# Water: What to use H3O+ or H+? [duplicate]

Why can we use $\ce{H+}$ and $\ce{H3O+}$ interchangeably?

I have seen in many places that first the reaction is written including Hydronium ion but then in bracket its written that we can write it as $\ce{H+}$ for simplicity. I am not clear with this.
$$\ce{H2O +H2O <=> H3O+ + OH-}$$ OR $$\ce{H2O(l) <=> H+(aq) + OH- (aq)}$$

Since your solvent is itself water, it makes no difference whether you use $\ce{H+}$ or $\ce{H3O+}$.
$\ce{H3O+}$ is basically the hydrated form of $\ce{H+}$. If you know, the oxygen atom in water contains two lone pairs. When it donates one of the lone pairs to the hydrogen atom which doesn't have any electrons, you get $\ce{H3O+}$.
So,
$\ce{H3O+}$ is not $\ce{H+}$
$\ce{H3O+}$ is $\ce{H+(aq)}$
This means that the aqueous form of $\ce{H+}$ is represented as $\ce{H3O+}$

In all cases, acids yield protons ( or hydronium ions H3O+) and bases yield OH- (hydroxide) ions in aqueous solutions.

The H3O+ ion is considered to be the same as the H+ ion as it is the H+ ion joined to a water molecule. The proton cannot exist in aqueous solution, due to its positive charge it is attracted to the electrons on water molecules and the symbol H3O+ is used to represent this transfer.

The equation can be written as:

H+ + H2O(l) → H3O+(aq).

This is hydrolysis as it is involving water as a reactant.

Consider the first equation in the question , the ionisation equation of water:

H2O(l) + H2O(l)→H3O+(aq) + OH-(aq)

The H3O+ is the conjugate acid of H2O. So H3O+ is used as a shorthand for a proton in aqueous solution. In a non-aqueous solution the proton would form a different structure.

The second equation:

H2O(l) → H+(aq) + OH-(aq)

Shows that H2O is made up of equal parts H+ and OH- ions and is amphoteric (can be an acid or a base) having a deprotonated form (OH-). The ionic component is at a very low concentration and a water molecule is generally considered covalent with a dipole moment favouring a slight positive charge.

The H3O+ ion concentration in pure water at 25° C is 10^-7 dm^-3. This can be written as:

[H3O+] = 10^-7

where the symbol [ ] means the "molarity of" (units in moles dm^-3).

The number of H3O+ and OH- ions formed by the ionisation of pure water must be equal ( from the equation):

[H3O+] = [OH-] = 10^-7).

This shows that pure water is neither acidic or basic, it is neutral. The product of [H3O+] = [OH-] is the ionic product of water.

[H3O+][OH-]=10^-7 × 10^-7 = 10^-14

shows that in aqueous (water) solutions, whether acidic, basic or neutral, the product of the ion concentrations equals 10^-14.

Acidic solutions contain more H3O+ ions than OH- ions. For basic solutions it is the reverse.

Therefore a water solution is : Neutral when [H3O+]= 10^-7. Acidic when [H3O+] > 10^-7. Basic when [H3O+] < 10^-7.

• What? I don't see how this answers the question about when to use $\ce{H+}$ or $\ce{H3O+}$. Have a look at the duplicate question to see a more in depth answer.
– bon
Oct 25 '15 at 10:26