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Chemistry
Equilibrium
the ionization constant of nitrous acid is 4 5 x 1...

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:

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

NO^-₂ + H₂O ↔ HNO₂ + OH^-

K_h = [ HNO₂ ] [ OH^-] / [NO^-₂]

⇒ K_w / k_a = 10^-14 / 4.5 x 10^-4= 0.22 x 10^-10

Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:

[NO^-₂ ] = 0.04 - x ; 0.04

[ HNO₂ ] = x

[ OH^-] = x

K_h = x² / 0.04 = 0.22 x 10^-10

x² = 0.0088 x 10^-10

x = 0.093 x 10^-5

∴ [ OH^-] = 0.093 x 10^-5M

[H₃O⁺] = 10^-14 / 0.093 x 10^-5= 10.75 x 10^-9M

⇒ pH = -log(10.75 x 10^-9)

= 7.96

Therefore, degree of hydrolysis

= x / 0.04 = (0.093 x 10^-5 ) / 0.04 = 2.325 x 10^-5

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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.

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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.

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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.