Question

Given that Kw for water is 2.4 Times 10-14 at 37 degree C, compute the pH of a neutral aqueous solution at 37 degree C, which is the normal human body temperature. PH = Is a pH = 7.00 solution acidic, basic, or neutral at 37 degree C? acidic basic neutral

Answer 1

[H+]=10^-7
[OH-]=2.4*10^-7
pOH=7-log2.4
so slightly basic

Answer 2

H2O <=> H+ + OH-

[H+] = [OH-]

Kw = [H+][OH-] = [H+]^2 = 2.4 x 10^(-14)

[H+] = (2.4 x 10^(-14)^(1/2) = 1.549 x 10^(-7) M

pH = -log[H+] = -log(1.549 x 10^(-7)) = 6.81

When pH = 7.00 => [H+] = 10^(-pH) = 10^(-7.00) = 1.0 x 10^(-7) M

[OH-] = Kw/[H+] = 2.4 x 10^(-14)/1.0 x 10^(-7) = 2.4 x 10^(-7) M

Since [OH-] > [H+] => solution is basic

Answer 3

concentration of [H+]=10^-7
[OH-]=2.4*10^-7
pOH=7-log2.4
so slightly basic
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