Question

What is the ¥w for pure water and for solution and for solute or water pot for solution with more solute present will be :?

Answer 1

`K_w(H_2O) = 1.00xx10^-14M^2`

Explanation:

The `K_w-"value"` for water comes from the equilibrium expression for autoionization of water; that is,

`H_2O + H_2O` => `H_3O^+ + OH^-` (Bronsted-Lowry Proton Transfer Rxn)

The equilibrium expression for autoionization of water is ...

`K_(eq) = ([H_3O^+][OH^-])/[H_2O]^2`

`[H_2O]^2` is so large compared to the concentrations of `[H_3O^+] and [OH^-]` that it is considered constant and moved to `K_(eq)` side of the equation. That is, ...

`K_(eq)[H_2O]^2` = `[H_3O^+][OH^-]` = `K_w` (Ionization Product Constant for Water)

At `25^oC` `[H_3O^+] = [OH^-] = 1.00xx10^-7M`

Substituting into `K_w` = `[H_3O^+][OH^-]`
=>`K_w` = `(1.00xx10^-7M)(1.00xx10^-7M)` = `1.00xx10^(-14)M^2` at `25^oC`