`0.15` mole of pyridinium chloride has been added into `500 cm^(3)` of `0.2M` pyridine solution. Calculate pH and hydroxyl ion contration in the resulting solution, assuming no change in volume. `(K_(b)"for pyridine" = 1.5xx10^(-9)M)`

Explanation

`["Pyridinium chlodide"]= (0.15//500)xx1000=0.3M`
`["Pyridine"]= 0.2M`
:. A mixture of pyridine and its salt pyridinium chloride forms a basic a buffer and therefore,
`pOH= -log K_(b)+log ["Salt"]//["Base"]`
or `pOH= -log 1.5xx10^(-9)+log(0.30//0.20)`
`= -log 1.5+9 log 10+log 1.5=9`
`:. [OH^(-)]=10^(-9)`
and `[H^(+)]=10^(-5)`
So `pH=5`