Ncert Exercises & Solutions

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

pH = 3.44

We know that,

pH = -log

[H^{+}]

∴ [H^{+}] = 3.63 \times 10^{- 4}

Then, K_{h} = \frac{{(3.63 \times 10^{- 4})}^{2}}{0.02} (∵ concentration = 0.02 M)

\Rightarrow K_{h} = 6.6 \times 10^{- 6}

Now, K_{h} = \frac{K_{W}}{K_{a}}

\Rightarrow K_{a} = \frac{K_{w}}{K_{h}} = \frac{10^{- 14}}{6.6 \times 10^{- 6}}

= 1.51 \times 10^{- 9}