What is the pH of pure water at 99^oC, if K_w(99^oC) is 8.836 x 10^-13?

Question

Quizplus · Accepted Answer

The pH of a solution is calculated using the formula pH = -log[H+], where [H+] is the concentration of hydrogen ions in the solution. For pure water, [H+] = [OH-], and the product of these concentrations, [H+][OH-], is equal to the water dissociation constant, Kw. At 99°C, Kw = 8.836 x 10^-13. Since the product of [H+] and [OH-] is equal to Kw and in pure water [H+] = [OH-], we can find [H+] by taking the square root of Kw. Thus, [H+] = sqrt(8.836 x 10^-13) ≈ 9.4 x 10^-7 M. The pH is then calculated as pH = -log(9.4 x 10^-7) ≈ 6.02.

What Is the PH of Pure Water at 99oC, If

What Is the PH of Pure Water at 99^oC, If