# In neutral water under standard conditions, what are [H_3O^+], and [HO^-]?

Apr 19, 2017

$\left[{H}_{3} {O}^{+}\right] = \left[H {O}^{-}\right] = {10}^{- 7} \cdot m o l \cdot {L}^{-} 1. \ldots \ldots \ldots \ldots \ldots \ldots$

#### Explanation:

By careful measurement at $298 \cdot K$, ${K}_{w} , \text{the ion product}$ $\equiv$

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$. Since pure water is neutral, $\left[H {O}^{-}\right] = \left[{H}_{3} {O}^{+}\right]$, and thus $\left[H {O}^{-}\right] = \left[{H}_{3} {O}^{+}\right] = \sqrt{{K}_{w}}$.

$= \sqrt{{10}^{-} 14} = {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$.

At higher temperatures, how do you think ${K}_{w}$ would evolve? Remember that the given equation is a BOND BREAKING reaction.