Question

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

Answer 1

Pyridinium hydrochloride is the salt of a weak base pyridine and a strong add HCl.

Answer 2

Given:

Molarity of solution = 0.02M

pH of the solution = 3.44

To find:

Ionization constant of pyridine ($K_a$)=?

Formula to be used:

$pH = -log[H^+]$

$K_a = \frac{K_w}{K_h}$

Calculation:

As per the formula $pH = -log[H^+]$, $[H^+]$ is calculated as $3.63\times 10^{-4}$.

As $[H^+]$ value and concentration of solution is known, with it $K_h$ value is calculated as

$K_h = \frac{(3.63\times 10^{-4})^2}{0.02} =6.6\times 10^{-6}$

As, $K_a = \frac{K_w}{K_h}$, substituting the value in the formula we get

$K_a = \frac{K_w}{K_h} = \frac{10^{-4}}{6.6\times 10^{-6}} = 1.51\times 10^{-9}$

Conclusion:

The ionization constant of pyridine $1.51\times 10^{-9}$.

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