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 k_{a} 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.

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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 = – log**_{10} [H^{+}]

or

**pH = log**_{10} **$\frac{1}{[H^+]}$ **

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 –**

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.

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+

**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

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

**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.

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**

**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 of some important acids are as follows.

**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- NH_{3}^{+} and10.79 for side chain NH_{3}^{+ } **Phosphate pKa Value**

Phosphoric acid H_{3}PO_{4} can undergo deprotonation in three steps. The values of pKa also vary for each step. They are as follows.

**H**_{3}PO_{4} →** H**_{2}PO_{4}^{-} + H^{+} pKa = 2

H_{2}PO_{4}^{-} →** HPO**_{4}^{2- }+ H^{+} pKa = 7

HPO_{4}^{2-} →** PO**_{4}^{3- }+ H^{+} pKa = 12

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 $\frac{[A^-]}{[AH]}$.**

or

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

The negative power to which 10 must be raised in order to express H

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.

Hence

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.**

Taking the log on both sides

By definition

Therefore

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

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

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.

or

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

Therefore

- 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

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

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

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

pKa = pH – log $\frac{[A^-]}{[AH]}$

$\frac{[A^-]}{[AH]}$ = 10

Acid |
Formula |
-pKa |

Hydrogen iodide | HI | -10 |

Hydrogen bromide |
HBr |
-9 |

Hydrogen chloride |
HCl |
-7 |

Sulfuric acid |
H_{2}SO_{4} |
-4.8 |

Ethyloxonium ion | CH_{3}CH_{2}OH_{2}^{+} |
-2 |

Hydronium ion |
H_{3}O^{+} |
-1.5 |

Hydrogen fluoride |
HF |
3.5 |

Acetic acid |
CH_{3}COOH |
4.7 |

Ammonium ion |
NH_{4}^{+} |
9.2 |

Water |
H_{2}O |
15.7 |

Methanol | CH_{3}OH | 16 |

Ethanol | C_{2}H_{5}OH | 16 |

Ammonia | NH_{3} | 36 |

Methane | CH_{4 } | 60 |

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- NH

Phosphoric acid H

H

HPO

We know that –

and if the dissociation is complete

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

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