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A solution contains `0.1 M` is `Cl^(-)` and `10^(-4) M CrO_(4)^(2-)`. If solid `AgNO_(3)` is gradually added to this solution, what will be the concentration of `Cl^(-)` when `Ag_(2)CrO_(4)` begins to precipitate? `(Ksp (AgCl) = 10^(-10) M^(2)`, `K_(sp) (Ag_(2)CrO_(4)) = 10^(-12)M^(3))`A. `10^(-6)M`B. `10^(-4) M`C. `10^(-5) M`D. `10^(-9) M` |
Answer» Correct Answer - A `[Ag^(+)] ` required for precipitation of `AgCl` `[Ag^(+)] = (ksp[AgCl])/([Cl^(-)]) = 10^(-9)` `[Ag^(+)]` required for precipitation of `Ag_(2)CrO_(2)` `[Ag^(+)] =sqrt((ksp[Ag_(2)CrO_(4)])/([CrO_(4)^(2-)])) = sqrt((10^(-12))/(10^(-4)) ) = 10^(-4)` At this point, the concentration of `Cl^(-)` ion in the solution can be calculated from `Ksp(AgCl)` `[Cl^(-)] = (ksp(AgCl))/([Ag^(+)]) = (10^(-10))/(10^(-4)) = 10^(-6) M` |
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