What is the pH of a 88103 M solution of KOH at 25 CSolutionK

What is the pH of a 8.8×10-3 M solution of KOH at 25 °C?

KOH ---> K+ + OH-

we know that

KOH is a very strong base , so it undergoes 100 % dissociation

now

from the above reaction , we get

concentration of OH- ( [OH-] ) = 0.0088 M KOH x 1 OH- / 1 KOH

concentration of OH- ([OH-] ) = 0.0088 M OH-

now

[H+] [OH-] = Kw = 1 x 10-14

[H+] x 0.0088 = 1 x 10-14

[H+] = 1.1364 x 10-12

now

pH = -log [H+]

pH = -log (1.1364 x 10-12)

pH = 11.944