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A saturated solution of sparingly soluble lead chloride on analysis was found to contain 11.84 g/ litre of the salt at room temperature. Calculate the solubility product constant at room temperature. (At. wt . : Pb = 207, Cl = 35.5 ) |
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Answer» Correct Answer - `3.09xx10^(-4)` Solubility of `PbCl_(2) = (11.84)/(207+2xx35.5) "mol" L^(-1) = 4.259 xx 10^(-2) "mol " L^(-1)` `PbCl_(2) rarr Pb^(2+)+2 Cl^(-), K_(sp)=[Pb^(2+)][Cl^(-)]^(2) = (4.259xx10^(-2))(2xx4.259xx10^(-2))^(2) = 3.09xx10^(-4)` |
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