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.
NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2).
NO-2 + H2O ↔ HNO2 + OH-
Kh = [ HNO2 ] [ OH-] / [NO-2]
⇒ Kw / ka = 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
Kh = 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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 61: The ionization constant of nitrous acid is 4.5 x 10-4. Calculate the pH of 0.04 M sodium nitrite sol....
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