Investigate the structure and properties of buffer systems, including their composition, how they resist pH change, and their importance in natural systems such as blood

What this dot point is asking

NESA wants you to describe what a buffer is, explain how the weak acid / conjugate base equilibrium resists pH change, use the Henderson-Hasselbalch equation for buffer pH calculations, and apply this to natural buffer systems (especially the carbonic acid / bicarbonate buffer in blood). This is a popular extended-response topic combining Brønsted-Lowry acid-base theory with Le Chatelier reasoning.

The answer

What a buffer is

A buffer solution resists changes in pH when small amounts of acid or base are added (or when the solution is moderately diluted). A buffer contains comparable concentrations of:

• a weak acid (HA) and
• its conjugate base ($A^-$).

Or equivalently, a weak base and its conjugate acid.

The buffer equilibrium

For an acetic acid / acetate buffer:

Both species are present in substantial concentration. The equilibrium has "reserves" in both directions.

Adding $H^+$. The conjugate base neutralises it: $CH_3COO^- + H^+ \rightarrow CH_3COOH$. Equilibrium shifts left. The pH drops only slightly.

Adding $OH^-$. The weak acid donates a proton: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$. Equilibrium shifts right. The pH rises only slightly.

Both responses are applications of Le Chatelier's principle to the buffer equilibrium.

The Henderson-Hasselbalch equation

For a buffer of weak acid HA with conjugate base $A^-$:

Useful results:

• When $[A^-] = [HA]$, $\log(1) = 0$ and pH = $pK_a$.
• A buffer is most effective within ±1 pH unit of its $pK_a$.
• The buffer pH depends on the ratio of $[A^-]$ to $[HA]$, so moderate dilution does not change pH much.

Buffer capacity

A buffer has a finite capacity. Once enough acid is added to consume all the conjugate base (or enough base to consume all the weak acid), the buffer fails and the pH changes rapidly. Capacity is maximised when $[HA]$ and $[A^-]$ are equal and both large.

The blood buffer system

Blood is buffered between pH 7.35 and 7.45 by the carbonic acid / bicarbonate system:

The buffer pair is $H_2CO_3$ ($pK_a$ ≈ 6.4) and $HCO_3^-$. At blood pH 7.4, the Henderson-Hasselbalch ratio $[HCO_3^-] / [H_2CO_3]$ is about 20:1.

The system is biologically powerful because both ends are open:

• Lungs regulate $CO_2$. Hyperventilating expels $CO_2$, shifting equilibrium left and raising pH.
• Kidneys regulate $HCO_3^-$. They can secrete or retain bicarbonate to compensate over hours to days.

When the buffer fails, the body experiences acidosis (low pH) or alkalosis (high pH), with consequences for enzyme activity, oxygen transport, and ionic balance.

Worked example 2: designing a buffer

Calculate the mole ratio of $CH_3COO^-$ to $CH_3COOH$ required to prepare a buffer with pH 5.20. $K_a$ of acetic acid is $1.8 \times 10^{-5}$ ($pK_a = 4.74$).

Step 1: Apply Henderson-Hasselbalch.

Step 2: Check suitability. The target pH 5.20 is within ±1 pH unit of the pKa (4.74), so acetic acid is a suitable buffer choice. For a buffer at pH 5.20, mix acetate and acetic acid in a 2.88:1 mole ratio. For 1.00 L of buffer, one recipe is 0.10 mol $CH_3COOH$ and 0.288 mol $CH_3COONa$ dissolved in water.

Step 3: Confirm by quick mental check. Above pKa means more conjugate base than acid (ratio > 1). Below pKa means more acid than base (ratio < 1). 5.20 > 4.74, so the ratio should exceed 1. 2.88 > 1. Sensible.

Examples in context

Example 1. Blood gas analysis at Royal Prince Alfred Hospital. Arterial blood gas results from RPA's emergency department report pH, $[HCO_3^-]$ and $pCO_2$ to within four decimal places. The patient's blood pH is set by the carbonic acid / bicarbonate buffer $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$, with $pK_a$ near 6.1. Plugging values into Henderson-Hasselbalch, $pH = 6.1 + \log([HCO_3^-] / [H_2CO_3])$, gives the normal arterial ratio of 20:1, putting blood pH at 7.4. A diabetic patient with ketoacidosis arrives with $[HCO_3^-]$ depleted to 12 mmol L$^{-1}$, dropping the ratio and the pH. Clinicians treat with intravenous bicarbonate, restoring the ratio and the pH.

Example 2. NSW HSC depth study acetate buffer. A common depth-study assignment is to prepare a 0.10 mol L$^{-1}$ acetate buffer at pH 4.74 by mixing equal moles of acetic acid and sodium acetate. Students add 1 mL of 0.10 mol L$^{-1}$ HCl to 50 mL of the buffer and measure the pH change with a calibrated electrode; it falls by less than 0.05 units. They repeat with unbuffered distilled water and watch the pH crash from 7 to about 3. The Henderson-Hasselbalch equation predicts the small change quantitatively, and students must justify the result using Le Chatelier and the conjugate-base reservoir.

Try this

Q1. Identify the two components of an effective buffer and explain why a strong acid and its conjugate base would not buffer effectively. [3 marks]

• Cue. Weak acid plus conjugate base in comparable concentrations; a strong acid is fully dissociated so there is no reservoir of undissociated acid for the equilibrium to shift back to.

Q2. A buffer is prepared by mixing 0.20 mol $CH_3COOH$ ($pK_a = 4.74$) with 0.10 mol $CH_3COONa$ in 1.0 L of water. Calculate the buffer pH. [3 marks]

• Cue. Apply Henderson-Hasselbalch: $pH = 4.74 + \log(0.10 / 0.20) = 4.74 - 0.30 = 4.44$.

Q3. A buffer at pH 7.4 is prepared from $H_2PO_4^-$ / $HPO_4^{2-}$ ($pK_a = 7.21$). (a) Calculate the ratio $[HPO_4^{2-}] / [H_2PO_4^-]$. (b) Write equations to show how the buffer responds to added $H_3O^+$ and added $OH^-$. (c) State one biological role of this phosphate buffer. [2+2+1 marks]

• Cue. (a) $\log(\text{ratio}) = 7.4 - 7.21 = 0.19$, so ratio $= 1.55$. (b) $HPO_4^{2-} + H_3O^+ \rightarrow H_2PO_4^- + H_2O$ and $H_2PO_4^- + OH^- \rightarrow HPO_4^{2-} + H_2O$. (c) Maintains intracellular pH in cytoplasm.