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 $\ce{H3O+}$.
$\ce{H3O+}$ 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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – bon
    Oct 25 '15 at 10:26
  • 1
    $\begingroup$ Please visit this page, this page and this ‎one on how to format your posts better.‎ Alternatively, visit this chatroom for further formatting guidance. $\endgroup$
    – M.A.R.
    Oct 25 '15 at 11:43
  • $\begingroup$ This post has triggered and automatic flag because you have edited it so many times, please avoid this in the future. The community would also very much appreciate if you could take a little time and read through the links that inɒzɘmɒЯ.A.M has posted. Edit summaries should comment on the things performed not work as hidden comments. $\endgroup$ Oct 26 '15 at 5:53
  • $\begingroup$ It is a correct answer. I also respect all of what your saying except for @Bon saying I copied it from the duplicate answer. $\endgroup$
    – Technetium
    Oct 26 '15 at 7:24

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