**Complete the Kw expression for the autoionization of water. Kw=1.00*10^-14**

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## Question “Complete the Kw expression for the autoionization of water. Kw=1.00*10^-14”

Complete the Kw expression for the autoionization of water.

K_{w}=1.00*10^-14

## Answer

### Answer 1

H2O <=> H+ + OH-

Kw = [H+][OH -]

### Answer 2

H2O (l) + H2O (l) <==> HTML3O+(aq), + OH

### Answer 3

The autoionization of water is

2H2O(l)↽−−⇀H3O+(aq)+OH−(aq)

$$2\phantom{\rule{thinmathspace}{0ex}}{\text{H}}_{2}^{\phantom{\rule[-.2em]{0pt}{0pt}}}\text{O}(\text{l})\stackrel{{\textstyle -\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\rightharpoonup}}{\leftharpoondown \phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-}{\text{H}}_{3}^{\phantom{\rule[-.2em]{0pt}{0pt}}}{\text{O}}^{+}(\text{aq})+{\text{OH}}^{-}(\text{aq})$$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−]

## Conclusion

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