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A `0.25 M` solution of pyridinium chloride `C_(5)H_(5)NH^(+)Cl^(-)` was found to have a `pH` of `2.75` What is `K_(b)` for pyridine, `C_(5)H_(5)N`? |
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Answer» `C_(5)H_(5)NH^(+)Cl^(-)`" " It is salt of `SAWB` Pyridinium ion on hydrolusis will produce `H_(3)O^(+)` `:.[H_(3)O^(+)]=antilog(-2.75)=Ch=0.25h` `h` comes negligible `(lt0.1)` `pH=7-(1)/(2)pK_(b)-(1)/(2)logCrArr pH=7-(1)/(2)pK_(b)-(1)/(2)log0.25` `rArr pK_(b)=9.1` `K_(b)=Antilog (-9.1)=8xx10^(-10)` |
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