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# pH and pKa

Most of the inorganic reactions between compounds formed out of ionic bonds are ionic in nature. The compound dissociates in a polar solvent and the reaction takes place between ions.

Some reactions are instantaneous and hence the progress of the reaction cannot be recorded. Certain reactions are very slow and it is difficult to record the completion of these reactions. Those reactions that are moderately fast can be observed and their speed can be calculated. These calculations are done in terms of rate of reaction ka which is called the dissociation constant. There exists an equilibrium between the reactants and products at any given time since the dissociation into ions are not complete.

The ratio of this equilibrium is the rate at that given time. In the case of acid-base reactions, the basic reaction is between the hydrogen ions(H+) and hydroxide ion(OH-) to form a molecule of water. In these reactions, the rate of reaction depends upon the concentration of H+ ions and is given by the value of potent H+ or pH.

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## Definition of pH

The pH of a solution is defined as the negative logarithm to the base 10 of the hydrogen ion concentration in g.ionsL-1.

pH = – log10 [H+]

or

pH = log10 $\frac{1}{[H^+]}$

Thus pH can also be defined as the logarithm to the base 10 of the reciprocal of H+ ion concentration.

or [H+] = 10 -pH

The negative power to which 10 must be raised in order to express H+ ion concentration of a solution in g.ionL-1 is equal to the pH of the solution.

Use the below widget to calculate the hydrogen ion concentration –

## pKa Definition

During the ionization of weak acid HA, where the ionization is not complete, the dissociation constant is the ratio of the dissociated and undissociated components.

Ka = $\frac{[H^+] [A^-]}{[HA]}$

Hence pKa is defined as the negative logarithm to the base 10 of the Ka in g ions/ L or as the logarithm to the base 10 of the reciprocal of Ka.

## pKa and pH

The pKa is defined as the pH at which a compound exists as 50% ionized and 50% non-ionized.

pKa = - log Ka

pH = - log H+
The relationship between pH and pKa is given by Henderson - Hasselbach equation.

pH = pKa + log $\frac{[A^-]}{[HA]}$
When a molecule AH is deprotonated the equation can be written as -

AH H+ + A-

Then
Ka = $\frac{[H^+] [A^-]}{[AH]}$

Taking the log on both sides

logKa = log [H+] + log $\frac{[A^-]}{[HA]}$

By definition

pKa = - log Ka and pH = - log H+

Therefore
pKa = pH – log $\frac{[A^-]}{[AH]}$
or
$\frac{[A^-]}{[AH]}$ = 10 (pH - pKa )

The above equation suggests that when pH = pKa of a solution, half of the molecules lose their protons.

## pH to pKa

Protonation decreases the nucleophilicity of a species, the pH of the medium affects the rate of many nucleophile reactions. At fixed pH, the most reactive group is usually the one with the lowest pKa.

Ka = $\frac{[H^+] [A^-]}{[ HA ]}$

Ka x $\frac{[HA]}{[A^-]}$ = [ H+ ]

log Ka + log $\frac{[HA]}{[A^-]}$ = log [H+]

- log Ka + log $\frac{[A^-]}{[HA]}$ = - log [H+]

pKa + log $\frac{[A^-]}{[HA]}$ = pH

The relationship between pKa and pH is given in the equation –

pKa = pH – log $\frac{[A^-]}{[AH]}$

or

pKa + log $\frac{[A^-]}{[HA]}$ = pH

Therefore pH = pKa + log $\frac{[A^-]}{[HA]}$ which is Henderson - Hasselbach equation.

## Henderson-Hasselbach Equation

• This relationship can be best explained when one uses the buffer. By definition, a buffer stops changes in the pH of the solution when added to the weak electrolytes. A buffer neutralizes the added amounts of acid or base.
• A buffer is a solution of a weak acid and its conjugate base or a weak base and its conjugate acid.
• A buffer is selected on the basis of its pKa value along with its chemical nature.
• The Henderson-Hasselbach equation can be written when the buffer is used as the ratio of the concentration of the salt and acid forms of the buffer.

When the concentration of the conjugate base and the undissociated acid are equal - [conjugate base] = [acid], the pH of the solution equals the pKa of the buffer.

When [conjugate base] = 10x[acid], then log ([conjugate base] / [acid]) = log 10 = 1 and thus

pH = pKa+ 1

When [conjugate base] = $\frac{1}{10}$ x [acid],

then log $\frac{[conjugate\ base]}{[acid]}$ = log $\frac{1}{10}$ = -1 and thus

pH = pKa – 1

pKa = pH – log $\frac{[A^-]}{[AH]}$

or

$\frac{[A^-]}{[AH]}$ = 10 (pH-pKa )

The above equation suggests that when pH = pKa of a solution, half of the molecules lose their protons. When pKa is less than 2, then it is a strong acid. When it is more than 2 and less than 7 it is a weak acid. When it is more than 7 but less than 10 it is a weak base and when it is between 10 and 14, it is a strong base.

## pKa Values

#### pKa values of some important acids are as follows.

 Acid Formula -pKa Hydrogen iodide HI -10 Hydrogen bromide HBr -9 Hydrogen chloride HCl -7 Sulfuric acid H2SO4 -4.8 Ethyloxonium ion CH3CH2OH2+ -2 Hydronium ion H3O+ -1.5 Hydrogen fluoride HF 3.5 Acetic acid CH3COOH 4.7 Ammonium ion NH4+ 9.2 Water H2O 15.7 Methanol CH3OH 16 Ethanol C2H5OH 16 Ammonia NH3 36 Methane CH4 60

Lysine pKa

Lysine has a structure COOH-CH(NH2)-CH2-CH2-CH2-CH2-NH2. It has an amino group on the side-chain also and hence the isoelectric point is the average of the amino group pK values. The concentration of the neutral species, lysine (LH), is a maximum at the isoelectric point, but the concentration is less than 100% because the difference in pKa values is only about 1.5. Lysine exists as a zwitterion in solution.

The pKa values of Lysine are 2.18 for α- COOH, 8.95 for α- NH3+ and10.79 for side chain NH3
Phosphate pKa Value

Phosphoric acid H3PO4 can undergo deprotonation in three steps. The values of pKa also vary for each step. They are as follows.

H3PO4 H2PO4- + H+ pKa = 2
H2PO4-
HPO42- + H+ pKa = 7
HPO42-
PO43- + H+ pKa = 12

## Calculate pKa

pH is an important variable in biological wastewater treatment processes, since microorganisms are only active in a certain often narrow, pH range. Calculation of the pH curve for a titration of a weak acid with a strong base really amounts to a series of buffer problems. In performing these calculations it is very important to remember that even though the acid is weak, it reacts essentially to completion with hydroxide ion a very strong base.

We know that –

pKa = - log Ka and Ka = $\frac{[H^+] [A^-]}{[ HA ]}$

and if the dissociation is complete Ka = [H+] [A-]

By knowing the molarity of the solution, pKa can be calculated.

The other method is by knowing the pH of the given solution and relating it to the pKa by using the equation pKa = pH – log $\frac{[A^-]}{[AH]}$.

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