Introduction
Ionic equilibrium is a fundamental concept in chemistry, especially pertinent to students preparing for the JEE Main examination. It deals with the equilibrium established between ions in an aqueous solution. This topic encompasses various subtopics, including the dissociation of weak acids and bases, the common ion effect, buffer solutions, and solubility product. Understanding these concepts is crucial for solving problems related to chemical reactions in solutions.
Dissociation of Weak Acids and Bases
Weak Acids
Weak acids do not completely dissociate in water. The equilibrium expression for the dissociation of a weak acid ( HA ) is given by:
$$ HA \rightleftharpoons H^+ + A^- $$
The equilibrium constant for this dissociation is known as the acid dissociation constant, ( K_a ):
$$ K_a = \frac{[H^+][A^-]}{[HA]} $$
The value of ( K_a ) indicates the strength of the acid; a higher ( K_a ) value means a stronger acid.
Weak Bases
Weak bases also do not completely ionize in water. The equilibrium expression for the dissociation of a weak base ( BOH ) is:
$$ BOH \rightleftharpoons B^+ + OH^- $$
The equilibrium constant for this dissociation is known as the base dissociation constant, ( K_b ):
$$ K_b = \frac{[B^+][OH^-]}{[BOH]} $$
Example: Calculation of ( K_a )
Given: A 0.1 M solution of acetic acid (( CH_3COOH )) has a ( pH ) of 2.88. Calculate the ( K_a ) of acetic acid.
Solution:
- Calculate the concentration of ( H^+ ) ions: $$ [H^+] = 10^{-pH} = 10^{-2.88} = 1.32 \times 10^{-3} \text{ M} $$
- For acetic acid dissociation: $$ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- $$ Initial concentration: ( [CH_3COOH] = 0.1 \text{ M} ) At equilibrium: $$ [CH_3COOH] = 0.1 - 1.32 \times 10^{-3} \approx 0.0987 \text{ M} $$ $$ [H^+] = [CH_3COO^-] = 1.32 \times 10^{-3} \text{ M} $$
- Calculate ( K_a ): $$ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{(1.32 \times 10^{-3})^2}{0.0987} \approx 1.77 \times 10^{-5} $$
Common Ion Effect
The common ion effect refers to the suppression of the ionization of a weak electrolyte by the presence of a common ion from a strong electrolyte. This principle is used to control the pH of solutions and is critical in buffer solutions.
Example of Common Ion Effect
Consider a solution of acetic acid (( CH_3COOH )) to which sodium acetate (( CH_3COONa )) is added. Sodium acetate dissociates completely, providing ( CH_3COO^- ) ions, which shift the equilibrium of acetic acid dissociation:
$$ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- $$
The presence of ( CH_3COO^- ) from sodium acetate suppresses the ionization of acetic acid, thereby reducing the concentration of ( H^+ ) ions.
A common mistake is to ignore the complete dissociation of the strong electrolyte in the solution. Always account for the added ions from the strong electrolyte.
Buffer Solutions
Buffer solutions resist changes in pH upon the addition of small amounts of acid or base. They are typically made from a weak acid and its conjugate base or a weak base and its conjugate acid.
Henderson-Hasselbalch Equation
For a buffer solution containing a weak acid ( HA ) and its conjugate base ( A^- ):
$$ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) $$
For a buffer solution containing a weak base ( BOH ) and its conjugate acid ( B^+ ):
$$ pOH = pK_b + \log \left( \frac{[B^+]}{[BOH]} \right) $$
The Henderson-Hasselbalch equation is extremely useful for calculating the pH of buffer solutions.
Example: Calculation of pH of a Buffer Solution
Given: A buffer solution contains 0.1 M acetic acid and 0.1 M sodium acetate. Calculate the pH of the buffer solution. ( K_a ) for acetic acid is ( 1.77 \times 10^{-5} ).
Solution:
- Calculate ( pK_a ): $$ pK_a = -\log (1.77 \times 10^{-5}) = 4.75 $$
- Use the Henderson-Hasselbalch equation: $$ pH = pK_a + \log \left( \frac{[CH_3COO^-]}{[CH_3COOH]} \right) = 4.75 + \log \left( \frac{0.1}{0.1} \right) = 4.75 $$
Solubility Product (Ksp)
Solubility product, ( K_{sp} ), is the equilibrium constant for the dissolution of a sparingly soluble salt. For a salt ( AB ) that dissociates as:
$$ AB \rightleftharpoons A^+ + B^- $$
The solubility product is given by:
$$ K_{sp} = [A^+][B^-] $$
Factors Affecting Solubility
- Common Ion Effect: The presence of a common ion decreases the solubility of the salt.
- pH: The solubility of salts containing basic anions increases with decreasing pH.
- Complex Ion Formation: The formation of complex ions can increase the solubility of certain salts.
A common mistake is to confuse ( K_{sp} ) with the solubility of the salt. ( K_{sp} ) is a constant, while solubility can vary with conditions.
Example: Calculation of Solubility Product
Given: The solubility of ( AgCl ) in water is ( 1.33 \times 10^{-5} ) M. Calculate the ( K_{sp} ) of ( AgCl ).
Solution:
- Write the dissociation equation: $$ AgCl \rightleftharpoons Ag^+ + Cl^- $$
- At equilibrium: $$ [Ag^+] = [Cl^-] = 1.33 \times 10^{-5} \text{ M} $$
- Calculate ( K_{sp} ): $$ K_{sp} = [Ag^+][Cl^-] = (1.33 \times 10^{-5})(1.33 \times 10^{-5}) = 1.77 \times 10^{-10} $$
Conclusion
Ionic equilibrium is a pivotal concept in understanding the behavior of ions in solution. Mastery of this topic requires a clear grasp of weak acid and base dissociation, the common ion effect, buffer solutions, and solubility product. By practicing problems and understanding the underlying principles, students can excel in this area and perform well in the JEE Main examination.
