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Nov 1, 2018

Fate of Proton in Aqueous Medium

Fate of Proton in Aqueous Medium :

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 A) 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  ( M.P.  112°C); HClO₄, H₂O ( M.P.  +50°C); HClO₄, 2H₂O ( M.P.  -17.8°C);
HClO₄, 3H₂O ( M.P.  -37°C); HClO₄, 3.5H₂O ( M.P.  -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 100°C 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 
Definition of pH and pOH in Aqueous Solution Mathematical Relationship between pH and pOH
Representation of Dissociation Constant 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⁻

Le-Chatelier's Principle: 

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 20°C, 25°C and 60°C 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 = -log₁₀[H⁺] = log₁₀1/[H⁺]
Thus for a solution having H⁺ concentration 10−¹M has a pH=1.
And for a solution having H⁺ concentration 10⁻¹⁴M 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.
0.1M H₂SO₄ is really 0.2N H₂SO₄ 
So that, pH = -log₁₀⁡[H⁺= -log₁₀⁡(0.2) = 0.699

Calculation of pH:

Since HCl is strong electrolyte and completely dissociated.
Thus, [H⁺] = [HCl] = 0.002M = 2×10⁻³M
So, pH = -log[H⁺] = -log(2×10⁻³) = (3-log2) =2.7 
(ii) For 0.002M H₂SO₄ solution:
For H₂SO₄, [H⁺] = 2[H₂SO₄] = 2×0.002 = 4×10⁻³M
Thus, pH = -log[H⁺] = -log(4×10⁻³) = (3 - log4) = 2.4
(iii) For 0.002M Acetic Acid Solution:
For Acetic Acid [H⁺] = √(Ka×[CH₃COOH] = √(2×10⁻⁵×2×10⁻³) = 2×10⁻⁴
Thus, pH = -log[H⁺] = -log2×10⁻⁴ = (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⁻]

It is follows from these relations that the lower the pH, the solution is  acidic.
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 Con-centration:
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 the concentrations of H⁺ and OH⁻ ions are equal.
Thus, [H⁺] = [OH⁻] = 10⁻⁷M

An acidic solution is one in which concentration of H⁺ exceeds that of  OH⁻ :
or, [H⁺] 〉10⁻⁷M and [OH⁻]〈 10⁻⁷M

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) 
An basic solution is one in which concentration of OH⁻ exceeds that of H⁺ : 

or, [OH⁻] 〉10⁻⁷M and [H⁺]〈 10⁻⁷M

In terms of pH we have the following relations:
Neutral Solution [H⁺] = 10⁻⁷M or, pH = 7 
Definition of pH and pOH in Aqueous Solution Mathematical Relationship between pH and pOH
The Scale of pH and pOH for Neutral Solution
Acidic Solution :  [H⁺] 〉10⁻⁷M or, pH〈 7.0 
Definition of pH and pOH in Aqueous Solution Mathematical Relationship between pH and pOH
The Scale of pH and pOH for Acidic Solution
Basic Solution : [H⁺]〈10⁻⁷M or, pH 〉7 
Definition of pH and pOH in Aqueous Solution Mathematical Relationship between pH and pOH
The Scale of pH and pOH for Basic Solution

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.