Investigate the application of buffer systems in natural and industrial contexts, including the bicarbonate buffer in blood and the Henderson-Hasselbalch description of buffer pH

What this dot point is asking

NESA wants you to consolidate buffer chemistry by applying it to natural and industrial contexts, especially the bicarbonate buffer in blood, and to use the Henderson-Hasselbalch equation quantitatively. You should be able to write the buffer equilibria, explain the response to added strong acid or base in equation form, and link the chemistry to physiological situations like exercise, hyperventilation, and acidosis. This builds on conjugate acid-base pairs and the Module 5 page on buffer systems.

The answer

Buffer composition and action (recap)

A buffer is a solution containing significant amounts of both a weak acid and its conjugate base (or a weak base and its conjugate acid). The two species sit in equilibrium:

Response to added strong acid (extra $H^+$). The conjugate base consumes it:

Response to added strong base (extra $OH^-$). The weak acid consumes it:

In each case, the strong reagent is converted into the corresponding member of the conjugate pair, so $[H^+]$ moves only slightly. The buffer fails when one component is essentially exhausted.

The Henderson-Hasselbalch equation

Take logarithms of the $K_a$ expression:

This is the Henderson-Hasselbalch equation. Three quick consequences:

• When $[A^-] = [HA]$ (equimolar buffer), $pH = pK_a$. Buffer capacity is maximised here.
• The useful range of a buffer is roughly $pK_a \pm 1$ (corresponding to a 10:1 ratio of components on either side).
• To make a buffer of a target pH, choose a weak acid with $pK_a$ within one unit of the target, then set the ratio.

The bicarbonate buffer in blood

Arterial blood is maintained at pH $7.40 \pm 0.05$ by a coupled buffer-respiratory-renal system. The dominant chemical buffer is the bicarbonate pair:

For this system $pK_{a1} = 6.10$ at body temperature, with typical concentrations $[HCO_3^-] \approx 24$ mmol/L and $[H_2CO_3] \approx 1.2$ mmol/L (in equilibrium with dissolved $CO_2$).

The ratio is far from 1:1, so chemically this is a poor buffer in isolation. What makes it physiologically powerful is that both components are continuously regulated:

• $[H_2CO_3]$ is controlled by breathing rate (the lungs expel $CO_2$).
• $[HCO_3^-]$ is controlled by the kidneys (which excrete or retain bicarbonate).

This "open" buffer can therefore reset its components in response to disturbances, something a closed buffer in a beaker cannot do.

Acid-base disturbances

In each case the body shifts the bicarbonate equilibrium to restore the 20:1 ratio.

Other physiological buffers

• Phosphate buffer ($H_2PO_4^-$/$HPO_4^{2-}$): $pK_a = 7.20$. Important inside cells where bicarbonate is less effective. Also the basis for laboratory buffers (PBS).
• Protein buffers (haemoglobin in particular): histidine side chains have $pK_a$ near 6, so they buffer near physiological pH. Haemoglobin doubles as the oxygen carrier and a major intracellular buffer.

Industrial and laboratory buffers

Examples in context

Example 1. Diabetic ketoacidosis presentation at Westmead ED. A 16-year-old with type-1 diabetes presents at Westmead emergency with rapid breathing and confusion. Arterial blood gases show pH 7.10, $[HCO_3^-]$ at 8 mmol L$^{-1}$ and $pCO_2$ at 18 mmHg. Plugging into Henderson-Hasselbalch with $pK_a = 6.1$, $pH = 6.1 + \log(8 / (0.03 \times 18)) = 6.1 + 1.17 = 7.27$, just above the patient's measured value, reflecting partial respiratory compensation. The patient's body is hyperventilating to blow off $CO_2$ and raise the ratio, but renal bicarbonate reserves are exhausted by ketoacid load. Treatment with intravenous saline plus insulin restores the ratio over 8 to 12 hours.

Example 2. Aquaculture water buffering at the Port Stephens prawn farm. Black Tiger prawn larvae from NSW DPI hatcheries require pond water held at pH 8.0 to 8.3. Operators buffer ponds with a sodium carbonate / sodium bicarbonate system, the $CO_3^{2-}$ / $HCO_3^-$ pair with $pK_a$ near 10.3. The Henderson-Hasselbalch ratio at pH 8.2 is therefore $10^{8.2 - 10.3} = 10^{-2.1} = 0.0079$, so $[HCO_3^-]$ overwhelms $[CO_3^{2-}]$. Adding 50 g of $Na_2CO_3$ per 1000 L raises buffering capacity without spiking pH. Without the buffer, algal photosynthesis during the day would push pond pH above 9.0 and dissolve larval shells.

Try this

Q1. State the Henderson-Hasselbalch equation and explain what it predicts for a buffer with equal concentrations of weak acid and conjugate base. [3 marks]

• Cue. $pH = pK_a + \log([A^-] / [HA])$; when concentrations are equal the log is 0 and $pH = pK_a$.

Q2. Calculate the volume of 0.10 mol L$^{-1}$ HCl that can be added to 250 mL of a phosphate buffer (containing 0.050 mol $HPO_4^{2-}$ and 0.050 mol $H_2PO_4^-$) before the pH drops by more than 0.1 unit. [3 marks]

• Cue. Use Henderson-Hasselbalch: the ratio change $\log(0.045/0.055) = -0.087$, so adding about 0.005 mol HCl corresponds to 50 mL of 0.10 mol L$^{-1}$ HCl.

Q3. Describe the bicarbonate buffer system and its compensation mechanisms. (a) Write the buffer equilibrium. (b) Explain respiratory compensation. (c) Explain renal compensation. [1+2+2 marks]

• Cue. (a) $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$. (b) Breathing rate adjusts $CO_2$ and hence $H_2CO_3$ within minutes. (c) Kidneys reabsorb or excrete $HCO_3^-$ over hours to days.