Henderson-Hasselbalch Equation: Derivation & Buffer pH Calculation

Henderson-Hasselbalch Equation: Derivation and pH Calculation This question tests the understanding of buffer solutions and their pH calculation using the Henderson-Hasselbalch equation. It requires deriving this equation from the acid dissociation equilibrium and then applying it to calculate the pH of a given buffer system. Mastery of this concept is crucial for quantitative problems involving weak acids/bases and their conjugate pairs.

Concept Overview

The Henderson-Hasselbalch equation is a fundamental tool for calculating the pH of a buffer solution. It relates the pH of the solution to the pKa of the weak acid and the ratio of the concentrations of the conjugate base and the weak acid. This equation is derived from the equilibrium expression for the dissociation of a weak acid and is invaluable for understanding buffer action and predicting how pH changes with the addition of acids or bases.

Step 1: Write the dissociation equilibrium for a weak acid. Let's consider a generic weak acid, HA, dissociating in water:

This equation represents the reversible reaction where the weak acid donates a proton to water, forming hydronium ions ($H_3O^+$) and its conjugate base ($A^-$).

Step 2: Write the acid dissociation constant ($K_a$) expression. The equilibrium constant for this reaction is the acid dissociation constant, $K_a$:

This expression quantifies the extent to which the weak acid dissociates. It is the ratio of the product of the concentrations of the conjugate base and hydronium ion to the concentration of the undissociated weak acid.

Step 3: Rearrange the $K_a$ expression to solve for $[H_3O^+]$. We want to relate pH to the concentrations, so let's isolate the hydronium ion concentration:

This step algebraically manipulates the $K_a$ expression to isolate the term that directly relates to pH.

Step 4: Take the negative logarithm of both sides of the equation. The pH is defined as the negative logarithm of the hydronium ion concentration ($pH = -\log[H_3O^+]$). Applying this to our equation:

Taking the negative logarithm is the key step that transforms the concentration-based equation into a pH-based equation.

Step 5: Apply logarithm properties to expand the right side. Using the logarithm property $\log(xy) = \log x + \log y$ and $\log(x/y) = \log x - \log y$:

This step uses fundamental logarithmic rules to break down the expression into simpler terms.

Step 6: Substitute pH and pKa definitions. We know that $pH = -\log[H_3O^+]$ and $pKa = -\log K_a$. Also, $-\log\frac{[HA]}{[A^-]} = \log\frac{[A^-]}{[HA]}$. Substituting these into the equation:

This is the Henderson-Hasselbalch equation. It elegantly connects the pH of a buffer solution to the acid's strength (pKa) and the relative amounts of the acid and its conjugate base.

Step 7: Apply the Henderson-Hasselbalch equation to calculate buffer pH. Let's consider an example: Calculate the pH of a buffer solution containing 0.1 M acetic acid ($CH_3COOH$) and 0.2 M sodium acetate ($CH_3COONa$). The $pK_a$ of acetic acid is 4.74. Here, $HA = CH_3COOH$ and $A^- = CH_3COO^-$. The concentrations are $[HA] = 0.1$ M and $[A^-] = 0.2$ M. Using the Henderson-Hasselbalch equation:

Plug in the values for $pK_a$ and the ratio of concentrations.

Step 8: Calculate the final pH.

The pH of the buffer solution is 5.04. This demonstrates how the equation is used in practice to determine the pH of a buffer system.

Key Takeaways:

• The Henderson-Hasselbalch equation is derived from the $K_a$ expression for a weak acid.
• It is used to calculate the pH of buffer solutions containing a weak acid and its conjugate base (or a weak base and its conjugate acid).
• The equation is $pH = pK_a + \log\frac{[A^-]}{[HA]}$, where $[A^-]$ is the concentration of the conjugate base and $[HA]$ is the concentration of the weak acid.
• When $[A^-] = [HA]$, $pH = pK_a$.

Answer: 5.04