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Ionization of Acids and Bases

The process in which neutral molecules get splits up into charged ions when exposed in a solution is referred to as the ionization of a compound. According to the Arrhenius theory, the acids are the compounds that dissociate in the aqueous medium in order to generate the hydrogen ions, H+ in the aqueous medium. Find interesting? Let’s learn more about the ionization of acids and bases in this section.

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Ionization of a Compound

While the bases are those compounds that furnish the hydroxyl ions, OH in the aqueous medium. The degree of ionization of the acids and bases helps determine its strength. On the basis of different acidic and basic compounds, the degree of ionization may differ.

Ionization of Acids

The degree of Ionisation refers to the strength of an acid or a base. A strong acid is said to completely ionize in water whereas a weak acid is said to only ionise partially. As there are different degrees of ionization of acids, there are also different levels of weakness for which there is a simple quantitative way to express.

Since the ionization of a weak acid is an equilibrium, the chemical equation and an equilibrium constant expression can be stated as :

HA ( aq ) + H2O ( l ) H3O+ ( aq ) + A

Ka = [ H3O+ ] [A] / [HA]

Equilibrium Constant for ionisation of an acid defines its Acid Ionisation Constant (Ka). However, the stronger the acid, the larger will be the acid ionisation constant (Ka). This means that a strong acid is a better proton donor. As a result of the concentration of the product in the numerator of the Ka, the stronger the acid the larger is the acid ionisation constant (Ka).

Ionization of Bases

Some bases like lithium hydroxide or sodium hydroxide get completely dissociated into their ions in an aqueous medium which is referred to as strong bases. Therefore, the ionisation of these bases yields hydrochloric ions, as (OH). A similar expression for the bases is:

A + H2O OH + HA+

Kb = [ OH ] [ HA+ ] / [ A ]

The base ionisation constant i.e Kb refers to the equilibrium constant for the ionisation of a base. Therefore, we can say that a strong base implies a good proton acceptor while a strong acid implies a good proton donor. The dissociation of weak acids or weak bases in water is:

CH3COOH + H2O ⇔ CH3COO‾ + H3O+

NH3 + H2O ⇔ NH4+ ( aq ) + OH‾( aq )

Solved Examples for You

Question: A 0.500 M solution of formic acid has a pH value of 2.04. Determine the Ka for formic acid.

Solution. Initial [HCOOH] = 0.500 M and pH = 2.04. Unknown, K= ?

$Ionization$

 Concentrations [HCOOH] [H + ] [HCOO − ] Initial 0.500 0 0 Change -9.12 × 10 -3 +9.12 × 10 -3 +9.12 × 10 -3 Equilibrium 0.491 9.12 × 10 -3 9.12 × 10 -3

Now substituting into the Ka expression gives:

$Ionization$

The value of Ka is consistent with that of a weak acid. Now, following the same steps, we find the value of Kb  of the base. If 0.750 M solution of the weak base ethylamine (C2H5 NH2) has a pH of 12.31.

$Ionization$

The pOH is 14 – 12.31 = 1.69. The [OH ] is found from 10 -1.69  = 2.04 × 10 -2  M. The ICE table shall be then as shown below.

 Concentrations [C2H5NH2] [C2H5NH3+] [OH−] Initial 0.750 0 0 Change -2.04 × 10 -2 +2.04 × 10 -2 +2.04 × 10 -2 Equilibrium 0.730 2.04 × 10 -2 2.04 × 10 -2

Substituting the value of Kb it yields the Kb for Ethylamine.

$Ionization$
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