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Lead chloride has a solubility productof 1.7xx10^(-5) at 298 K. Calculate itssolubility at this temperature. |
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Answer» Solution :The SOLUBILITY equilibrium for `PbCl_(2)` may be represented as : `PbCl_(2)(s) HARR Pb^(2+) (aq) + 2Cl^(-) (aq)` Let the solubility of `PbCl_(2) ` be s moles/litre. Then the solution will contain moles of `Pb^(2+)` ions and 2s moles `Cl^(-)` ions respectively per litre. Hence, the solubility product, `K_(sp)` of `PbCl_(2)` WOULD be given by the expression, `K_(sp) = [ Pb^(2+)][Cl^(-)]^(2) = sxx (2s)^(2)=4S^(3)` But the value of `K_(sp) = 1.7xx10^(-5) ` (Given) `4s^(3)=1.7xx10^(-5) = (1.7xx10^(-5))/(4) = 0.425xx10^(-5) or s = (0.425xx10^(-5))^(1//3)=(4.25xx10^(-6))^(1//3)` Let` x= (4.25)^(1//3) :. log x = 1//3 log 4.25 = 1//3 (0.6284) = 0.2095` x= Antilog 0.2095 = 1.620 Hence, `s=1.620 xx 10^(-2) "mol L"^(-1)` |
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