What Happens When a Buffer’s Conjugate Base Is Used Up?

A buffer contains a weak acid and its conjugate base. You add strong acid, and the conjugate base reacts with it. Add a little more, and the same thing happens. But what if you keep adding acid until there is no conjugate base left to react?

The buffer loses its ability to resist further pH changes. Once its available conjugate base has been consumed, the buffer has reached its acid-neutralising capacity. Further added acid is no longer effectively absorbed by the buffer, so the pH can fall much more rapidly.

There is one important detail, though: the conjugate base is not necessarily absent from the solution at the molecular level. The remaining weak acid can still dissociate slightly and regenerate some conjugate base. That does not restore the buffer’s original capacity.

Understanding the difference between exhausting the original conjugate base and reaching a new chemical equilibrium is the key to answering this question correctly.

01What actually happens when the conjugate base runs out?

Consider a buffer containing ethanoic acid, \(\ce{CH3COOH}\), and its conjugate base, the ethanoate ion, \(\ce{CH3COO-}\).

When hydrochloric acid is added, it supplies hydronium ions to the solution. The ethanoate ions react with those hydronium ions:

\[ \ce{CH3COO- + H3O+ -> CH3COOH + H2O} \]

The conjugate base removes the added acid by accepting a proton. Each mole of ethanoate ions can react with one mole of hydronium ions.

At first, this works well. The added hydronium ions are consumed, and the pH changes only slightly.

But the ethanoate ions are a finite resource. Every mole of added strong acid consumes approximately one mole of the available ethanoate ions.

The three stages of buffer exhaustion

The behaviour is easiest to understand by following the solution as more acid is added.

Stage 1: The buffer is working.

There are substantial amounts of both ethanoic acid and ethanoate ions. Added hydronium ions react with ethanoate ions, converting them into ethanoic acid.

The concentration ratio changes, but the pH remains relatively stable.

Stage 2: The conjugate base is running low.

More ethanoate ions have been converted into ethanoic acid. The buffer can still neutralise added acid, but it has less capacity remaining.

The pH begins changing more noticeably as the ratio of conjugate base to weak acid decreases.

Stage 3: The original conjugate base has been consumed.

The available ethanoate ions have been used up by the added acid. The solution is now predominantly a weak acid solution rather than an effective buffer.

Further hydrochloric acid adds hydronium ions that are no longer neutralised by a substantial supply of ethanoate ions. The pH can fall sharply.

02Why the equilibrium shift cannot save an exhausted buffer

The weak acid in the buffer exists in equilibrium with its conjugate base:

\[ \ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)} \]

When hydrochloric acid is added, the concentration of \(\ce{H3O+}\) increases.

According to Le Chatelier’s principle, the equilibrium shifts to the left, consuming hydronium ions and producing more weak acid.

But notice what else happens: the equilibrium also consumes \(\mathrm{A}^{-}\).

The equilibrium shift explains how the buffer responds to added acid. It does not provide an unlimited supply of conjugate base.

This is an important distinction. Le Chatelier’s principle predicts the direction of a shift, but it does not guarantee that a system can resist an unlimited disturbance.

The mole comparison that tells you when the buffer fails

For a monoprotic weak acid buffer, the neutralisation reaction has a 1:1 mole ratio:

\[ \ce{A- + H3O+ -> HA + H2O} \]

Let:

• \(n_0(\mathrm{A}^{-})\) be the initial moles of conjugate base.
• \(n(\ce{H3O+})_{\mathrm{added}}\) be the moles of hydronium ions added.

The amount of conjugate base remaining after the initial neutralisation is:

\[ n(\mathrm{A}^{-})_{\mathrm{remaining}} = n_0(\mathrm{A}^{-})-n(\ce{H3O+})_{\mathrm{added}} \]

This gives three possible situations.

The middle case is particularly important. You cannot keep using a buffer equation after one component has been consumed.

03When does the Henderson-Hasselbalch approximation fail?

For a weak acid buffer, the Henderson-Hasselbalch equation is:

\[ pH=pK_a+\log_{10} \left(\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}\right) \]

Here, \(pK_a=-\log_{10}(K_a)\), where \(K_a\) is the acid dissociation constant.

The equation connects pH to the ratio of conjugate base to weak acid.

As strong acid is added:

• The concentration of \(\mathrm{A}^{-}\) decreases.
• The concentration of \(\ce{HA}\) increases.
• The ratio \([\mathrm{A}^{-}]/[\ce{HA}]\) decreases.
• The pH decreases.

The approximation works when both buffer components are present in appreciable amounts and the equilibrium changes are small compared with their concentrations.

A useful warning: the buffer can become ineffective before it is exhausted

A common practical buffer range is:

\[ pH\approx pK_a\pm1 \]

This corresponds approximately to:

\[ 0.1\leq \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} \leq10 \]

This is a useful guideline, not an exact boundary.

A solution can still contain both components outside this range, but it has limited capacity in one direction.

For example, suppose a buffer has 0.001 mol of conjugate base and 0.020 mol of weak acid remaining. It technically contains both components.

However, its conjugate base supply is already quite small. Adding more acid will quickly make the situation worse.

There is another limitation worth understanding: the Henderson-Hasselbalch approximation can become inaccurate before the stoichiometric amount of conjugate base reaches zero. This happens when the amount of conjugate base becomes comparable to the amount produced by weak acid dissociation.

At exact stoichiometric exhaustion, the equation cannot be used at all because its logarithm would contain a zero concentration.

The actual solution still has a finite pH. You must calculate it using the weak acid equilibrium instead.

04What would an experiment show?

We can investigate buffer exhaustion by gradually adding hydrochloric acid to a propanoic acid/propanoate buffer and measuring its pH.

The experiment is useful because it connects the mole calculations to something we can actually measure.

Setting up the experiment

Consider a beaker containing 100.0 mL of a propanoate buffer.

The buffer initially contains:

• 10.0 mmol of propanoic acid, \(\ce{CH3CH2COOH}\).
• 4.00 mmol of propanoate ions, \(\ce{CH3CH2COO-}\).

A burette above the beaker contains 0.400 mol L⁻¹ hydrochloric acid.

The beaker sits on a magnetic stirrer. A pH electrode is immersed in the solution and connected to a digital pH meter.

As acid is released from the burette, it enters the stirred buffer. The pH electrode measures the resulting pH after the solution has mixed.

Before starting, the pH meter should be calibrated, and the temperature should be kept constant.

What should we predict?

The initial conjugate base supply is 4.00 mmol.

The hydrochloric acid concentration is 0.400 mol L⁻¹.

Using \(n=cV\), the volume of acid required to consume all the original conjugate base is:

\[ \begin{aligned} V&=\frac{n}{c}\\ &=\frac{0.00400}{0.400}\\ &=0.0100\text{ L}\\ &=10.0\text{ mL} \end{aligned} \]

So the predicted stoichiometric exhaustion point is 10.0 mL of hydrochloric acid.

Before this point, the buffer should resist changes in pH. Near this point, its ability to resist added acid should become much weaker.

After this point, additional strong acid remains in the solution.

What would the measurements look like?

The following values are illustrative equilibrium-model predictions, not measurements from an actual experiment.

Assume:

• Initial buffer volume: 100.0 mL.
• Initial propanoic acid: 10.0 mmol.
• Initial propanoate: 4.00 mmol.
• HCl concentration: 0.400 mol L⁻¹.
• \(K_a\) for propanoic acid: \(1.3\times10^{-5}\).
• Temperature: 25°C.

The predicted pattern is clear.

Initially, the pH decreases gradually. The propanoate ions are consuming the added hydronium ions.

As the remaining propanoate supply becomes small, the pH changes more rapidly.

At 10.0 mL, the original propanoate has been consumed stoichiometrically. The solution is predominantly propanoic acid.

Beyond 10.0 mL, additional hydrochloric acid contributes to the hydronium concentration. The weak acid equilibrium still operates, but the solution is no longer an effective buffer.

Notice that the curve does not have to show a perfectly vertical jump at 10.0 mL. Chemical equilibria change continuously. The transition between effective buffering and weak-acid or excess-strong-acid behaviour is not an instantaneous switch.

What was the limitation of the original experimental design?

Suppose the student initially adds hydrochloric acid in 1.00 mL increments.

Each addition supplies:

\[ n=0.400\times0.00100 =0.000400\text{ mol} \]

That is 0.400 mmol of acid per addition, or 10% of the original conjugate base supply.

Large additions make it difficult to identify exactly where the pH begins changing rapidly. A student could easily pass through the important region without collecting enough readings.

A better design would use 1.00 mL increments initially, then reduce to 0.10 mL increments as the predicted exhaustion point approaches.

Each smaller addition supplies only 0.040 mmol of acid.

The student should also:

• Allow the solution to mix before recording each pH.
• Keep the temperature constant.
• Use the same initial buffer composition between repeated trials.
• Use a calibrated pH electrode.
• Record the exact volume of acid added.

One final experimental limitation matters: a pH meter measures pH, not the number of propanoate ions remaining.

The titration curve provides evidence of changing buffer behaviour. The mole calculations and equilibrium model explain the chemical changes responsible for it.

05Worked examples

06What if the conjugate acid is used up instead?

The same finite-resource principle applies when strong base is added.

This time, the weak acid is the component that reacts:

\[ \ce{HA + OH- -> A- + H2O} \]

The weak acid donates protons to neutralise the hydroxide ions.

As more strong base is added:

• The amount of \(\ce{HA}\) decreases.
• The amount of \(\mathrm{A}^{-}\) increases.
• The buffer becomes less effective at resisting further additions of base.

If all the original weak acid is consumed, the solution is no longer an effective buffer.

At exact stoichiometric exhaustion, the remaining conjugate base can react with water:

\[ \ce{A- + H2O <=> HA + OH-} \]

The resulting solution is basic.

If even more strong base is added, excess hydroxide ions remain and increasingly control the pH.

The key distinction is:

• Added strong acid consumes the conjugate base.
• Added strong base consumes the weak acid.

Each buffer component protects against a different disturbance, and both have finite amounts available.

For a broader explanation of how this behaviour operates in natural systems, see Buffers in Natural Systems: HSC Chemistry Explained.

For the standard mole calculations when both buffer components remain, see How to Calculate Buffer Composition After Adding Strong Acid or Base.

07Questions and solutions

Use these values unless a question provides different information:

• \(K_a\) for propanoic acid = \(1.3\times10^{-5}\).
• \(pK_a\) for propanoic acid = 4.89.
• \(K_a\) for ammonium ions = \(5.6\times10^{-10}\).
• Temperature = 25°C.

Assume ideal dilute solutions and complete dissociation of strong electrolytes unless otherwise stated.

Question 1

A propanoate buffer contains 2.40 mmol of propanoate ions.

A student adds 0.150 mol L⁻¹ hydrochloric acid.

Calculate the volume of hydrochloric acid required to consume all the original propanoate ions.

Explain what happens at this point.

Solution 1

The required volume is 16.0 mL of hydrochloric acid. At this point, the original conjugate base has been consumed stoichiometrically, and the solution must be treated as a weak acid system.

Step 1: Identify the neutralisation reaction.

\[ \ce{CH3CH2COO- + H3O+ -> CH3CH2COOH + H2O} \]

The reaction has a 1:1 mole ratio.

Therefore, 2.40 mmol of hydronium ions are needed to consume 2.40 mmol of propanoate ions.

Step 2: Calculate the required volume.

Use:

\[ V=\frac{n}{c} \]

Substitute:

\[ \begin{aligned} V&=\frac{0.00240}{0.150}\\ &=0.0160\text{ L}\\ &=16.0\text{ mL} \end{aligned} \]

Step 3: Explain the chemical change.

At exactly 16.0 mL, all the original propanoate ions have reacted with the added hydronium ions.

The reaction has converted them into propanoic acid.

The solution can no longer operate as an effective propanoate buffer. The pH must now be determined using the weak acid equilibrium.

Importantly, this does not mean the pH becomes zero or that all propanoate ions are permanently absent. A small amount is regenerated through equilibrium.

Question 2

An ammonia/ammonium buffer is prepared by mixing:

• 50.0 mL of 0.200 mol L⁻¹ ammonia, \(\ce{NH3}\).
• 25.0 mL of 0.300 mol L⁻¹ ammonium chloride, \(\ce{NH4Cl}\).

A student adds 50.0 mL of 0.200 mol L⁻¹ hydrochloric acid.

Given that \(K_a\) for ammonium ions is \(5.6\times10^{-10}\), calculate the approximate final pH.

Assume volumes are additive and the small-\(x\) approximation is valid.

Solution 2

The final pH is approximately 5.05. All the original ammonia has been consumed, leaving an ammonium ion solution that behaves as a weak acid.

Step 1: Calculate the initial amount of ammonia.

\[ \begin{aligned} n(\ce{NH3})&=cV\\ &=0.200\times0.0500\\ &=0.0100\text{ mol} \end{aligned} \]

This is 10.0 mmol.

Step 2: Calculate the initial amount of ammonium ions.

Ammonium chloride dissociates completely:

\[ \ce{NH4Cl -> NH4+ + Cl-} \]

Therefore:

\[ \begin{aligned} n(\ce{NH4+})&=0.300\times0.0250\\ &=0.00750\text{ mol} \end{aligned} \]

This is 7.50 mmol.

Step 3: Calculate the moles of hydrochloric acid added.

\[ \begin{aligned} n(\ce{HCl})&=0.200\times0.0500\\ &=0.0100\text{ mol} \end{aligned} \]

This produces 10.0 mmol of hydronium ions.

Step 4: Write the neutralisation reaction.

\[ \ce{NH3 + H3O+ -> NH4+ + H2O} \]

The 10.0 mmol of hydronium ions consume all 10.0 mmol of ammonia.

The new ammonium amount is:

\[ \begin{aligned} n(\ce{NH4+})&=7.50+10.0\\ &=17.5\text{ mmol} \end{aligned} \]

Step 5: Calculate the final ammonium concentration.

The total volume is:

\[ V=50.0+25.0+50.0=125.0\text{ mL} \]

Therefore:

\[ [\ce{NH4+}] =\frac{0.0175}{0.1250} =0.140\text{ mol L}^{-1} \]

Step 6: Calculate the pH using the weak acid equilibrium.

The ammonium ion dissociates as follows:

\[ \ce{NH4+ + H2O <=> NH3 + H3O+} \]

Let \(x\) be the equilibrium concentration of hydronium ions.

\[ K_a=\frac{x^2}{0.140-x} \]

Using the small-\(x\) approximation:

\[ \begin{aligned} 5.6\times10^{-10} &\approx\frac{x^2}{0.140}\\ x&\approx\sqrt{5.6\times10^{-10}\times0.140}\\ &=8.85\times10^{-6}\text{ mol L}^{-1} \end{aligned} \]

Therefore:

\[ \begin{aligned} pH&=-\log_{10}(8.85\times10^{-6})\\ &\approx5.05 \end{aligned} \]

The final pH is approximately 5.05.

Notice that the original conjugate base, ammonia, has been consumed. The remaining ammonium ions still establish an equilibrium that produces a small amount of ammonia, but the solution is no longer an effective buffer.

Question 3

A student investigates the exhaustion of a propanoate buffer.

The initial solution contains:

• 10.0 mmol of propanoic acid.
• 4.00 mmol of propanoate ions.
• A total volume of 100.0 mL.

The student adds 0.400 mol L⁻¹ hydrochloric acid using a burette.

The following illustrative pH readings are obtained:

The pH reading at 10.00 mL was lost due to an electrode fault.

(a) Calculate the theoretical volume of hydrochloric acid required to consume all the original propanoate ions.

(b) Explain why the pH decreases before this volume is reached.

(c) Explain why the solution can still contain some propanoate ions after the original amount has been consumed.

(d) Evaluate the reliability of using the pH readings alone to determine the exact exhaustion point. Suggest one improvement to the experimental method.

Solution 3

The theoretical exhaustion volume is 10.00 mL. The pH readings support a loss of buffering capacity, but they cannot independently establish the exact concentration of propanoate ions.

Step 1: Calculate the required moles of hydrochloric acid.

The initial propanoate amount is 4.00 mmol.

From the 1:1 reaction:

\[ \ce{CH3CH2COO- + H3O+ -> CH3CH2COOH + H2O} \]

We need 4.00 mmol of hydronium ions.

Step 2: Calculate the theoretical volume.

\[ \begin{aligned} V&=\frac{n}{c}\\ &=\frac{0.00400}{0.400}\\ &=0.0100\text{ L}\\ &=10.00\text{ mL} \end{aligned} \]

Step 3: Explain the pH decrease before exhaustion.

The pH decreases because added hydronium ions react with propanoate ions.

This decreases the conjugate base-to-acid ratio.

As the ratio decreases, the pH moves further below the \(pK_a\) of propanoic acid.

The buffer is still functioning, but its ability to absorb additional acid is becoming weaker.

Step 4: Explain why propanoate ions can still exist after exhaustion.

After the original propanoate supply has been consumed, the remaining propanoic acid establishes its own equilibrium:

\[ \ce{CH3CH2COOH + H2O <=> H3O+ + CH3CH2COO-} \]

A small amount of propanoate is regenerated through acid dissociation.

However, this equilibrium concentration is much smaller than the original buffer supply.

Step 5: Evaluate the experimental evidence.

The pH readings show a substantial downward change around the predicted exhaustion point.

However, the pH electrode measures hydronium ion activity, not the amount of propanoate present.

The exact exhaustion volume is more reliably established by calculating the moles of hydrochloric acid required to react with the initial propanoate supply.

A useful improvement would be to add smaller increments of hydrochloric acid near 10.00 mL.

For example, 0.10 mL increments would provide much finer resolution.

The student should also calibrate the pH meter, maintain constant temperature, and allow the solution to mix fully before recording each reading.

Question 4

A propanoate buffer reaches its stoichiometric exhaustion point after the addition of hydrochloric acid.

The remaining solution contains 14.0 mmol of propanoic acid.

A student suggests adding water to restore the original buffering capacity.

Another student suggests adding sodium propanoate, \(\ce{CH3CH2COONa}\).

The target pH is 4.50, and the \(pK_a\) of propanoic acid is 4.89.

(a) Explain why adding water alone does not restore the original buffer capacity.

(b) Calculate the amount of sodium propanoate that must be added to produce a buffer with an approximate pH of 4.50.

Assume complete dissociation of the salt and negligible volume change from adding the solid.

Solution 4

Adding water alone does not restore buffer capacity. Approximately 5.70 mmol of sodium propanoate must be added to establish the required conjugate base-to-acid ratio.

Step 1: Explain the effect of dilution.

Adding water increases the solution volume.

This decreases the concentrations of the dissolved species, but it does not increase the number of moles of propanoate ions available to react with added acid.

Dilution can change the pH of a weak acid solution slightly, but it does not recreate the original conjugate base reserve.

Step 2: Identify what the added salt supplies.

Sodium propanoate dissociates:

\[ \ce{CH3CH2COONa -> CH3CH2COO- + Na+} \]

It supplies the missing conjugate base.

The solution can therefore function as a buffer again.

Step 3: Rearrange the Henderson-Hasselbalch equation.

\[ pH=pK_a+ \log_{10}\left(\frac{n(\mathrm{A}^{-})}{n(\ce{HA})}\right) \]

Rearranging:

\[ \frac{n(\mathrm{A}^{-})}{n(\ce{HA})} =10^{pH-pK_a} \]

Step 4: Substitute the target pH.

\[ \begin{aligned} \frac{n(\mathrm{A}^{-})}{n(\ce{HA})} &=10^{4.50-4.89}\\ &=10^{-0.39}\\ &=0.407 \end{aligned} \]

Step 5: Calculate the required amount of propanoate ions.

The remaining propanoic acid amount is 14.0 mmol.

Therefore:

\[ \begin{aligned} n(\mathrm{A}^{-})&=0.407\times14.0\\ &=5.70\text{ mmol} \end{aligned} \]

The student needs to add approximately 5.70 mmol of sodium propanoate.

This restores both members of the conjugate pair in appreciable amounts.

The important distinction is that dilution changes concentrations, while adding the salt supplies new conjugate base particles.

Question 5

A student prepares a propanoate buffer containing:

• 10.0 mmol of propanoic acid.
• 4.00 mmol of propanoate ions.

The initial volume is 100.0 mL.

The student adds exactly 10.0 mL of 0.400 mol L⁻¹ hydrochloric acid.

After the reaction, the student waits for the solution to reach equilibrium.

They then add a further 0.250 mL of the same hydrochloric acid.

Given:

• \(K_a=1.3\times10^{-5}\).
• Temperature = 25°C.
• Assume ideal solution behaviour.
• Neglect water autoionisation.
• Treat hydrochloric acid as fully dissociated.

Two students make different claims:

Student A: “All the propanoate was used up in the first addition. Therefore, none exists when the second addition is made, and all the extra acid must remain as hydronium ions.”

Student B: “Propanoic acid regenerates propanoate ions at equilibrium. Therefore, the buffer has effectively restored itself, and the second acid addition should have little effect on pH.”

(a) Calculate the pH after the first acid addition and subsequent equilibration.

(b) Determine the equilibrium concentration of propanoate ions at this point.

(c) Calculate the pH after the second acid addition.

(d) Evaluate both students’ claims. Explain why neither is completely correct.

Solution 5

Neither student is completely correct. The weak acid regenerates a small amount of propanoate ions at equilibrium, but this does not restore the original buffer capacity. The second acid addition still produces a noticeable pH decrease.

This question tests the difference between stoichiometric neutralisation and the equilibrium that follows it.

Step 1: Calculate the moles of hydrochloric acid in the first addition.

\[ \begin{aligned} n(\ce{HCl})&=cV\\ &=0.400\times0.0100\\ &=0.00400\text{ mol} \end{aligned} \]

This is exactly 4.00 mmol.

All the original propanoate ions are consumed.

The amount of propanoic acid becomes:

\[ 10.0+4.00=14.0\text{ mmol} \]

Step 2: Calculate the pH after the first addition.

The final volume is 110.0 mL.

The propanoic acid concentration is:

\[ \begin{aligned} C&=\frac{0.0140}{0.110}\\ &=0.1273\text{ mol L}^{-1} \end{aligned} \]

The remaining weak acid establishes the equilibrium:

\[ \ce{HA + H2O <=> H3O+ + A-} \]

Let \(x\) be the equilibrium concentration of hydronium ions.

Because the acid produces equal concentrations of hydronium and propanoate ions:

\[ K_a=\frac{x^2}{C-x} \]

Using the small-\(x\) approximation:

\[ \begin{aligned} x&\approx\sqrt{K_aC}\\ &=\sqrt{1.3\times10^{-5}\times0.1273}\\ &=1.28\times10^{-3}\text{ mol L}^{-1} \end{aligned} \]

Therefore:

\[ \begin{aligned} pH&=-\log_{10}(1.28\times10^{-3})\\ &\approx2.89 \end{aligned} \]

The pH after the first addition is approximately 2.89.

Step 3: Determine how much propanoate has been regenerated.

At equilibrium:

\[ [\mathrm{A}^{-}]=[\ce{H3O+}] \]

Therefore:

\[ [\mathrm{A}^{-}]\approx1.28\times10^{-3} \text{ mol L}^{-1} \]

The number of moles regenerated is:

\[ \begin{aligned} n(\mathrm{A}^{-})&=cV\\ &=1.28\times10^{-3}\times0.110\\ &=1.41\times10^{-4}\text{ mol} \end{aligned} \]

This is approximately 0.141 mmol.

So Student A is incorrect: the equilibrium concentration of propanoate is not zero.

However, compare the regenerated amount with the original 4.00 mmol.

Only a small fraction of the original conjugate base supply has been regenerated.

Step 4: Calculate the additional hydrochloric acid.

The second addition is 0.250 mL of 0.400 mol L⁻¹ hydrochloric acid.

\[ \begin{aligned} n(\ce{HCl})&=0.400\times0.000250\\ &=0.000100\text{ mol}\\ &=0.100\text{ mmol} \end{aligned} \]

The new total volume is:

\[ V=110.0+0.250=110.25\text{ mL} \]

Step 5: Recognise that the weak acid equilibrium must still be considered.

This is the important part.

We now have:

• Total propanoic acid species: 14.0 mmol.
• Excess chloride ions relative to sodium ions: 0.100 mmol.
• Final volume: 110.25 mL.

Define:

\[ C_T=\frac{0.0140}{0.11025} =0.12698\text{ mol L}^{-1} \]

Here, \(C_T\) is the total concentration of propanoic acid and propanoate species.

The excess strong acid concentration is:

\[ C_s=\frac{0.000100}{0.11025} =9.07\times10^{-4}\text{ mol L}^{-1} \]

Let \(h=[\ce{H3O+}]\).

Using charge balance, with water autoionisation neglected:

\[ [\mathrm{A}^{-}]=h-C_s \]

The total propanoic acid concentration gives:

\[ [\ce{HA}]=C_T-h+C_s \]

Substitute these expressions into the acid dissociation equation:

\[ K_a= \frac{h(h-C_s)} {C_T-h+C_s} \]

Rearranging:

\[ h^2+(K_a-C_s)h -K_a(C_T+C_s)=0 \]

Step 6: Solve for the positive hydronium concentration.

Substituting the values:

\[ \begin{aligned} h^2 &-8.94\times10^{-4}h\\ &-1.66\times10^{-6}=0 \end{aligned} \]

Using the quadratic formula and taking the positive root:

\[ h\approx1.81\times10^{-3} \text{ mol L}^{-1} \]

Step 7: Calculate the new pH.

\[ \begin{aligned} pH&=-\log_{10}(1.81\times10^{-3})\\ &\approx2.74 \end{aligned} \]

The final pH is approximately 2.74.

Step 8: Evaluate the two claims.

Student A is wrong because propanoate ions are regenerated by the weak acid equilibrium.

Student B is also wrong because the regenerated amount is small. It does not recreate the original buffer reserve.

The pH has decreased from approximately 2.89 to 2.74 after the additional acid was introduced.

There is another useful check here.

If we calculated the pH using only the excess hydrochloric acid concentration, we would obtain approximately:

\[ pH=-\log_{10}(9.07\times10^{-4}) \approx3.04 \]

This is not accurate because the remaining propanoic acid also contributes hydronium ions.

The exact equilibrium calculation gives the lower pH of approximately 2.74.

The deeper lesson is that stoichiometric exhaustion does not mean a chemical species can never reappear. Equilibrium continues to operate, but a small regenerated amount is not equivalent to the substantial reserve that made the original solution an effective buffer.

When a buffer reaches its limit, the next step is to identify what controls the pH instead: the equilibrium of the remaining weak acid, the equilibrium of the remaining conjugate base, or excess strong acid or base. That decision determines which calculation you should use.