Answer to Question #310621 in Physical Chemistry for atenz

Question

Given that at body temperature, Kw = 10^-13.6, what is the pH of neutral water at body temperature? 

Answer 1

K_w = 10^-13.6;

H₂O = H⁺ + OH^-;

K_w = [H⁺] * [OH^-] =10^-13.6;

If to use neutral water then: [H⁺] = [OH^-];

Therefore, 10^-13.6 = x²;

x = SQRT(10^-13.6) = 10^-6.8;

[H⁺] = 10^-6.8;

pH = -log[H⁺] = 6.8.

Answer: 6.8.