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 analogous equation to the Henderson-Hasselbalch equation for weak base and its conjugate acid systems, relating pOH and pKb, is given by: $$\mathrm{pOH}=\mathrm{p}{K}_b+\log \frac{[B{H}^{+}]}{[B]}$$

Step-by-step solution

Write the base dissociation equilibrium expression

Write the base equilibrium dissociation expression for a weak base B, which accepts a proton from the solvent water to yield its conjugate acid BH+ and hydroxide ion OH-: $$B+H_{2}O\leftrightarrows B{H}^{+}+O{H}^{-}$$

Write the dissociation constant expression

Write the equilibrium constant expression, Kb, for the base dissociation equilibrium: $$K_b=\frac{[B{H}^{+}][O{H}^{-}]}{[B]}$$

Rewrite equation using pOH and pKb

Convert the equation to pOH and pKb by taking the negative base-10 logarithm of both sides. This allows us to define pOH as -log10[OH-] and pKb as -log10Kb: $$\mathrm{p}{K}_b=\mathrm{pOH}-\log \frac{[B{H}^{+}]}{[B]}$$

Rearrange the equation to solve for pOH

Rearrange the equation to calculate the pOH value, which is needed to derive an equation analogous to the Henderson-Hasselbalch equation: $$\mathrm{pOH}=\mathrm{p}{K}_b+\log \frac{[B{H}^{+}]}{[B]}$$ Our final equation analogous to the Henderson-Hasselbalch equation, relating pOH, pKb, and the concentrations of the weak base and its conjugate acid, is: $$\mathrm{pOH}=\mathrm{p}{K}_b+\log \frac{[B{H}^{+}]}{[B]}$$

Key concepts

pOH

Understanding $\mathrm{pOH}$ is crucial when discussing buffer solutions involving weak bases. It is the negative logarithm of the hydroxide ion concentration in a solution. Calculated as $\mathrm{pOH}=-{\log }_{10}[{\mathrm{OH}}^{-}]$, it provides insight into the basicity of a solution. In simpler terms, a lower $\mathrm{pOH}$ indicates a more basic solution, while a higher value suggests an acidic environment. $\mathrm{pOH}$ is complementary to pH because together they define the neutrality of a solution: $\mathrm{pH}+\mathrm{pOH}=14$ in pure water at 25°C.

In a reaction involving a weak base like ammonia (${\mathrm{NH}}_3$), calculating the $\mathrm{pOH}$ helps quantify how much the base increases the hydroxide ion concentration, relative to its conjugate acid form, ${\mathrm{NH}}_4^{+}$. This paves the way for understanding the Henderson-Hasselbalch-like equation that involves weak bases.

pKb

The $\mathrm{p}{K}_b$ value of a weak base is a key factor in evaluating its strength and behavior in a buffer solution. It is defined as the negative base-10 logarithm of the base dissociation constant $K_b$, such that $\mathrm{p}{K}_b=-{\log }_{10}{K}_b$. Lower values of $\mathrm{p}{K}_b$ indicate stronger bases, as they dissociate more readily to produce hydroxide ions in solution.

The $\mathrm{p}{K}_b$ is analogous to the $\mathrm{p}{K}_a$ value for acids, and understanding this parameter helps in buffering calculations. The Henderson-Hasselbalch-like equation for bases leverages $\mathrm{p}{K}_b$ to express how the concentration of the base and its conjugate acid determine the $\mathrm{pOH}$ of the solution. A precise $\mathrm{p}{K}_b$ value is crucial to making accurate predictions about the buffering capacity and pH balance of the solution.

Buffer Solution

A buffer solution plays a vital role in maintaining the pH of a chemical environment, containing a weak base and its conjugate acid. For instance, in a mixture of ammonia (${\mathrm{NH}}_3$) and ammonium ions (${\mathrm{NH}}_4^{+}$), the buffer resists changes in pH upon addition of small amounts of acids or bases.

A buffer solution stabilizes pH by neutralizing added acids or bases through two complementary reactions. When an acid is added, the base part (such as ${\mathrm{NH}}_3$) reacts to form more of its conjugate acid ${\mathrm{NH}}_4^{+}$, whereas the conjugate acid will donate its proton when a base is added, regenerating more of the base. This balance ensures the overall pH doesn't change drastically, which is essential for many biochemical applications and industrial processes.

Weak Base Dissociation

Weak base dissociation is the process by which a weak base reacts with water to form a conjugate acid and hydroxide ions. In the equilibrium expression involving a weak base like ammonia, we write:
$${\mathrm{NH}}_3+{\mathrm{H}}_{2}O\leftrightarrows {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}$$
This reversible reaction highlights that weak bases do not fully dissociate in solution, meaning they only partially convert into hydroxide ions and their conjugate acid.

Understanding this concept is fundamental to grasping how buffer solutions work. By knowing the point of equilibrium, chemists can predict how much of the base remains in its undissociated form and how much has converted into its conjugate acid, enabling precise calculations of $\mathrm{pOH}$ and related concentrations. This equilibrium circumstance forms the basis for applying the Henderson-Hasselbalch equation to buffer solutions.

Conjugate Acid-Base Pair

In chemistry, conjugate acid-base pairs are pivotal in understanding buffer systems and equilibrium reactions. A conjugate acid-base pair differs by a hydrogen ion. When a base like ammonia (${\mathrm{NH}}_3$) gains a proton, it becomes its conjugate acid, ammonium ion ${\mathrm{NH}}_4^{+}$. Conversely, the loss of a proton from ${\mathrm{NH}}_4^{+}$ regenerates ${\mathrm{NH}}_3$. These pairs exhibit a dynamic equilibrium, crucial for buffering.

The significance of conjugate acid-base pairs emerges in the buffer equation, as they balance the concentrations of acids and bases under small pH shifts. Their relationship is vital for the derivation and understanding of the Henderson-Hasselbalch-like equation for bases, illustrating how they neutralize added substances to maintain pH stability in solutions.