Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | Ionic product of water at 310 K is 2.7 x

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Equilibrium. This page offers a step-by-step solution to the specific question from Excercise ".$ex_no." , Question 65: ionic product of water at 310 k is 2 7 x 10 14 wh....
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.

More Questions From Class 11 Chemistry - Chapter Equilibrium

Recently Viewed Questions of Class 11 Chemistry

Write a Comment: