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.

Given:

K_a of nitrous acid is 4.5 × 10^–4

[NO₂] = 0.04 M

To calculate the degree of hydrolysis, we apply the formula:

K_h = K_w / K_a

Where K_h is a degree of hydrolysis of a salt which is defined as the fraction of one mole of the salt which is hydrolyzed, when the equilibrium is attained. Hence for a salt (acid and base),

K_h = K_w / K_a

NaNO₂ is the salt of a strong base (NaOH) and a weak acid (HNO₂)

Using the formula, we write

K_h =

As K_w = 10^-14 (same value)

K_a = 4.5 × 10^–4 (given)

∴ K_h =

⇒K_h = 0.22 × 10^-10

Thus, the degree of hydrolysis is 0.22 × 10^-10

NO₂ + H₂O ↔ HNO₂ + OH^-

[NO₂] = 0.04 M (given)

Now, if x moles of the salt undergo hydrolysis, then the concentration of the various species will be:

[NO₂] = 0.04 M – x ≈0.04 M

[HNO₂] = x

[OH] = x

Using the formula,

K_h =

⇒K_h =

⇒K_h =

As K_h == 0.22 × 10^-10 (calculated)

∴ 0.22 × 10^-10 M× 0.04 M = x²

⇒X =

⇒X = 0.93 × 10^-5

As [OH] = x

⇒[OH^-] = 0.93 × 10^-5

Now,

K_w = [OH^-][H⁺]

Using the formula, we write

[H⁺] =

As the value of K_w is taken as 10^-14

⇒[OH^-]= 0.93 × 10^-5(calculated)

∴ [H⁺] =

⇒[H⁺] = 10.75 × 10^-9

To calculate pH of the solution, we apply the formula,

pH= -log(H⁺)

⇒pH= -log (10.75 × 10^-9)

⇒pH = -log 10.75 – (-9) log 10

⇒pH= 7.96

Thus, the pH of the solution is 7.96