Question 65

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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