Question

starting from the relationship kw = 5.47 * 10^-14 at 323 k, calculate the ph of water at this temperature. explain any assumptions you make in arriving at your answer.

Answer 1

Kw = 1*10^-14 mol²*dm^-6 at 25 Celsius degrees

Kw is also called the ionic product of water and it varies with temperature

Kw = [H⁺]*[OH^-]

But [H⁺]=[OH^-] => Kw = [H⁺]²

Replace the value of Kw:

5.47*10^-14 = [H⁺]²

Take square root both sides:

[H⁺] = sqrt 5.47*10^-14

[H+] = 2.33*10^-7

pH = -lg 2.33*10^-7

pH = lg (1/2.33) - lg 10^-7

pH = 7 - 0.36

pH = 6.64

Therefore pH of water at this temperature is 6.64

Since pH is smaller than 7, then an acid solution is dissolved in water.