An amount of 0.15 mole of pyridinium chloride has been added into 500 ml of 0.2 M pyridine solution. — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

$[\text{Salt}] = 0.15\text{ mol} / 0.5\text{ L} = 0.3\text{ M}$. $[\text{Base}] = 0.2\text{ M}$. Using the basic buffer equation: $[\text{OH}^-] = K_b \times \frac{[\text{Base}]}{[\text{Salt}]} = 1.5 \times 10^{-9} \times \left(\frac{0.2}{0.3}\right) = 1.0 \times 10^{-9}\text{ M}$. The $\text{pOH} = -\log(10^{-9}) = 9.0$. Therefore, $\text{pH} = 14 - 9.0 = 5.0$.