Buffer systems explained: HSC Chemistry Module 5

A 5 mark answer needs the buffer composition, the equilibrium, and equations showing how the buffer responds to both added acid and added base.

Buffer composition. The buffer contains a weak acid ($CH_3COOH$) and its conjugate base ($CH_3COO^-$, from $CH_3COONa$) in comparable concentrations. The equilibrium is:

Adding a small amount of acid (extra $H^+$). The $CH_3COO^-$ acts as a base and accepts the added protons:

By Le Chatelier's principle, the equilibrium shifts left, removing most of the added $H^+$. The pH falls only slightly.

Adding a small amount of base (extra $OH^-$). The $CH_3COOH$ donates protons to neutralise the added hydroxide:

The equilibrium shifts right, replacing the consumed acid. The pH rises only slightly.

Why it works. Both the weak acid and the conjugate base are present in significant concentration. The buffer has the capacity to neutralise either an added acid or an added base by shifting the equilibrium.

Markers reward (1) explicit identification of the buffer pair, (2) the equilibrium equation, (3) separate equations for the response to acid and to base, (4) referring to Le Chatelier or the shift in equilibrium, (5) a sentence on capacity (buffering only works for small additions).

Blood is buffered at pH 7.35 to 7.45 by the carbonic acid / bicarbonate equilibrium:

The buffer pair is $H_2CO_3$ (weak acid) and $HCO_3^-$ (conjugate base). When metabolism produces excess $H^+$ (for example, from lactic acid during exercise), $HCO_3^-$ accepts the proton to form $H_2CO_3$, which dissociates into $CO_2$ and $H_2O$. The $CO_2$ is exhaled by the lungs. When $H^+$ falls, the reverse shift occurs.

Importance. Blood pH outside the 7.35 to 7.45 range causes acidosis (pH below 7.35) or alkalosis (pH above 7.45). These conditions disrupt enzyme function (enzymes are highly pH-sensitive), oxygen transport by haemoglobin, and ionic balance. Severe deviations can be fatal.

Markers reward (1) the equation showing both equilibria, (2) explanation of buffering in both directions, (3) mention of acidosis/alkalosis and a physiological consequence (enzyme function, oxygen transport).

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

Comparable concentrations are needed so the buffer has "reserves" on both sides of the equilibrium: enough $A^-$ to neutralise added acid, and enough HA to neutralise added base.

Marking criteria: 1 mark for correctly naming the weak acid / conjugate base pair, 1 mark for the reasoning about comparable concentrations giving capacity in both directions.

When $[CH_3COO^-] = [CH_3COOH]$, the ratio $[A^-]/[HA] = 1$.

So the buffer pH equals the $pK_a$ of acetic acid, pH = 4.74.

Marking criteria: 1 mark for recognising the ratio is 1, 1 mark for correctly evaluating $\log_{10}(1) = 0$, 1 mark for concluding pH = $pK_a$ = 4.74.

Step 1: moles of ethanoic acid (the weak acid, HA).

Step 2: moles of sodium ethanoate (source of the conjugate base, $A^-$).

Since both species share the same 250.0 mL volume, the ratio of moles equals the ratio of concentrations, so the volume cancels in the Henderson-Hasselbalch ratio.

Step 3: $pK_a$.

Step 4: apply Henderson-Hasselbalch.

Step 5: round to 2 decimal places (matching the precision implied by the data).

Marking criteria: 1 mark for correct moles of ethanoic acid, 1 mark for correct moles of sodium ethanoate, 1 mark for correct $pK_a$, 1 mark for correctly substituting into Henderson-Hasselbalch, 1 mark for the final pH to appropriate precision. Note the pH is below the $pK_a$ because there is more acid than conjugate base present (ratio < 1).

(a) Why the curve is flat. In the marked region, the flask contains a mixture of unreacted $CH_3COOH$ and the $CH_3COO^-$ produced by the NaOH added so far, i.e. a buffer. As more NaOH is added, $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$ consumes the acid and produces more conjugate base, but because both species remain present in comparable amounts throughout this region, the equilibrium continuously replenishes the buffering capacity and the pH rises only gradually. Only once the acid is nearly used up (near the equivalence point at 25 mL) does one component run out and the pH rise steeply.

(b) pH at half-equivalence. At the half-equivalence point (12.5 mL added), exactly half of the original $CH_3COOH$ has been converted to $CH_3COO^-$, so $n(CH_3COO^-) = n(CH_3COOH)$ and the ratio $[A^-]/[HA] = 1$.

So the pH at half-equivalence equals the $pK_a$ of acetic acid, 4.74. This is why the half-equivalence point is used experimentally to determine an unknown acid's $pK_a$ directly from a titration curve.

Marking criteria: (a) 1 mark for identifying the mixture as a buffer, 1 mark for the neutralisation equation, 1 mark for explaining that comparable concentrations of both species maintain buffering across the region. (b) 1 mark for identifying the ratio is 1 at half-equivalence, 1 mark for concluding pH = $pK_a$.

Step 1: find $pK_b$.

Step 2: convert to $pK_a$ of the conjugate acid using $pK_a + pK_b = 14$ (at 25 degrees C).

Step 3: apply Henderson-Hasselbalch with $NH_4^+$ as HA and $NH_3$ as $A^-$.

Since $[NH_3] = [NH_4^+]$, the ratio is 1 and $\log_{10}(1) = 0$.

Marking criteria: 1 mark for correct $pK_b$, 1 mark for correctly converting to $pK_a$ using the 14 relationship, 1 mark for identifying the ratio is 1, 1 mark for the final pH = 9.26.

This is a 7-mark ASSESS: markers reward a judgement backed by chemistry, not just a description of the buffer.

Band 6 PLAN.

• Thesis: the carbonic acid / bicarbonate buffer is far more effective than a single-component excretion system because it provides an immediate, reversible chemical response at the moment lactic acid is produced, backed up by two independent physiological control points (lungs and kidneys), whereas relying solely on excretion is too slow to prevent a dangerous pH swing during exercise.
• Mechanism: during vigorous exercise, anaerobic respiration produces lactic acid, releasing $H^+$. The bicarbonate buffer immediately neutralises it: $HCO_3^- + H^+ \rightarrow H_2CO_3 \rightarrow CO_{2} + H_2O$, converting a strong-acid threat into extra $CO_2$.
• Fast compensation: the increased $CO_2$ is detected by chemoreceptors, triggering faster/deeper breathing (hyperventilation), which exhales $CO_2$ and shifts the equilibrium further to the right, continuously regenerating buffering capacity within seconds.
• Slow compensation: over hours, the kidneys can adjust $HCO_3^-$ reabsorption to restore the buffer's reserve ready for the next exertion.
• Contrast: a hypothetical single-component system that relies only on excreting excess acid (e.g. via the kidneys alone, with no chemical buffer) would leave $H^+$ free in the blood for minutes to hours before the kidneys respond, during which enzyme function and haemoglobin's oxygen affinity would be significantly disrupted; the buffer's near-instant chemical response prevents this.
• Judgement: the two-tier design (chemical equilibrium acting in milliseconds, respiratory compensation acting in seconds, renal compensation acting over hours) makes the carbonic acid / bicarbonate system far more effective than any single-component approach for a tissue that cannot tolerate even brief large pH swings.

Model paragraph (excerpt). The carbonic acid / bicarbonate buffer is markedly more effective than a hypothetical single-component excretion system because it intercepts excess $H^+$ chemically, within the equilibrium $HCO_3^- + H^+ \rightleftharpoons H_2CO_3 \rightleftharpoons CO_2 + H_2O$, at the instant lactic acid is produced during vigorous exercise. This immediate buffering prevents the sharp pH drop that would otherwise impair enzyme activity and haemoglobin's oxygen-binding affinity. The lungs then reinforce the chemical response within seconds by exhaling the extra $CO_2$, continually shifting the equilibrium right and regenerating the buffer's capacity for further exertion, a feedback loop that a single-component excretory system, acting only over minutes to hours through the kidneys, cannot match.

Marker's note: top-band answers (1) give the buffer equation showing the direction of shift under exercise-induced acid load, (2) explicitly name BOTH the fast (respiratory) and slow (renal) compensation mechanisms, (3) give a concrete physiological consequence of buffer failure (enzyme function, oxygen affinity), and (4) end with an explicit comparative judgement, not a neutral summary of the buffer alone.