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`K_("sp")` of `PbBr_(2)` (Molar mass `=367`) is `3.2xx10^(-5)`. If the salt is `80%` dissociated in solution, caculate the solubility of salt in `g` per `litre`. |
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Answer» Correct Answer - `00009.175` `00009.175` Let solubility of `PbBr_(2)` is `S` mol `"litre"^(-1)` `PbBr_(2)(s)+"Aq" hArr PbBr_(2)("aq")hArr underset ((Sxx80)/100)(Pb^(2))+underset((2Sxx80)/100)(2Br^(-)` Since, `PbBr_(2)` ionises to `80%` only. Now `K_("sp")=[Pb^(2+)][Br^(-)]^(2)` or `3.2xx10^(-5)=[(Sxx80)/100][(2Sxx80)/100]^(2)` `S=0.025 "mol litre"^(-1)` `=0.025xx367g "litre"^(-1)` `=19.175g "litre"^(-1)` |
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