SELECT * FROM question_mgmt as q WHERE id=3295 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=48 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

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-

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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