The relationship between Ka and Kb for a substance

1. Key Concept

• The strength of an acid is inversely proportional to the strength of its conjugate base.
• Weak acids produce strong conjugate bases, and strong acids produce weak conjugate bases.

2. Equilibrium Expressions

For Acids:

A weak acid $(HA)$ ionises in water:

$HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$

The equilibrium constant expression:

$K_a = \frac{[H_3O^+][A^-]}{[HA]}$

For Bases:

A weak base $(A⁻)$ ionises in water:

$A^-(aq) + H_2O(l) \rightleftharpoons HA(aq) + OH^-(aq)$

The equilibrium constant expression:

$K_b = \frac{[HA][OH^-]}{[A^-]}$

3. Relationship Between Ka and Kb

• By multiplying the equilibrium expressions for $K_a$ and $K_b$, we get: $K_a \times K_b = \frac{[H_3O^+][A^-]}{[HA]} \times \frac{[HA][OH^-]}{[A^-]}$

• Simplifies to: $K_a \times K_b = [H_3O^+][OH^-]$

• Since $[H_3O^+][OH^-] = K_w = :highlight[1.0 \times 10^{-14}]$ at 25°C, the final relationship is: $K_a \times K_b = K_w$

4. Key Takeaways

• If $K_a$ is large, $K_b$ is small, meaning strong acids have weak conjugate bases.
• If $K_a$ is small, $K_b$ is large, meaning weak acids have strong conjugate bases.
• The pH scale (1 – 14) relates to these values, with $K_a$ and $K_b$ helping determine pH and $pOH.$