Defining an Acid

This lesson covers: 

  1. The Brønsted-Lowry definitions of acids and bases
  2. Proton transfer in acid-base reactions
  3. Reactions of acids with metals and bases
  4. The ionic product of water, Kw

Brønsted-Lowry acids are proton donors, bases are proton acceptors

According to the Brønsted-Lowry theory:

  • An acid is defined as a substance that donates a proton (H+).
  • A base is defined as a substance that accepts a proton.


For instance, when a Brønsted-Lowry acid (HA) is dissolved in water, it donates a proton to a water molecule, forming a hydronium ion (H3O+):

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)


Conversely, when a Brønsted-Lowry base (B) is mixed with water, it accepts a proton from a water molecule, forming a hydroxide ion (OH-):

B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)

Proton transfer in acid-base reactions

Acids can only release their protons in the presence of a base that accepts them.


This proton transfer process is represented by the following equilibrium reaction:

HA(aq) + B(aq) ⇌ BH+(aq) + A-(aq)


Where:

  • HA is the acid (proton donor)
  • B is the base (proton acceptor)

Shifting the acid-base equilibrium

The position of the acid-base equilibrium can be influenced by changing the concentrations of reactants or products:

  1. Increasing the concentration of acid (HA) or base (B) shifts the equilibrium to the right, favouring the formation of products (BH+ and A-).
  2. Increasing the concentration of conjugate acid (BH+) or conjugate base (A-) shifts the equilibrium to the left, favouring the formation of reactants (HA and B).

Acid-base reactions in water

When an acid is added to water, water acts as the base, accepting the proton from the acid:

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)


The position of this equilibrium depends on the strength of the acid:

  1. For weak acids, the equilibrium lies far to the left, with more reactants (HA and H2O) present.
  2. For strong acids, the equilibrium lies far to the right, with more products (H3O+ and A-) present.

Water can act as an acid or a base

Water is an amphiprotic substance, meaning it can behave as both an acid and a base depending on the reaction.


1. Water as a base:

  • When reacting with acids, water accepts a proton to form a hydronium ion (H3O+).
  • Example:  HCl(aq) + H2O(l) ⇌ Cl-(aq) + H3O+(aq)


2. Water as an acid:

  • When interacting with bases, water donates a proton to form a hydroxide ion (OH-).
  • Example:  NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

Acids react with metals and bases

1. Acids react with reactive metals to produce a salt and hydrogen gas:

  • Metal + acid ➔ salt + hydrogen
  • Example:  Mg(s) + 2HCl(aq) ➔ MgCl2(aq) + H2(g)


2. Acids react with metal carbonates to produce a salt, water, and carbon dioxide gas:

  • Metal carbonate + acid ➔ salt + water + carbon dioxide
  • Example:  CaCO3(s) + 2HCl(aq) ➔ CaCl2(aq) + H2O(l) + CO2(g)


3. Acids react with alkalis (soluble bases) to yield a salt and water:

  • Acid + alkali ➔ salt + water
  • Example:  HCl(aq) + NaOH(aq) ➔ NaCl(aq) + H2O(l)


4. Acids react with insoluble bases (usually metal oxides) to form a salt and water:

  • Acid + metal oxide ➔ salt + water
  • Example:  2HCl(aq) + CuO(s) ➔ CuCl2(aq) + H2O(l)

The ionic product of water (Kw)

A small fraction of water molecules spontaneously transfer protons between themselves, a process called autoionisation:

2H2O(l) ⇌ H3O+(aq) + OH-(aq)


Or, more simply:

H2O(l) ⇌ H+(aq) + OH-(aq)


The equilibrium constant (Kc) for this reaction is:

Kc​=[H2​O][H+][OH−]​

However, because water is present in vast excess compared to H+ and OH-, its concentration [H2O] is essentially constant.


Multiplying both sides of the Kc expression by [H2O] gives the ionic product of water (Kw):

Kw = [H+][OH-]


At 298 K, Kw​=1.00×10−14 mol2 dm−6

The value of Kw varies with temperature.

H+ and OH- concentrations in pure water

In pure water, the concentrations of H+ and OH- are equal:

Kw​=[H+]2=[OH−]2=1.00×10−14 mol2 dm−6


Therefore, in pure water at 298 K:

[H+]=[OH−]=√(1.00×10−14)​=1.00×10−7 mol dm−3