The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Asked by Pragya Singh | 1 year ago |  121

##### Solution :-

Sodium nitrite is a salt of NaOH (strong base) and HNO2 (weak acid).

$$NO_{2}^{-} + H_{2}O \leftrightarrow HNO_{2} + OH^{-}$$

$$K_{h} = \dfrac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]}$$

$$\Rightarrow \;\dfrac{K_{w}}{K_{a}}$$

$$= \dfrac{10^{-14}}{4.5\times 10^{-4}} = 0.22\times 10^{-10}$$

Let, y mole of salt has undergone hydrolysis, then the concentration of various species present in the solution will be:

$$[NO_{2}^{-}] = 0.04 – y; 0.04$$

$$[HNO_{2}] = y$$

$$[OH^{-}] = y$$

$$K_{h} = \dfrac{y^{2}}{0.04} = 0.22\times 10^{-10}$$

$$y^{2} = 0.0088\times 10^{-10}$$

$$y = 0.093\times 10^{-5}y$$

$$[OH^{-}]= 0.093\times 10^{-5}M$$

$$[H_{3}O^{+}] = \dfrac{10^{-14}}{0.093\times 10^{-5}}$$

$$= 10.75\times 10^{-9}$$

Thus, $$pH = -\log (10.75\times 10^{-9})= 7.96$$

Thus, the degree of hydrolysis is

$$= \dfrac{y}{0.04} = \dfrac{0.093\times 10^{-5}}{0.04}$$

$$2.325\times 10^{-5}$$

Answered by Abhisek | 1 year ago

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