A weak acid HA has degree of dissociation x . Which option gives the correct expression of $$(pH pK_{a})$$ ?

Find the expression for $$(pH - pK_a)$$ for a weak acid HA with degree of dissociation $$x$$. Let the initial concentration of HA be $$c$$. At equilibrium for HA $$\rightleftharpoons$$ H$$^+$$ + A$$^-$$, the concentrations are $$[HA] = c(1-x)$$, $$[H^+] = cx$$ and $$[A^-] = cx$$.

The acid dissociation constant is given by $$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{cx \cdot cx}{c(1-x)} = \frac{cx^2}{1-x}$$.

Since $$pH = -\log[H^+] = -\log(cx)$$ and $$pK_a = -\log K_a = -\log\left(\frac{cx^2}{1-x}\right)$$, their difference follows from the logarithm rule $$\log A - \log B = \log\frac{A}{B}$$.

Thus, $$pH - pK_a = -\log(cx) - \left(-\log\frac{cx^2}{1-x}\right) = -\log(cx) + \log\frac{cx^2}{1-x} = \log\frac{cx^2/(1-x)}{cx} = \log\frac{x}{1-x}$$.

The correct answer is Option D: $$\log\left(\frac{x}{1-x}\right)$$.