water ionisation (1 Viewer)

sssona09

Member
Joined
Oct 2, 2015
Messages
90
Gender
Female
HSC
2018
can someone please explain tis equation in terms of water ionisation?
H2O (l) -->H+ (aq) + OH-(aq)

also, if acid ionises in water, then it gives up a proton - so how do we get H3O+? shouldn't it be H3o-
 

captainhelium

water enthusiast
Joined
Mar 24, 2016
Messages
160
Location
Tokyo
Gender
Undisclosed
HSC
2017
can someone please explain tis equation in terms of water ionisation?
H2O (l) -->H+ (aq) + OH-(aq)

also, if acid ionises in water, then it gives up a proton - so how do we get H3O+? shouldn't it be H3o-
The Bronsted-Lowry definition of acids and bases states that acids are considered as proton donors and bases are considered as proton acceptors.

For example, consider the acid ionisation equation for hydrochloric acid in water:

HCl(aq) + H2O --> H3O+ +Cl-

In the above reaction, HCl donates a proton (H+) to water. Remember that when HCl is an aqueous solution, it dissociates into its respective ions, that is

HCl --> H+ + Cl-.

Essentially, this H+ ion from the dissociation of HCl will be given to the water molecule. Thus, water gains an extra hydrogen ion, which is positively charged (since hydrogen ions are essentially protons), which is why water is considered a base in the acid ionisation equation since it accepts a proton.

So basically, with H2O accepting a proton (H+) from HCl, you get hydronium which is H3O+. Notice how hydronium has an extra hydrogen atom compared to water and is positively charged? This is due to water accepting the H+ from HCl.

Now since HCl has donated a proton, all it has left is a chloride ion (from the equation regarding the dissociation of HCl). Thus, the products of an acid ionisation equation are hydronium and some random anion.

Thus we get the equation,

HCl + H2O --> H3O+ + Cl-

Hopefully that made more sense about why H3O+ is formed.

Also, regarding your question about explaining water ionisation, essentially water ionises with itself in water, which sounds kinda funny when you first hear about it. Consider it like an interaction between an acid and a base -

H2O + H2O --> H3O+ + OH-

So in the above equation, one of the water molecules will act as an acid and the other will act as a base. For example, let's say that the first water molecule is an acid.

This means that it will 'lose' one of its protons (H+ ion) and give it to the other water molecule. Thus, when one of the H2O molecule loses and donates a positive H+, it becomes OH- (since OH- has one less H+ than water).

This donated proton gets accepted by the other H2O molecule and so becomes H3O+. Thus we get the equation,

H2O + H2O --> H3O+ + OH-.

Essentially, one of the water molecules acts as an acid (since it donates a proton) and the other water molecule acts as a base (since it is accepting a proton).

The equation H2O --> H+ + OH- is really just a simplification of the above equation without considering the Bronsted-Lowry definitions I think.

Also, just remember that the self-ionisation of water is a reversible reaction

Hope that helped! Haven't touched Chemistry since my HSC exam so I might be a bit fuzzy on the info.
 
Last edited:

dan964

what
Joined
Jun 3, 2014
Messages
3,473
Location
South of here
Gender
Male
HSC
2014
Uni Grad
2019
Can agree, captain helium is on point with regards to the theory.
Water ionization would be an equilibrium reaction


Remember that when we talk about donating a proton, it is equivalent to talking about donating a ion.
The equation would be



Combining these two equations gives the full ionization equation for water which is reversible.
 
Last edited:

Ployixi

Penrith Represent.
Joined
Feb 2, 2014
Messages
8
Gender
Male
HSC
2016
can someone please explain tis equation in terms of water ionisation?
H2O (l) -->H+ (aq) + OH-(aq)

also, if acid ionises in water, then it gives up a proton - so how do we get H3O+? shouldn't it be H3o-
Protons have a positive charge so that's why the hydronium ion (h3o+) is positive

Sent from my Pixel using Tapatalk
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top