Question

At 90'C, pure water has [H30+]=10-6mole/liter. the value of kW at 90'C is:


10-6

10-8

10-12

10-14

Answer 1

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

Solution : Given,

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

The ionization reaction of water is,

$H_2O\rightleftharpoons H^++OH^-$

The expression for equilibrium constant is,

$K=\frac{[H^+][OH^-]}{[H_2O]}$

$K\times [H_2O]=[H^+][OH^-]$

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

$K_w$ = 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 $10^{-12}$

Answer 2

it is 10^-6
[H3O+]=[OH-]=10^-6
so,Kw=[H3O+][OH-]
...........=[10^-6][10^-6]
.......... =10^-12