A silver wire dipped in `0.1M HCI` solution saturated with `AgCI` develops a potential of `-0.25V`. If `E_(Ag//Ag^(+))^(@) =- 0.799V`, the `K_(sp)` of `AgCI` in pure water will beA. `2.95 xx 10^(-11)`B. `5.1 xx 10^(-11)`C. `3.95 xx 10^(-11)`D. `1.95 xx 10^(-11)`

A silver wire dipped in `0.1M HCI` solution saturated with `AgCI` deve

1.	A silver wire dipped in `0.1M HCI` solution saturated with `AgCI` develops a potential of `-0.25V`. If `E_(Ag//Ag^(+))^(@) =- 0.799V`, the `K_(sp)` of `AgCI` in pure water will beA. `2.95 xx 10^(-11)`B. `5.1 xx 10^(-11)`C. `3.95 xx 10^(-11)`D. `1.95 xx 10^(-11)`
Answer» Correct Answer - B `Ag +CI^(-) rarr AgCI +e^(-)` `E = E^(@) + 0.0591 log [CI^(-)]` `-0.25 = E^(@) +0.0591 log 0.1 E^(@) =- 0.1909` Now for reaction `AgCI +CI^(-)rarrAgCI +e^(-)` `{:(Ag^(+)+e^(-)rarrAg),(bar(Ag^(+)+I^(-)rarrAgCI)):}` `E = E_(Ag//AgCI//CI^(-))^(@) +E_(Ag^(+)//Ag)^(@) +0.0591 log K_(sp)` `O =- 0.1909 +0.799 +0.0591 log K_(sp)` `K_(sp) = 5.13 xx 10^(-11)`

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A silver wire dipped in `0.1M HCI` solution saturated with `AgCI` develops a potential of `-0.25V`. If `E_(Ag//Ag^(+))^(@) =- 0.799V`, the `K_(sp)` of `AgCI` in pure water will beA. `2.95 xx 10^(-11)`B. `5.1 xx 10^(-11)`C. `3.95 xx 10^(-11)`D. `1.95 xx 10^(-11)`

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