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How will you calculate solubility product of AgCl which is a sparingly soluble salt? |
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Answer» Solution :Substances like AgCl, `PbSO_(4)` etc., are sparingly soluble in water. The solubility product of such substances can be determined using conductivity measurements. Let us consider AgCl as an example `AgCl iff Ag^(+)+Cl^(-)` `K_(SP)=[Ag^(+)][Cl^(-)]` Let the CONCENTRATION of `[Ag^(+)]` be .C. mol `L^(-1)`. As per the stoichiometry, if `[Ag^(+)]=C`, then `[Cl^(-)]` ALSO equal to .C. mol `L^(-1)`. `K_(sp)=C.C` `rArr K=C^(2)` We know that the concentration (in mol `dm^(-3)`) is related to the MOLAR and specific conductance by the following expressions `""wedge_(0)=(kappa times 10^(-3))/(C("in mol "L^(-1)))` `""` (or) `""C=(kappa times 10^(-3))/wedge` Substitute the concentration value in the relation `K_(sp)=C^(2)` `""K_(sp)=((kappatimes 10^(-3))/wedge)^(2)` |
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