Ionic EquilibriumHard
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).
Options
A.9.0
B.5.0
C.8.64
D.5.36
Solution
$\left\lbrack OH^{-} \right\rbrack = \frac{K_{6}.\left\lbrack \text{Pyridine} \right\rbrack}{\left\lbrack \text{Pyridinium ion} \right\rbrack} = \frac{1.5 \times 10^{- 9} \times 0.2}{\left( \frac{0.15}{500} \times 1000 \right)} = 1 \times 10^{- 9\text{ M}}$
$\therefore P^{OH} = - \log\left( 10^{- 9} \right) = 9.0 \Rightarrow P^{H} = 5.0$
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