1.

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`?

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)`


Discussion

No Comment Found

Related InterviewSolutions