# In a NEUTRAL solution, how does [HO^-] compare to [H_3O^+]?

Apr 10, 2017

How do you think it compares...........?

#### Explanation:

In aqueous solution under standard conditions, the following equilibrium operates:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

At $298 \cdot K$, the ion-product, $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$,

we know that $\left[{H}_{3} {O}^{+}\right] = \left[H {O}^{-}\right]$ by specification (we have a neutral solution), and thus,

$\left[{H}_{3} {O}^{+}\right] = \left[H {O}^{-}\right] = \sqrt{{10}^{-} 14} \cdot m o l \cdot {L}^{-} 1 = {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$.