Question

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

Answer 1

NaNO₂ when dissolved in water gets dissociated to give Na⁺ and NO₂^-. Nitrite ion (NO₂^-) undergoes hydrolysis in accordance with the reaction.

If α is the degree of hydrolysis, then

Then, [OH^-] = Cα = 0.04 mol L^-1 x 2.36 x 10^-5 = 9.43 x 10^-7

and [H⁺] = K_w/[OH^-] = \(\frac{1\times 10^{-14}}{9.43\times 10^{-7}}\) = 1.06 x 10^-8

pH = -log[H⁺] = -log(1.06 x 10^-8) = 7.975