Not sure if this is right, pure water so its [H] squared
Kw = [H] squared so I square rooted and took the -log of this to get the pH
Kw = [H+][OH-]
since water is neutral,
[H+] = [OH-]
So,
Kw = [H+][H+]
4.52*10^-15 = [H+]^2
[H+] = 6.72*10^-8 M
use:
pH = -log [H+]
= -log (6.72*10^-8)
= 7.1726
Answer: 7.173
Not sure if this is right, pure water so its [H] squared Kw = [H] squared...
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