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Like all equilibrium constants, the value of kw depends on temperature. at body temperature (37∘c), kw=2.4⋅10−14. part a what is the [h3o+] in pure water at body temperature?

Respuesta :

superman1987
first, we have to know that we can put [H+] instead of [H3O+]

so, according to the reaction equation:

by using ICE table:

             H2O ↔  H+  +  OH-

initial                    0           0

change                +X         +X

Equ                       X             X

when Kw = [H+] [OH-]

and when we have Kw = 2.4 x 10^-14 

and when [H+] = [OH-] = X

∴ 2.4 x 10^-14 = X^2

∴ X = √(2.4 x 10^-14)

      = 1.55 x 10^-7

∴[H+] = 1.55 x 10^-7


pstnonsonjoku

The concentration of  [ H3O^+] is 1.6 × 10^−7.

We know that the auto-ionization of water occurs as follows;

2H2O(l) ⇄ H3O^+(aq) + OH^-(aq)

The equilibrium constant for the auto-ionization of water is given as follows;

Kw = [ H3O^+] [ OH^-]/[H2O]^2

But water is a pure liquid hence;

Kw = [H3O^+] [ OH^-]

But  [H3O^+] =  [ OH^-] =

so Kw = [H3O^+] ^2

In this case Kw = 2.4 × 10^−14

[H3O^+] = √Kw

[H3O^+] = √2.4 × 10^−14

[H3O^+] = 1.6 × 10^−7 M

Learn more: https://brainly.com/question/18956703

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