The proton plays key role in the acid-base functions. In aqueous medium it is usually represented as H3O+
A naked hydrogen ion has a vanishingly small size ( radius ~10-13 cm =10-15 Å) and therefore has a very high (charge/radius) ratio(~105 and is expected to be the most effective in polarizing other ions or molecules according to Fajan’s rules. In H3O+ there are assumed to coordinate bonds from water oxygen to the proton, thus giving the proton a helium electronic configuration.
Experimental evidence of such formulations:
Perchloric acid (HClO4) reacts vigorously with water and It gives a series of hydrates which are :
HClO4 112°C
HClO4, H2O +50°C
HClO4, 2H2O  -17.8°C
HClO4, 3H2O  -37°C
HClO4, 3.5H2O -41.4°C
Of these hydrates the most remarkable is the mono hydrate, melting at the much higher temperature then the covalent anhydrous acid. It is very stable and can be heated to around 1000C without decomposition. The mono-hydrate is about ten times viscous as the anhydrous acid. It has the same crystal lattice as the ionic ammonium perchlorate, showing that it is a ionic compound, [H3O+][ClO4].
Water as an Acid and as a Base:
We know that water dissociates weakly to H+ and OH- ions. Regardless of what other ions are present in water, there will always be equilibrium between H+ and OH- ions.
H2O H+ + OH-
The proton, however, will be solvated and is usually written as [H3O+] . For simplicity we will write H+only. The above equilibrium will have its own equilibrium constant:
K = ([H+] × [OH-])/[H₂O]
or, K × [H2O] = [H+] × [OH-]
The square brackets indicate concentrations. Recognizing the fact that in any dilute aqueous solution, the concentration of water molecules (55.5 moles / liter) greatly exceed that of any other ion, [H2O] can be taken as a constant. Hence,
K×[H2O]=KW=[H+]×[OH-]
Where KW is the dissociation constant of water (ionic product of water). 
The value of H+ in pure water has been determined as 10-7 M, so that KW becomes,
KW = [H+] × [OH-
= 10-7× 10-7 
= 1.0 × 10-14 M
Fate of Proton in Aqueous Medium (Concept of pH and pOH)
Ionic product of water
The above relation tells us that in aqueous solution the concentration of H+and OH- are inversely proportional to each other. 
If H+ concentration increases 100 fold, that of OH- has to decrease 100 fold to maintain Kw constant. 
Dissociation of H2O into H+ and OH- ions:
Dissociation of water into H+ and OH- ions are an endothermic reaction.
Endothermic Reaction:
The Reactions in which heat is absorbed by the system from the surroundings are known as endothermic reactions.
H2O+13.7 kcal H++OH-
If a system is in equilibrium, a change in any factors that determine the condition of equilibrium will cause the equilibrium to shift in such way as to minimize the effect of this change. 
Thus according to Le-Chatelier’s principle, increasing temperature will facilitate dissociation, thus giving higher values of Kw. The value of Kw at 200C, 250C and 600C are 0.68×10-14, 1.00×10-14 and 9.55×10-14 respectively.
The Concept of pH:
The dissociation of water, Kw, has such low value that expressing the concentrations of H+ and OH- ions of a solution in terms of such low figures is not much convenient and meaningful. Such expressions necessarily have to involve negative power of the base 10. Sorensen proposed the use of a term known as pH , defined as:
pH = - log[H+] = log(1/[H+])
Thus for a solution having H+ concentration 10-1 M has a pH = 1
And for a solution having H+ concentration 10-14 M has a pH = 14.
For such solutions having H⁺ concentration in the range of 10-1 M to 10-14 M is more convenient and meaningful to express the acidity in terms of pH rather then H+ concentrations. The use of small fractions or negative exponents can thus be avoided. For a mono basic acid molarity and normality are the same while they are different for poly-basic acid.
Thus 0.1 M H2SO4 is really 0.2 N H2SO4 and the the pH of the solution is, 
pH = -log[H+] = -log(0.2) = 0.699.
It follows from these relations that the lower the pH, the more acidic the solution is. If the acidity of a solution goes down 100 fold its pH goes up by the two units.For example a solution of pH 1 has [H+] which is 100 times greater than that of pH 3.
Calculation of pH:
  1. For 0.002 M HCl Solution:
  2. Since HCl is strong electrolyte and completely dissociated. 
    Thus, [H+] = [HCl] = 0.002 M
    = 2×10-3
    So, pH = - log[H+] = - log(2×10-3
    = (3 - log2)
    2.7
  3. For 0.002 M H2SO4 solution:
  4. For H2SO4, [H+] = 2[H2SO4
    = 2 × 0.002
    = 4 × 10-3 M
    Thus, pH = -log[H+]
    = - log(4×10-3)
    = (3 - log4)
    = 2.4
  5. For 0.002 M Acetic Acid Solution:
  6. For Acetic Acid [H+]
    = (Ka×[CH3COOH]
    = (2×10-5×2×10-3)
    = 2×10⁻⁴
    Thus, pH = -log[H+]
    = -log(2 × 10-4)
    = (4 - log2)
    = 3.7
pH of a solution is 4.5. Calculate the concentration of H+ ion.
We have from the definition, 
pH = - log[H+] = 4.5 
or, log[H+] = -4.5 
∴ [H+] = 3.16×10-5
The Concept of pOH :

The corresponding expression for the hydroxide ion is, 
pOH = - log[OH-]
If the acidity of a solution goes down 100 fold its pH goes up by two units. A solution of pH  1 has [H+] which is 100 times greater than that of pH 3. 
Taking the case of OH- ions, the pOH will go down by two unit(from 13 to 11).
The product of [H+] and [OH-] is 10⁻¹⁴ and that this has to remain constant. 
Thus, [H+] [OH-] = 10-14
or, log[H+] [OH-] = log10-14
or, log[H+] + log[OH-] = -14
or, -log[H+] - log[OH-] = 14
or, pH + pOH = 14
We have from the definition, pH = -log[H+] and pOH = -log [OH-].
Calculate the [H+], [OH-] and pH of a solution prepared by diluting 20 ml of 0.1M HCl to one lit. 
[H+] = (20×0.1)/1000 =0.002 = 2×10-3 
[OH-] = (1×10-14)/[H+
= (1×10-14)/(2×10-3
= 0.5×10-11 
pH = -log[H+] = -log2×10-3 
= 3-log2 
= 2.7
Difference Between Acidic, Basic and Neutral solution on the basis of Concentration:
We can now proceed to differentiate between neutral, acidic or basic on the basis of relative concentrations of H+ and OH- ions on the one hand, and on the basis of pH on the other. 
A Neutral solution is one in which concentration of H+ exceeds that of OH-:
A neutral solution is one in which the concentrations of H+ and OH- ions are equal. 
Thus, [H+] = [OH-] = 10-7 M
In terms of pH we have the following relations, 
[H+] = 10-7
or, pH = 7
[H+] pH [OH-] pOH
10-7 7 10-7 7
An acidic solution is one in which concentration of H⁺ exceeds that of OH⁻
[H+]  〉[OH-]
or, [H+]  〉 10-7
and [OH-] 〈  10-7 M
In terms of pH we have the following relations, 
[H+] 〉 10-7 M 
or, pH 〈  7
[H+] pH [OH-] pOH
100 0 10-14 14
10-1 1 10-13 13
10-2 2 10-12 12
10-3 3 10-11 11
10-4 4 10-10 10
10-5 5 10-9 9
10-6 6 10-8 8
An basic solution is one in which concentration of OH- exceeds that of H+
[H+]〈 [OH⁻
or, [OH-] 〉10⁻⁷M 
and [H+]〈 10⁻⁷M
In terms of pH we have the following relations, 
[H+] 〈 10-7 M 
or, pH 〉 7
[H+] pH [OH-] pOH
10-8 8 10-6 6
10-9 9 10-5 5
10-10 10 10-4 4
10-11 11 10-3 3
10-12 12 10-2 2
10-13 13 10-1 1
10-14 14 100 0
Mathematical definition of pH provides a negative value when [H+] exceeds 1M. However pH measurements of such concentrated solutions are avoided as these solutions are not likely to be dissociated fully. Concentration of such strongly acid solutions is best expressed in terms of molarity than in terms of pH.
Calculate the [H+], [OH-] and pH of a solution obtained by dissolving 28 gm KOH to make 200 ml of a solution.
[OH- ] = (28×1000)/(56×200) 
= 2.5M 
(Molecular Weight of KOH = 56gm) 
[H+] = (1×10⁻¹⁴)/[OH-
= (1 × 10-14)/(2.5) 
= 4×10-15 
pH = -log[H+] = -log4×10-15 
= (15 - log4)