Question

Question 7.65 Ionic product of water at 310 K is 2.7 × 10^(–14). What is the pH of neutral water at this temperature?

Class XI Equilibrium Page 230

Answer 1

Ionic product { Kw } = [H3O+][OH-]
Kw = [H3O+][OH-] = 2.7 × 10^-14 at 310K

we know,
H2O + H2O <=> [H3O+][OH-]
[H3O+] = [OH-]
therefore, [H3O+] = √{2.7 × 10^-14}
[H3O+] = 1.643 × 10^-7 M
PH = -log[H3O+] = -log(1.643 × 10^-7)
= -{log10^-7 + log(1.643)}
= -{ -7 + 0.2156}
= 7 - 0.2156 = 6.7844

hence, PH = 6.7844

Answer 2

Hi

Please see the attached file!

Hope it helps you !