Question:
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine
Solution:
$\mathrm{pH}=3.44$
We know that,
$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$
$\therefore\left[\mathrm{H}^{+}\right]=3.63 \times 10^{-4}$
Then, $K_{h}=\frac{\left(3.63 \times 10^{-4}\right)^{2}}{0.02}$
$(\because$ concentration $=0.02 \mathrm{M})$
$\Rightarrow K_{h}=6.6 \times 10^{-6}$
Now, $K_{b}=\frac{K_{w}}{K_{a}}$
$\Rightarrow K_{a}=\frac{K_{w}}{K_{h}}=\frac{10^{-14}}{6.6 \times 10^{-6}}$
$=1.51 \times 10^{-9}$
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