Question

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Answer 1

kw = [H+][OH-]. Kw =[10^-6][10^-6]. So therefor Kw = 10^-12. Hope it help you

Answer 2

Answer: The value of K_w at 90^oC is 10^{-12}

Solution:

Concentration of H_3O^+ = 10^{-6}mole/L

The ionization reaction of water is,

H_2 O = H^++OH^-

The expression for equilibrium constant is,

K={[H^+][OH^-]}{[H_2O]}  / [H_2O]=[H^+][OH^-]

K_w=[H^+][OH^-]

K_w = equilibrium constant of water

Pure water means the concentration of hydronium ion and hydroxide ion are equal.

[H^+]=[OH^-]=10^{-6}mole/L

Now put all the given values in the expression of equilibrium constant of water, we get

K_w=[H^+][OH^-]

K_w=(10^{-6})\times (10^{-6})=10^{-12}

therefore, the value of K_w at 90^oC is 10^{-12}