Question

At 100 degrees Celsius, the ion product for pure water is Kw= 51.3*10^-14. What is the pOH of water at this temperature? A) 7.00 B) 7.27 C) 6.14 D) 6.63 E) 7.86

Answer 1

Dissociation of water as

H2O  $\rightleftharpoons$ H+(aq) + OH-(aq)

Kw = [H+] [OH-]

when 1 molecule of water dissociate to form 1 molecule of H+ and 1 molecule of OH-

therefore [H+] = [OH-]

hence,

Kw = [OH-]2

51.3 X 10-14 =  [OH-]2

[OH-] = $\sqrt{}$ (51.3 X 10-14

[OH-] = 7.1624 X 10-7

pOH = -log [ OH- ] = -log(7.1624 X 10-7) = 6.14

Ans = C) 6.14