Ionic product of water at 310 K is 2.7 x 10-14. What is the pH of neutral water at this temperature?
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.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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 65: Ionic product of water at 310 K is 2.7 x 10-14. What is the pH of neutral water at this temperature?....
Comments