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A solution is a mixture of 0.05 M NaCl and 0.05 M AgI. The concentration of iodide in the solution when AgCl just starts precipitating is equal to: `(K_(sp)AgCl=1xx10^(-10)M^(2), K_(sp)AgI=4xx10^(-16)M^(2))`A. `4xx10^(-6)M`B. `2xx10^(-8)M`C. `2xx10^(-7)M`D. `8xx10^(-15)M` |
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Answer» Correct Answer - C `[Ag^(+)]=(K_(SP)AgCI)/([CI^(-)])` For AgCI precipitation `= (10^(-10))/(0.05)=2xx10^(-9)` `:. [Ag^(+)]=(K_(SP)AgI)/([I^(-)])` For AgI precipitation `= (4xx10^(-16))/(0.05)=8xx10^(-15)` Thus, AgI will precipitate first. AgCI will precipitate only, when `[Ag^(+)]=2xx10^(-9)`, Thus `[I^(-)]_(l eft)=(4xx10^(-16))/(2xx10^(-9))=2xx10^(-7)M` |
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