1. Write the dissociation equilibrium

For a weak acid $HA$:

$HA \rightleftharpoons H^+ + A^-$

Let the initial concentration of acid be $C$ and degree of dissociation be $x$.

Then at equilibrium:

• $[HA] = C(1-x)$
• $[H^+] = Cx$
• $[A^-] = Cx$

2. Write the acid dissociation constant

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

Substitute the equilibrium concentrations:

$K_a = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^2}{1-x}$

3. Write $[H^+]$ and pH

Since $[H^+] = Cx$,

$\mathrm{pH} = -\log[H^+] = -\log(Cx)$

Also,

$\mathrm{p}K_a = -\log K_a = -\log\left(\frac{Cx^2}{1-x}\right)$

4. Compute $(\mathrm{pH} - \mathrm{p}K_a)$

$\mathrm{pH} - \mathrm{p}K_a = -\log(Cx) + \log\left(\frac{Cx^2}{1-x}\right)$

$= \log\left(\frac{Cx^2}{(1-x)Cx}\right)$

$= \log\left(\frac{x}{1-x}\right)$

5. Match with options

Thus,

$\boxed{\mathrm{pH} - \mathrm{p}K_a = \log\left(\frac{x}{1-x}\right)}$

This corresponds to Option D.

6. Verification with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.