Buffer Solutions: The Henderson-Hasselbalch Equation and Buffer Capacity

Imagine trying to drive a car where a slight tap on the gas pedal sends you flying at 100 mph, or a tap on the brake brings you to an instant standstill. It would be uncontrollable. In chemistry, many aqueous solutions—especially in biological systems—are just as sensitive to the addition of acids or bases. A single drop of strong acid could drastically alter the pH, denaturing proteins or stopping chemical reactions.

Buffer solutions are the "shock absorbers" of chemistry. They maintain a stable pH environment, neutralizing added intruders to keep the system running smoothly. Whether you are studying for a university exam or prepping for the MCAT/DAT, mastering buffers, the Henderson-Hasselbalch equation, and buffer capacity is non-negotiable.

What is a Buffer Solution? (Composition and Function)

A buffer solution is an aqueous system that resists changes in pH when small amounts of strong acid or strong base are added.

How It Works

Think of a buffer as a reservoir of two components:

• The Acid Component ($HA$): Neutralizes added strong base ($OH^-$).
• The Base Component ($A^-$): Neutralizes added strong acid ($H^+$).

For example, consider a buffer made of Acetic Acid ($CH_3COOH$) and Sodium Acetate ($CH_3COONa$).

• If you add HCl (acid), the acetate ions react with it: $$CH_3COO^- + H^+ \rightarrow CH_3COOH$$
• If you add NaOH (base), the acetic acid reacts with it: $$CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$$

In both cases, the strong acid or base is replaced by a weaker counterpart, resulting in a negligible change in pH.

Calculating Initial Buffer pH: The Henderson-Hasselbalch Equation

To find the pH of a buffer, we use the standard calculation derived from the acid dissociation constant ($K_a$). This derivation results in the Henderson-Hasselbalch equation:

Where:

• $pK_a = -\log(K_a)$ of the weak acid.
• $[A^-]$ is the concentration of the conjugate base.
• $[HA]$ is the concentration of the weak acid.

Illustrative Example: Basic pH Calculation

Let's calculate the pH of a 1.00 L buffer solution containing 1.09 M Sodium Acetate ($CH_3COONa$) and 1.10 M Acetic Acid ($CH_3COOH$). The $K_a$ for acetic acid is $1.8 \times 10^{-5}$.

1. Determine the $pK_a$: $$pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$$
2. Identify Concentrations:
   • Base $[A^-] = 1.09 \text{ M}$
   • Acid $[HA] = 1.10 \text{ M}$
3. Apply the Henderson-Hasselbalch Equation: $$pH = 4.74 + \log \left( \frac{1.09}{1.10} \right)$$ $$pH = 4.74 + \log(0.991)$$ $$pH = 4.74 + (-0.004) = 4.736$$

Result: The pH of the buffer is approximately 4.74 (rounded).

Preparing a Buffer at a Target pH

Often in the lab, you aren't given a buffer; you have to make one. If you know the target pH, you can rearrange the Henderson-Hasselbalch equation to find the required ratio of base to acid.

Case Study: Reaching pH 5.30

Problem: You have 100.0 mL of 0.10 M acetic acid ($pK_a = 4.74$). How many millimoles of sodium acetate ($CH_3COONa$) must you add to achieve a pH of 5.30?

Step 1: Set up the equation for the target pH.

Note: Since both components are in the same volume, we can use moles directly in the log term instead of molarity.

Step 2: Solve for the ratio.

Step 3: Calculate initial moles of acid ($HA$).

Step 4: Solve for moles of Base ($A^-$).

Conclusion: You must add approximately 36.3 mmol of sodium acetate. (Specific values may vary slightly based on significant figure rounding in the $pK_a$ calculation).

Effect of Added Acid or Base (Buffer Capacity)

The true test of a buffer is how well it holds up under stress. When calculating the pH change after adding strong acid or base, you must perform the calculation in two distinct phases:

1. Stoichiometry Phase (ICF Table): The strong acid/base reacts completely with the buffer component. Calculate the new mole amounts.
2. Equilibrium Phase (H-H Equation): Use the new mole amounts to calculate the new pH.

Practice Problem: Adding Strong Acid

Scenario: Consider 1.00 L of a buffer with 1.09 M Acetate ($A^-$) and 1.10 M Acetic Acid ($HA$). We add 0.129 mol of HCl. Calculate the new pH.

1. Stoichiometry (The Reaction):
The $H^+$ from HCl reacts with the base $A^-$ to form more acid $HA$.

• Initial moles: $A^- = 1.090$ mol, $HA = 1.100$ mol.
• Change: $-0.129$ mol $A^-$, $+0.129$ mol $HA$.
• Final moles:
  • $A^- = 1.090 - 0.129 = 0.961$ mol
  • $HA = 1.100 + 0.129 = 1.229$ mol

2. Equilibrium (Henderson-Hasselbalch):
Use the new mole values ($pK_a = 4.7447$).

Observation: The pH dropped from ~4.74 to 4.64. If this were pure water, adding 0.129 mol of HCl would have crashed the pH to roughly 0.89! The buffer successfully minimized the change.

Advanced Capacity: Calculating Limits

Question: A buffer contains 0.100 M $NH_3$ and 0.135 M $NH_4Br$ (100 mL total). The $pK_b$ of $NH_3$ is 4.75. What mass of HCl can this buffer neutralize before the pH falls below 9.00?

This is a "breaking point" calculation.

1. Find $pK_a$ of $NH_4^+$: $14 - 4.75 = 9.25$.
2. Determine the Ratio at pH 9.00: $$9.00 = 9.25 + \log \left( \frac{[Base]}{[Acid]} \right)$$ $$-0.25 = \log(\text{Ratio}) \implies \text{Ratio} = 10^{-0.25} \approx 0.562$$
3. Set up Stoichiometry with variable $x$ (moles HCl added):
   • Initial $NH_3$ (Base) = $0.010$ mol.
   • Initial $NH_4^+$ (Acid) = $0.0135$ mol.
   • After adding $x$ mol HCl: Base becomes $0.010 - x$; Acid becomes $0.0135 + x$.
4. Solve for $x$: $$\frac{0.010 - x}{0.0135 + x} = 0.562$$ $$0.010 - x = 0.562(0.0135 + x)$$

   Solving this yields $x \approx 0.0015$ mol HCl.

5. Convert to Mass: $0.0015 \text{ mol} \times 36.46 \text{ g/mol} \approx 0.055 \text{ g}$.

Selecting the Best Buffer System

Not all buffers are created equal. If you need to buffer a reaction at pH 5.0, a buffer with a $pK_a$ of 9.0 is useless.

The Two Golden Rules for Selection:

1. Ideal $pK_a$: Choose a weak acid with a $pK_a$ close to your target pH. The most effective buffering occurs when $pH = pK_a$.
2. The 10:1 Limit: A buffer is generally effective only within the range $pH = pK_a \pm 1$. This corresponds to a base/acid ratio between 1:10 and 10:1.

Selection Example

Suppose you need to maintain a pH between 4.0 and 5.0. Which system is best?

• Acetate ($pK_a \approx 4.75$): Range $\approx 3.75 - 5.75$. (Excellent match)
• Phosphate ($pK_a \approx 7.20$): Range $\approx 6.20 - 8.20$. (Too basic)
• Ammonia ($pK_a \approx 9.25$): Range $\approx 8.25 - 10.25$. (Far too basic)

Verdict: The Acetate system is the correct choice because the target pH falls squarely within its buffering capacity range.