Question

A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Answer 1

pH = 3.44
We know that, pH = -log [H+]
∴ [H+] = 3.63 × 10-4
Then, Kh =$\frac{(3.63 × 10^{-4})^2}{0.02}$ (∵ concentration = 0.2M)
⇒ Kh = 6.6 × 10-6
Now, Kh = $\frac{K_w}{K_\alpha}$ ⇒ Kα = $\frac{K_w}{K_h}$ = $\frac{10^{-14}}{6.6 × 10^{-6}}$ = 1.51 × 10-9