Question

What is the equilibrium constant of pure water at 25°C?

Answer 1

Answer:The equilibrium constant of pure water is
`K_(eq)=1* 10^(-14)`

Step-by-step explanation:

`H_2O+H_2O\rightleftharpoons H_3O^++OH^-`

`K_(eq)=([H_3O^+][OH^-])/([H_2O][H_2O])`

`K_i=K_(eq)* [H_2O]}=([H_3O^+][OH^-])/([H_2O])`

`K_i=\text{Ionization constant of water}`

Since,the water is found to be poorly ionized ,concentration of pure water practically remains the same. So, concentration of water can be combined with
`K_i` to give new constant known as ionic product of water that is
`K_w`.

`K_w=K_i* [H_2O]=[H_3O^+][OH^-]`

In pure water:

`[H_3O^+]=1* 10^(-7) M]=[OH^-]`

`K_w=1* 10^(-14)M^2`

`K_w=K_(eq)=1* 10^(-14)`