The ionization constant of water at 100oC is 1.0 x 10-12. What is the pH of pure water at 100oC?

Question

a) 5
b) 6
c) 7
d) 8
e) 9

Answer 1

The correct answer is (b) 6

pH=-log[H+]

The ionization of constant of water is Kw=[H+][OH-].

Since we are talking about pure water [H+]=[OH-] and thus
Kw=[H+]^2 => [H+]=SQRT(Kw) (SQRT stands for square-root)

and pH= -log(SQRT(Kw))=
=-log(SQRT(10^-12))=
=-log(10^-6)=6

Answer 2

Kw = Ka x Kb = a million.0 x 10^-13 In organic water, Ka = Kb the place Ka is the Acid Dissociation consistent And Kb is the backside Dissociation consistent replace Ka for Kb interior the 1st equation: Ka x Ka = a million x 10^-13 Ka^2 = a million x 10^-13 Ka = 3.sixteen x 10^-7 pH = -log(Ka) = -log(3.sixteen x 10^-7) = 6.5 many times (at 25 levels Celsius), Kw is a million x 10^-14, so: Ka = a million x 10^-7 and pH = -log(a million x 10^-7) = 7 it incredibly is why we many times think of of water as having a pH of seven

Answer 3

the answer is 6 but the pH of pure water is really 7

Answer 4

pure water at 100C has a Ph of 7