A 0.02M solution of pyridinium hydrochloride $$(C_5H_6ClN)$$  is having pH = 3.44. Calculate the ionization constant of $$C_5H_5N$$ (pyridine).

Asked by Abhisek | 2 years ago |  127

##### Solution :-

pH = 3.44As

we know,

pH = $$\log [H^{+}]$$

$$[H^{+}]$$ = $$3.63\times 10^{-4}$$

Now, $$K_{h} = \dfrac{3.63\times 10^{-4}}{0.02}$$

(Given that concentration = 0.02M)

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

As we know that,

$$K_{h} = \dfrac{K_{w}}{K_{a}}=K_{a}$$

$$= \dfrac{K_{w}}{K_{h}}$$

$$= \dfrac{10^{-14}}{6.6\times 10^{-6}}$$

$$1.51×10^{−9}$$

Answered by Abhisek | 2 years ago

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