Derive the Henderson-Hasselbalch equation for both acids and bases.

The acid dissociation constant, $Ka$, describes the extent of dissociation of a weak acid. $pKa$ is defined as the negative logarithm of $Ka$.

$$pKa = -\log Ka$$

For a weak acid $HA$ dissociating into $H^+$ and its conjugate base $A^-$:

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

The equilibrium constant $Ka$ is given by:

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

Taking the negative logarithm of both sides:

$$-\log Ka = -\log \left(\frac{[H^+][A^-]}{[HA]}\right)$$ $$pKa = -\log [H^+] - \log \left(\frac{[A^-]}{[HA]}\right)$$

Since $pH = -\log [H^+]$, we can substitute this into the equation:

$$pKa = pH - \log \left(\frac{[A^-]}{[HA]}\right)$$

Rearranging to solve for $pH$, which is the Henderson-Hasselbalch equation for acids:

$$pH = pKa + \log \left(\frac{[A^-]}{[HA]}\right)$$

In a buffer solution, $[A^-]$ is often represented as the concentration of the salt (conjugate base) and $[HA]$ as the concentration of the acid.

$$\boxed{pH = pKa + \log \frac{[Salt]}{[Acid]}}$$

For a weak base $BOH$ dissociating into $B^+$ and $OH^-$:

$$BOH \rightleftharpoons B^+ + OH^-$$

The base dissociation constant $Kb$ is given by:

$$Kb = \frac{[B^+][OH^-]}{[BOH]}$$

Similar to $pKa$, $pKb$ is defined as:

$$pKb = -\log Kb$$

Taking the negative logarithm of the $Kb$ expression:

$$-\log Kb = -\log \left(\frac{[B^+][OH^-]}{[BOH]}\right)$$ $$pKb = -\log [OH^-] - \log \left(\frac{[B^+]}{[BOH]}\right)$$

Since $pOH = -\log [OH^-]$, we substitute this into the equation:

$$pKb = pOH - \log \left(\frac{[B^+]}{[BOH]}\right)$$

Rearranging to solve for $pOH$, which is the Henderson-Hasselbalch equation for bases:

$$pOH = pKb + \log \left(\frac{[B^+]}{[BOH]}\right)$$

In a buffer solution, $[B^+]$ is often represented as the concentration of the salt (conjugate acid) and $[BOH]$ as the concentration of the base.

$$\boxed{pOH = pKb + \log \frac{[Salt]}{[Base]}}$$

The ion product of water, $Kw$, is a constant that describes the autoionization of water. At $25^\circ C$, its value is $1.0 \times 10^{-14}$.

$$Kw = [H^+][OH^-] = 1.0 \times 10^{-14}$$

Taking the negative logarithm of $Kw$:

$$-\log Kw = -\log ([H^+][OH^-])$$ $$pKw = -\log [H^+] - \log [OH^-]$$

Since $pH = -\log [H^+]$ and $pOH = -\log [OH^-]$:

$$pKw = pH + pOH$$

At $25^\circ C$, $pKw = -\log (1.0 \times 10^{-14}) = 14$.
Therefore, the relationship between $pH$ and $pOH$ is:

$$\boxed{pH + pOH = 14}$$