An amount of 0.15 mole of pyridinium chloride has been added into 500 ml of 0.2 M pyridine solution.

Question

An amount of 0.15 mole of pyridinium chloride has been added into 500 ml of 0.2 M pyridine solution. Calculate pH of the resulting solution assuming no change in volume. The value of Kb for pyridine = 1.5 × 10–9 (log 2 = 0.3, log 0.3 = 0.48).

Accepted Answer

$\left\lbrack OH^{-} ightbrack = \frac{K_{6}.\left\lbrack 	ext{Pyridine} ightbrack}{\left\lbrack 	ext{Pyridinium ion} ightbrack} = \frac{1.5 	imes 10^{- 9} 	imes 0.2}{\left( \frac{0.15}{500} 	imes 1000 ight)} = 1 	imes 10^{- 9	ext{ M}}$$	herefore P^{OH} = - \log\left( 10^{- 9} ight) = 9.0 \Rightarrow P^{H} = 5.0$

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