Henderson–Hasselbalch equation Knowledge-Based Study

Study notes

1. Origin in Weak Acid Equilibrium

   A buffer solution containing a weak acid, HA, and its conjugate base, A−, establishes the following dynamic equilibrium in aqueous solution:

   $$\text{HA}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{A}^-(aq)$$

   The acid dissociation constant, $K_a$, for this system is expressed as:

   $$K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$$

   Rearranging this expression to isolate the hydronium ion concentration yields:

   $$[\text{H}_3\text{O}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]}$$

   Taking the negative logarithm (base 10) of both sides gives:

   $$-\log[\text{H}_3\text{O}^+] = -\log K_a - \log\left(\frac{[\text{HA}]}{[\text{A}^-]}\right)$$

   Using the definitions $\text{pH} = -\log[\text{H}_3\text{O}^+]$ and $\text{p}K_a = -\log K_a$, along with logarithmic properties to invert the quotient, we obtain the standard Henderson–Hasselbalch equation:

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

   The Relationship Between pH, pKa, and Component Ratio

   The Henderson–Hasselbalch equation reveals how the ratio of conjugate base to weak acid governs the solution's pH relative to the acid's $\text{p}K_a$:

   • Equimolar Conditions: When $[\text{A}^-] = [\text{HA}]$, the ratio is 1, and since $\log(1) = 0$, the equation simplifies to $\text{pH} = \text{p}K_a$.
   • Predominance of Base: When $[\text{A}^-] > [\text{HA}]$, the logarithmic term is positive, resulting in $\text{pH} > \text{p}K_a$.
   • Predominance of Acid: When $[\text{A}^-] < [\text{HA}]$, the logarithmic term is negative, resulting in $\text{pH} < \text{p}K_a$.

   Because both components exist in the same final solution volume, molar concentrations can be directly replaced by molar quantities:

   $$\text{pH} = \text{p}K_a + \log\left(\frac{n_{\text{A}^-}}{n_{\text{HA}}}\right)$$

   An effective buffer typically maintains a component ratio between 0.1 and 10, defining the effective buffering range as $\text{pH} = \text{p}K_a \pm 1$.

   Sample Calculation: Preparing a Buffer

   Consider a buffer solution prepared by mixing 0.15 mol of acetic acid ($\text{CH}_3\text{COOH}$, $K_a = 1.8 \times 10^{-5}$) and 0.20 mol of sodium acetate ($\text{CH}_3\text{COONa}$) in enough water to make 1.0 L of solution.

   Step 1: Calculate $\text{p}K_a$

   $$\text{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74$$

   Step 2: Identify Conjugate Pair Concentrations

   Because sodium acetate dissociates completely, $[\text{CH}_3\text{COO}^-] = 0.20\text{ M}$, while $[\text{CH}_3\text{COOH}] = 0.15\text{ M}$.

   Step 3: Apply the Equation

   $$\text{pH} = 4.74 + \log\left(\frac{0.20}{0.15}\right)$$

   $$\text{pH} = 4.74 + \log(1.33) \approx 4.74 + 0.12 = 4.86$$

   The calculated pH of 4.86 is slightly above the $\text{p}K_a$, consistent with the higher concentration of the conjugate base.

   Valid Operating Conditions

   The Henderson–Hasselbalch equation relies on two major chemical approximations:

   • Negligible Acid Dissociation: The equilibrium concentration of unionized acid, $[\text{HA}]$, is assumed equal to its initial analytical concentration, $[\text{HA}]_0$.
   • Negligible Base Hydrolysis: The equilibrium concentration of conjugate base, $[\text{A}^-]$, is assumed equal to its salt's initial concentration, $[\text{A}^-]_0$.

   When Does the Approximation Fail?

   These assumptions break down under specific conditions:

   1. Moderately Strong Weak Acids: If $K_a > 10^{-3}$, significant dissociation occurs, making $[\text{HA}] < [\text{HA}]_0$.
   2. Very Dilute Buffers: When component concentrations drop below approximately $10^{-3}\text{ M}$, autoionization of water and significant relative ionization alter the assumed equilibrium concentrations.
   3. Extreme Ratios: When the ratio $[\text{A}^-]/[\text{HA}]$ is beyond the 0.1 to 10 window, buffer capacity is exhausted toward one of the species, and full equilibrium calculations must be employed instead.