Question

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\).

Short answer

The final equation analogous to the Henderson-Hasselbalch equation, relating pOH and pKb for a buffered solution composed of a weak base and its conjugate acid, such as NH3 and NH4+, is: \[ \mathrm{pOH} = \mathrm{p}K_{\mathrm{b}} + \log \frac{[\mathrm{BH}^{+}]}{[\mathrm{B}]} \]

Step-by-step solution

Recall Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to describe the pH of a buffer solution composed of a weak acid and its conjugate base, and can be written as: \[ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]} \] Where pH is the pH of the solution, pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

Write the Expression for Kb

Now, let's write the expression for the equilibrium constant Kb for the dissociation of a weak base, B, and its conjugate acid, BH+: \[ \mathrm{B} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{BH}^{+} + \mathrm{OH}^{} \] The equilibrium constant, Kb, for this reaction can be written as: \[ K_{\mathrm{b}} = \frac{[\mathrm{BH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{B}]} \]

Find pOH and pKb

In order to create an equation analogous to the Henderson-Hasselbalch equation, we have to relate the pOH and pKb to this equilibrium constant. To do so, we first find pOH and pKb: pOH = -log[OH-] pKb = -log(Kb)

Substitution and Manipulation of Equations

Now, let's rearrange the equation of Kb so that it is in terms of [OH-]: \[ [\mathrm{OH}^{-}] = \frac{\mathrm{K}_{\mathrm{b}}[\mathrm{B}]}{[\mathrm{BH}^{+}]} \] Then, calculate the logarithm of the equation above: \[ -\log [\mathrm{OH}^{-}] = -\log \frac{\mathrm{K}_{\mathrm{b}}[\mathrm{B}]}{[\mathrm{BH}^{+}]} \] Since pOH = -log[OH-] and pKb = -log(Kb), we can make a substitution: \[ \mathrm{pOH} = \mathrm{p}K_{\mathrm{b}} + \log \frac{[\mathrm{BH}^{+}]}{[\mathrm{B}]} \]

Final Equation

The final equation analogous to the Henderson-Hasselbalch equation, but relating pOH and pKb for a buffered solution composed of a weak base and its conjugate acid, such as NH3 and NH4+, is: \[ \mathrm{pOH} = \mathrm{p}K_{\mathrm{b}} + \log \frac{[\mathrm{BH}^{+}]}{[\mathrm{B}]} \]