`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

[ Pyridium chloride ] = ` (0.15)/(500) xx 1000 = 0.3 M `
Pyridine = `0.2 M `
A mixture of pyridine and salt pyridinium chloride forms a basic buffer ,
`:. pOH = pK_(b) + log. (["Salt"])/(["Base"]) `
`rArr pOH = - log 1.5 xx 10^(-9) + log . ([0.30])/([0.20])`
` = - log 1.5 + 9 log _(10)^(10) + log 1.5 = 9 `
` :. [ OH ] = 10^(-9) `
`[H^(+)] = 10^(-5)`
so , pH = 5