Question

Given that Kw for water is 2.4 x 10-14 at 37 °C, calculate the pH of a neutral aqueous solution at 37 °C, which is the normal
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Answer #1

i)

Answer

pH = 6.81

Explanation

H2O(l) + H2O(l) <------> H3O+(l) + OH-(aq)

Kw = [H3O+][OH-] = 2.4×10-14

[H3O+] = [OH-]

x2 = 2.4×10-14

x = 1.55×10-7

[H3O+] = 1.55×10-7M

pH = - log(H3O+)

pH = -log(1.55×10-7) = 6.81

ii)

Basic

at 37℃ , a solution will be neutral at pH = 6.81. So, pH=7 solution at 37℃ will be basic

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