# Acid - base strenght

Referring to the acid-base definition of Bronsted and Lowry , a substance is not acidic or basic as absolutely; we can just say if it behaves as acid or base towards another substance . To say which of the two substances is more acidic , however, you can put them in comparison with the same base.

The substance taken as reference for the measurement of the acidity / basicity of substances is water , H2O . The water is suitable for this role because many reactions which we deal with take place in aqueous solution. The reaction of an acid with H2O will be of the type :

$HA&space;+&space;H_2O&space;\rightleftharpoons&space;H_3O^+&space;+&space;A^-$

The equilibrium constant of this reaction will be:

$K&space;=&space;\frac{[H_3O^+][A^-]}{[H_2O][HA]}$

for the definition we have given above, the acid will be stronger the more dissociated it is (and so very inclined to donate a proton). We can even say that it will be stronger the greater the equilibrium constant is.

This expression can be further simplified. In fact, operating in ideal conditions or in very diluted solutions (large volume of water), you can consider the constant concentration of H2O.

$[H_2O]_{initial}&space;-&space;[H_2O]_{reacted}&space;=&space;[H_2O]_{final}$

but:

$[H_2O]_{reacted}&space;<<&space;[H_2O]_{initial}$

so we can consider that:

$[H_2O]_{initial}&space;\approx&space;[H_2O]_{final}$

Assuming that the aqueous solution has density = 1, we can calculate the molarity as:

$M&space;=&space;\frac{d}{PM}&space;=&space;\frac{1000g/l}{18g/mol}&space;=&space;55,5&space;mol/l$

Being a constant value, it can be incorporated in the equilibrium constant:

$K\cdot[H_2O]&space;=&space;Ka&space;=&space;\frac{[H_3O^+][A^-]}{[HA]}$

This new constant, Ka, is called dissociation constant (and ionization constant) of the acid. The larger the numerical value of Ka , the stronger will be the acid in question, with a greater dissociated component than the undissociated one.
The smaller the numerical value of the Ka the weaker will be the acid in question,  the greater the undissociated component with respect to that dissociated.
Those that we commonly call strong acids, such as HNO3, HCl, H2SO4 etc. in aqueous solution they are completely dissociated into their constituent ions.
Take for example HCl:

HCl + H2O  $\rightleftharpoons$  H3O+ + Cl

Since it is a very strong acid, this balance will be completely shifted to the right;  the whole HCl in solution is dissociated into H+ and Cl-. Also according to Bronsted and Lowry, we can deduce that Cl- is a very weak base, because he has no tendency to accept the proton from H3O+.

HCl and Cl- are a conjugate acid-base pair. From the example we have just brought, it follows that the stronger the acid the weaker it is the conjugated base. It 'also true the contrary: the stronger the base, the weaker it is the conjugated acid.

The argument that we made to find a measure of the acid strength can be similarly made to a base:

$B&space;+&space;H_2O&space;\rightleftharpoons&space;BH^+&space;+&space;OH^-$

A base is stronger as it is more likely to accept a proton from another substance. Even in this case, the relative strength of two bases, can be compared by comparing them with the same acid. We choose once again to find a base strength scale referring to water. The same considerations that we made earlier to the acids are still valid.

$K\cdot[H_2O]&space;=&space;Kb&space;=&space;\frac{[BH^+][OH^-]}{[B]}$

The constant Kb allows us to express the strength of a base. Is also called the basic hydrolysis constant or basic dissociation constant.

Notice that, when a base tears a water proton, are obtained the protonated base and the OH- ion.

The strong bases are therefore those who accept a proton more easily, or which release in solution the highest concentration of OH- ions.

Strong bases for common use are the hydroxides, because free quantitatively OH- ions into solution. For the hydroxides, we can still avail ourselves of the acid-base theory of Arrhenius, since we are in aqueous solution.

Eg:

$NaOH&space;\xrightarrow[]{H_2O}&space;Na^+&space;+&space;OH^-$

$Kb&space;=&space;\frac{[Na^+][OH^-]}{[NaOH]}&space;=&space;great$

Weak bases however, do not release quantitatively OH- ions, because they do not get fully protonated. Example:

$NH_3&space;+&space;H_2O&space;\rightleftharpoons&space;NH_4^+&space;+&space;OH^-$

$Kb&space;=\frac{&space;[NH_4^+][OH^-]}{[NH_3]}&space;=&space;1,8&space;\cdot&space;10^{-5}$

Since the value of Kb is 1,8 ·10-5, means that the denominator, [NH3], is greater than the numerator [NH4+], [OH-], or that the amount of not protonated base is greater than that protonated.