Question

What is the K_w of pure water at 51.4ºC if the pH is 6.78? Enter your answer in scientific notation using "e" instead of "×10^" (1.23×10^-7 = 1.23e-7) and round to three sig figs.

please show work

Answer 1

Kw is called as Ionic product of water

The product of Molar Concentration of H+ ion and Molar concentration of OH- is called as ionic product of water.

Kw = [H+] [OH-]

The given   PH = 6.78

according to the definiation of PH

PH = -log[H+]

6.78 = -log[H+]

[H+] = 10^-6.78

[H+] = 1.66 x10^-7

Concentration of H+ = 1.66 x10^-7M

but , In a pure water

Concentration of H+ = concentration of OH-

i,e [H+] = [OH-]

SO, the concentration of OH- = 1.66 x10^-7M

Kw = [H+][OH-]

Kw =[ 1.66 x10^-7 ] [ 1.66 x10^-7]

Kw = 2.76 x10^-14

Kw = 2.76 e^-14