Derive an equation for $pK_{a}$ for carboxylic acid compounds. Also,give the relation between $pK_{a}$ and acidic strength.

(N/A) $(i)$ Carboxylic acids dissociate in water to give resonance-stabilized carboxylate anions and hydronium ions.
$RCOOH + H_{2}O \rightleftharpoons RCOO^{-} + H_{3}O^{+}$
This carboxylate ion is stable due to its resonance activity.
$(ii)$ If the equilibrium constant for the aqueous acidic solution is $K_{eq}$,then:
$K_{eq} = \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH][H_{2}O]}$
$\therefore K_{eq}[H_{2}O] = \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH]}$
$\therefore K_{a} = \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH]} \quad (\because K_{eq}[H_{2}O] = K_{a})$
Where,$K_{a}$ is the dissociation constant for the acid.
Taking the negative logarithm on both sides:
$-\log K_{a} = -\log \left( \frac{[H_{3}O^{+}][RCOO^{-}]}{[RCOOH]} \right)$
$\therefore pK_{a} = -\log K_{a}$
Relation: $K_{a} \propto \text{acidic strength}$ and $pK_{a} \propto \frac{1}{\text{acidic strength}}$. For convenience,the strength of an acid is generally indicated by its $pK_{a}$ value rather than its $K_{a}$ value.