Given that Kw for water is 2.4× 10–14 M2 at 37 °C, compute the
pH of a neutral aqueous solution at 37 °C, which is the normal
human body temperature.
PH?
Kw = [H3O+][OH-]
2.4*10^-14 = x*x
x = [H3O+] = [OH-] = 1.55*10^-7 M
pH = -log(H3O+)
= -log(1.55*10^-7)
= 6.81
Given that Kw for water is 2.4× 10–14 M2 at 37 °C, compute the pH of...
Given that Kw for water is 2.4× 10–14 at 37 °C, compute the pH of a neutral aqueous solution at 37 °C, which is the normal human body temperature.
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