`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

Pyridine is a weak base. Thus, pyridine + pyridine chloride solution is a basic buffer. Hence,
`pOH = pK_(b) + log .(["Salt"])/(["Base"])`
`pK_(b) = - log K_(b) = - log (1.5xx10^(-9))=9-0.1761=8.8239`
[Pyridine ] = 0.23 M (Given ), [ Pyridinium chloride ] = `(0.15)/(500) xx 1000 = 0.30 M`
`:. pOH = 8.82 + log. (0.30)/(0.20) = 8.82 + 0.1761 = 8.896`
i.e., `- log [OH^(-)]= 8.896 or log [OH^(-)] = - 8.896 = bar(9) . 104 or [OH^(-) ] = 1.271xx10^(-9)`
`[OH^(-)] ` from `H_(2)O = 10^(-7)` M cannot be neglected.
Hence, total `[OH^(-)]=1.27xx10^(-9) + 10^(-7)= 10^(-9) (1.27+100) = 111.27 xx 10^(-9) M = 1.1127 xx 10^(-7) M`
`[H^(+)] = (K_(w))/([OH^(-)])=(10^(-14))/(1.1127xx10^(-7))=8.987 xx 10^(-8)M`
`pH = - log [H^(+)] = - log (8.987 xx 10^(-8))=8-0.9536= 7.0464`.