Ionization of Acids and Bases
Ionization refers to the process where neutral molecules split into charged ions when dissolved in a solution. This concept is crucial in understanding the behavior of acids and bases in aqueous environments, especially in the context of the Arrhenius theory.
Arrhenius Theory of Ionization
According to the Arrhenius theory, acids are compounds that dissociate in an aqueous solution to generate hydrogen ions (H+). Conversely, bases are compounds that dissociate to produce hydroxide ions (OH–). The strength of an acid or a base is determined by the degree of ionization, which varies depending on the compound.
Key Concepts:
- Ionization of Acids:
- Strong acids completely ionize in water, meaning they dissociate fully into their constituent ions.
- Weak acids only partially ionize in water, resulting in an equilibrium between the undissociated molecules and the ions.
For a weak acid, the ionization can be represented by the equation:
HA ( aq ) + H2O ( l )H3O+( aq ) + A–
The strength of the acid is quantified by its Acid Ionization Constant (Ka), defined as:
Ka= [ H3O+ ] [A–] / [HA]
- Ionization of Bases:
- Strong bases like sodium hydroxide (NaOH) completely dissociate in water to yield hydroxide ions (OH–).
- Weak bases only partially ionize in water, forming an equilibrium between the base and its ions.
The ionization of a weak base can be represented as:
A + H2OOH–+ HA+
Kb= [ OH–] [ HA+] / [ A ]
- A higher Kb value indicates a stronger base, which is a better proton acceptor.
Example:
Question: Consider a 0.500 M solution of formic acid with a pH of 2.04. Determine theKafor formic acid.
Solution. Initial [HCOOH] = 0.500 M and pH = 2.04.Unknown,Ka= ?
|
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 theKaexpression gives:
The value ofKais 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 (C2H5NH2) has a pH of 12.31.
The pOH is 14–12.31=1.69. The [OH−] is found from 10-1.69=2.04×10-2M. 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 ofKbit yields theKbfor Ethylamine.