = 2.048 S^(3)` |
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