SELECT * FROM question_mgmt as q WHERE id=3297 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) CBSE Free NCERT Solution of 11th chemistry Equilibrium ionic product of water at 310 k is 2 7 x 10 14 wh

Question:

Ionic product of water at 310 K is 2.7 x 10-14. What is the pH of neutral water at this temperature?

Answer:

Ionic product,

Kw   =  [H+] [ OH-] 

Let [H+]   =  x.

Since [H+]  =  [ OH-] , Kw  =  x2

⇒ Kw at 310K is 2.7x 10-14

∴   2.7x 10-14 = x2

⇒ x  =  1.64 x 10-7

⇒ [H+]   = 1.64 x 10-7

⇒ pH  = -log [H+] 

= -log (1.64 x 10-7)

= 6.78

Hence, the pH of neutral water is 6.78.


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