# When an acid dissolves in water, what ion does water form?

Aug 27, 2017

Well, we conceive it to be the $\text{hydronium ion.......}$

#### Explanation:

Water is conceived to undergo an autoprotolysis reaction as shown.....

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

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ under standard conditions.

When an acid, say $H X$, is dissolved in water, we conceive it to INCREASE the concentration of the characteristic cation, ${H}_{3} {O}^{+}$, and accordingly reduce concentrations of the characteristic anion, $H {O}^{-}$, so that the autoprotolysis equilibrium is maintained.

And a base likewise increases concentrations of the characteristic anion, $H {O}^{-}$.

We can quantify the acid-base equilibrium in water under standard conditions by the equation......

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

${K}_{w} = {10}^{-} 14 = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$

And taking $- {\log}_{10}$ of both sides.....

$- {\log}_{10} {K}_{w} = - {\log}_{10} {10}^{-} 14 = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left[H {O}^{-}\right]$

And thus $14 = {\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{\text{pH"+underbrace(-log_10[HO^-])_"pOH}}$

And thus our working equation.....$14 = p H + p O H$.

See here for practical issues with respect to the uses of acids in water.