Question

Complete the Kw expression for the autoionization of water.

K_w=1.00*10^-14

Answer 1

H2O <=> H+ + OH-

Kw = [H+][OH-]

Answer 2

H₂O_(l) + H₂O_(l) <==> H₃O⁺_(aq) + OH

Answer 3

The autoionization of water is

2H2O(l)↽−−⇀H3O+(aq)+OH−(aq)$$2{\text{H}}_2^{}\text{O}(\text{l})\stackrel{-\rightharpoonup }{\leftharpoondown -}{\text{H}}_3^{}{\text{O}}^{+}(\text{aq})+{\text{OH}}^{-}(\text{aq})$$

Recall that pure liquids are not included in the ?w$K_{\text{w}}$ expression. The concentrations of the products are in the numerator of the expression, and the concentrations of the reactants are in the denominator. Therefore, the ?w$K_{\text{w}}$ expression is

?w=[H3O+][OH−]