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`PbCl_(2)` has a solubility product of `1.7xx10^(-8)`. Will a precipitate of `PbCi_(2)` form when 0.010 mole of lead nitrate and 0.010 mole of potassium chloride are mixed and water added upto 1 litre ? |
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Answer» Correct Answer - Yes As final volume of the solution = 1 L, therefore `[Pn^(2+)] = 0.01 M = 10^(-2) M, [ Cl^(_)] = 0.01 M = 10^(-2) M` Ionic product of `PbCl_(2) = [Pb^(2+)][Cl^(-)]^(2) (10^(-2))^(2) = 10^(_6)` which is greater than `K_(sp)`. Hence , ppt. of `PbCl_(2) ` will be formed. |
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