InterviewSolution
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InterviewSolution
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At `18^(@)C`, the solubility product of AgCI is `1.8xx10^(-10)`. In the solution, the value of `Ag^(+)` is `4xx10^(-3)` mol `L^(-1)`. The value of `[CI^(-)]` to precipitate AgCI from this solution should be greater thanA. `4.5xx10^(-8)mol L^(-1)`B. `7.2xx10^(-12)mol L^(-1)`C. `4.0xx10^(-3)mol L^(-1)`D. `4.5xx10^(-7)mol L^(-1)` |
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Answer» Correct Answer - A Equilibrium expression is `AgCl(s)hArr Ag^(+)(aq.)+Cl^(-)(aq.)` `K_(sp)=C_(Ag^(+))C_(Cl^(-))` `1.8xx10^(-10)=(4xx10^(-3))C_(CI^(-)` or `C_(CI^(-))= (1.8xx10^(-10))/(4.0xx10^(-3))` `= 0.45xx10^(-7)mol L^(-1)` `=4.5xx10^(-8)mol L^(-1)` Since precipitation takes place whenever ionic product just exceeds solubility product, `C_(CI^(-))` should be greater than `4.5xx10^(-8)mol L^(-1)` to precipitate AgCI. |
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