Relation Between $K_a$ and $K_b$
Conjugate Acid-Base Pairs | Ionic Equilibrium
1. Conjugate Acid-Base Pairs
According to the Brønsted-Lowry concept, an acid and a base that differ by a single proton ($H^+$) form a conjugate pair.
Example: $NH_4^+$ (Acid) $\rightleftharpoons$ $NH_3$ (Conjugate Base) + $H^+$
2. Mathematical Derivation
Let's consider a weak acid $HA$ and its conjugate base $A^-$.
Reaction 1: Ionization of Acid ($HA$)
$$ HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq) $$ $$ K_a = \frac{[H_3O^+][A^-]}{[HA]} \quad \dots(1) $$Reaction 2: Hydrolysis of Conjugate Base ($A^-$)
$$ A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^-(aq) $$ $$ K_b = \frac{[HA][OH^-]}{[A^-]} \quad \dots(2) $$Multiplying (1) and (2):
$$ K_a \times K_b = \left( \frac{[H_3O^+][A^-]}{[HA]} \right) \times \left( \frac{[HA][OH^-]}{[A^-]} \right) $$ $$ K_a \times K_b = [H_3O^+][OH^-] $$
$$ K_a \times K_b = K_w $$
Where $K_w$ is the ionic product of water ($10^{-14}$ at 25°C).
3. Logarithmic Relationship ($pK_a$ & $pK_b$)
Taking negative logarithm ($-\log_{10}$) on both sides of $K_a \times K_b = K_w$:
$$ -\log(K_a \times K_b) = -\log(K_w) $$ $$ (-\log K_a) + (-\log K_b) = -\log K_w $$
$$ pK_a + pK_b = pK_w $$
At 25°C:
$$ pK_a + pK_b = 14 $$
4. Significance & Interpretation
This relationship implies an inverse relationship between the strength of an acid and its conjugate base.
- Strong Acid (Large $K_a$): Has a very Weak Conjugate Base (Small $K_b$).
- Weak Acid (Small $K_a$): Has a Strong Conjugate Base (Large $K_b$).
5. Worked Example
Problem: The $K_a$ of Acetic Acid ($CH_3COOH$) is $1.8 \times 10^{-5}$. Calculate the $K_b$ of Acetate ion ($CH_3COO^-$).
Solution:
$$ K_b = \frac{K_w}{K_a} $$ $$ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} $$ $$ K_b = 0.55 \times 10^{-9} = 5.5 \times 10^{-10} $$Practice Quiz
Test your ability to link Acid and Base constants.
Thank you sir.
ReplyDelete