How to Calculate Buffer Composition After Adding Strong Acid or Base

Two beakers contain acetate buffers, and both have an initial pH of 4.74. One contains 20.0 mmol of acetic acid and 20.0 mmol of acetate ions. The other contains just 2.00 mmol of each. If you add 1.00 mmol of hydrochloric acid to both, will their pH values change by the same amount?

No. The smaller buffer loses half its acetate ions, while the larger buffer loses only 5% of its supply. The smaller buffer experiences a much greater pH change, even though both started at exactly the same pH.

This is the key to calculating buffer composition after adding a strong acid or base: calculate the new mole amounts first, then calculate the pH from the updated composition. You cannot simply use the original buffer concentrations.

01The calculation starts with a chemical reaction

A buffer contains appreciable amounts of a weak acid and its conjugate base.

For a general weak acid, HA, the equilibrium in water is:

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

The two important components are:

• HA: the weak acid, which can donate a hydrogen ion.
• A⁻: its conjugate base, which can accept a hydrogen ion.

Think of these as two chemical reserves. The weak acid can react with added hydroxide ions, while the conjugate base can react with added hydronium ions.

Neither reserve is unlimited.

What happens when you add a strong acid?

Imagine adding hydrochloric acid to an acetic acid/acetate buffer.

Hydrochloric acid ionises essentially completely in water:

\[ \ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl^-(aq)} \]

The added hydronium ions react with the acetate ions:

\[ \ce{CH3COO^-(aq) + H3O+(aq) -> CH3COOH(aq) + H2O(l)} \]

Notice what happens to the buffer composition:

• Acetate ions are consumed.
• Acetic acid is produced.
• The amount of weak acid increases.
• The amount of conjugate base decreases.

The added strong acid is converted into the weak acid component of the buffer.

This also explains the equilibrium shift. Adding hydronium ions increases the concentration of a product in the weak acid equilibrium. According to Le Chatelier’s principle, the equilibrium shifts to the left, favouring the formation of HA.

What happens when you add a strong base?

Now imagine adding sodium hydroxide to the same buffer.

Sodium hydroxide dissociates completely:

\[ \ce{NaOH(aq) -> Na+(aq) + OH^-(aq)} \]

The hydroxide ions react with the weak acid:

\[ \ce{OH^-(aq) + CH3COOH(aq) -> CH3COO^-(aq) + H2O(l)} \]

The changes are now reversed:

• Acetic acid is consumed.
• Acetate ions are produced.
• The amount of weak acid decreases.
• The amount of conjugate base increases.

The added hydroxide ions remove hydronium ions from the equilibrium system through neutralisation. The weak acid equilibrium shifts to the right, producing more hydronium ions and acetate ions.

The buffer therefore resists the pH increase.

The important distinction is that the strong acid or base reacts first. You must account for this neutralisation before considering the weak acid equilibrium.

02The four-step method for calculating buffer composition

Use this sequence whenever a question asks you to calculate the composition or pH of a buffer after adding a strong acid or strong base.

Step 1: Calculate the moles of strong acid or base added

Use:

\[ n=cV \]

where:

• \(n\) is the amount in moles (mol).
• \(c\) is the molar concentration in mol L⁻¹.
• \(V\) is the volume in litres (L).

Convert millilitres to litres before substituting.

Step 2: Write the neutralisation reaction

Identify which buffer component reacts with the added substance.

For a monoprotic acid buffer, the reactions have a 1:1 mole ratio.

This means one mole of added hydronium ions consumes one mole of conjugate base, and one mole of added hydroxide ions consumes one mole of weak acid.

Step 3: Calculate the new mole amounts

Set up a simple table showing the initial amounts, changes, and final amounts.

For example, when adding a strong acid:

For a strong base, reverse the changes.

Check that neither component has a negative final amount. If one is completely consumed, you no longer have an ordinary buffer calculation.

Step 4: Calculate the new pH

Once you have the updated composition, use the acid dissociation equilibrium.

The acid dissociation constant is:

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

Rearranging and taking logarithms gives the Henderson-Hasselbalch equation:

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

Here, \(pK_a=-\log_{10}(K_a)\).

Because both buffer components are in the same final solution, they share the same volume. Their concentration ratio is therefore equal to their mole ratio:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} = \frac{n_{\mathrm{A}^{-}}}{n_{\ce{HA}}} \]

This gives the particularly useful form:

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

You can use this directly with your updated mole amounts.

The Henderson-Hasselbalch equation is derived from the \(K_a\) expression. Its usual concentration-based application assumes a sufficiently dilute solution and that equilibrium ionisation is small compared with the amounts of both buffer components. OpenStax explains the derivation and assumptions.

05When does the buffer approximation stop working?

This is one of the most important parts of buffer calculations.

A buffer can only neutralise as much strong acid or base as its components allow. Once the relevant component is exhausted, you must change your calculation method.

Case 1: Less strong acid than conjugate base

Suppose the buffer initially contains:

• 0.0100 mol HA.
• 0.00600 mol A⁻.

You add 0.00500 mol of strong acid.

The reaction consumes 0.00500 mol of A⁻:

\[ \begin{aligned} n(\mathrm{A}^{-})&=0.00600-0.00500\\ &=0.00100\text{ mol} \end{aligned} \]

The weak acid increases:

\[ \begin{aligned} n(\ce{HA})&=0.0100+0.00500\\ &=0.0150\text{ mol} \end{aligned} \]

Both components remain, so the Henderson-Hasselbalch equation can still be used.

Case 2: Exactly enough strong acid to consume all conjugate base

Now suppose you add 0.00600 mol of strong acid.

All 0.00600 mol of A⁻ is consumed.

The final composition is:

• 0 mol A⁻.
• 0.0160 mol HA.

You no longer have a buffer containing both members of the conjugate pair.

You cannot use the Henderson-Hasselbalch equation because the conjugate base concentration is zero.

Instead, you must calculate the pH of the remaining weak acid using its equilibrium expression.

For example, if the final solution volume is 100.0 mL:

\[ [\ce{HA}]=\frac{0.0160}{0.100}=0.160\text{ mol L}^{-1} \]

The equilibrium is now:

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

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

\[ K_a=\frac{x^2}{0.160-x} \]

Using the small-\(x\) approximation:

\[ \begin{aligned} 1.8\times10^{-5}&\approx\frac{x^2}{0.160}\\ x&\approx\sqrt{1.8\times10^{-5}\times0.160}\\ &=1.70\times10^{-3}\text{ mol L}^{-1} \end{aligned} \]

Therefore:

\[ pH=-\log_{10}(1.70\times10^{-3}) \]

\[ \boxed{pH\approx2.77} \]

You have switched from a buffer calculation to a weak acid equilibrium calculation.

Case 3: More strong acid than conjugate base

Suppose you add 0.00800 mol of strong acid to the original buffer.

Only 0.00600 mol can react with A⁻.

That leaves:

\[ n(\ce{H3O+})_{\mathrm{excess}} =0.00800-0.00600 \]

\[ =0.00200\text{ mol} \]

If the final volume is 100.0 mL:

\[ [\ce{H3O+}]=\frac{0.00200}{0.100} \]

\[ =0.0200\text{ mol L}^{-1} \]

The excess strong acid now controls the pH:

\[ pH=-\log_{10}(0.0200) \]

\[ \boxed{pH\approx1.70} \]

The remaining weak acid contributes very little additional hydronium because the solution already contains a substantial concentration of strong acid.

The important point is that you must not try to force a buffer calculation when the conjugate base has been completely consumed.

What if you add too much strong base?

The same reasoning applies in reverse.

For a strong base, compare the moles of hydroxide ions added with the initial moles of weak acid.

At 25°C:

\[ pH+pOH=14.00 \]

If the weak acid is completely consumed but no hydroxide remains in excess, the remaining conjugate base undergoes hydrolysis. You must use its base dissociation constant:

\[ K_b=\frac{K_w}{K_a} \]

If hydroxide remains in excess, calculate its concentration using the excess moles and final volume, then calculate pOH and pH.

A useful check: Is the buffer still effective?

There is a difference between having both buffer components present and having an effective buffer.

A common practical range is:

\[ pH\approx pK_a\pm1 \]

This corresponds approximately to a conjugate base-to-acid ratio between 0.1 and 10.

Outside this range, the solution may still contain both components, but it has limited ability to resist further addition in one direction.

This is not an exact boundary at which the Henderson-Hasselbalch equation suddenly becomes invalid. Its accuracy still depends on the equilibrium assumptions. However, it is a useful warning that one component is becoming relatively scarce.

Also remember that two buffers with the same pH can have very different capacities. Their mole ratios may be identical, but the buffer containing larger absolute amounts of both components can neutralise more added acid or base.

06How would you test these predictions experimentally?

The calculations describe what should happen to the composition of a buffer. A pH meter allows you to test the predicted change in pH.

An appropriate experimental arrangement would contain:

• A beaker containing a measured volume of buffer.
• A magnetic stirrer to mix the solution.
• A pH electrode immersed in the buffer and connected to a digital meter.
• A burette containing a known concentration of strong acid or base.

Record the initial pH. Add a measured volume of strong acid, allow the solution to mix, and record the new pH.

Repeat with a buffer containing a different total amount of the same conjugate pair, keeping the initial pH, solution volume, temperature, and added amount of strong acid constant.

The important experimental control is the amount of acid added. If the two buffers receive different quantities, the comparison becomes less useful.

The expected result is that the more concentrated buffer experiences a smaller pH change. As its conjugate base is gradually consumed, the solution eventually loses its ability to resist further acid addition.

There is one limitation to this experiment: a pH meter measures the pH of the solution, not the individual amounts of acetic acid and acetate ions.

The mole calculations provide the predicted composition. The pH measurement tests whether the predicted equilibrium behaviour is consistent with the actual solution.

This distinction between calculated composition and measured pH is important when evaluating experimental results.

07Questions and solutions

The following questions progress from standard buffer calculations to situations where you must decide whether the buffer model is still appropriate.

Use the following data unless otherwise stated:

• \(K_a\) for acetic acid = \(1.8\times10^{-5}\).
• \(pK_a\) for acetic acid = 4.74.
• \(K_a\) for ammonium ions = \(5.6\times10^{-10}\).
• \(pK_a\) for ammonium ions = 9.25.
• Temperature = 25°C.

Question 1

A buffer contains 0.0200 mol of acetic acid and 0.0150 mol of acetate ions.

A student adds 10.0 mL of 0.100 mol L⁻¹ hydrochloric acid.

Calculate:

(a) The new mole amounts of acetic acid and acetate ions.

(b) The final pH of the buffer.

Solution 1

The final composition is 0.0210 mol of acetic acid and 0.0140 mol of acetate ions, giving a pH of approximately 4.57.

Step 1: Calculate the moles of hydrochloric acid added.

\[ \begin{aligned} n&=cV\\ &=0.100\times0.0100\\ &=0.00100\text{ mol} \end{aligned} \]

Step 2: Write the neutralisation reaction.

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

The added acid consumes acetate ions and produces acetic acid.

Step 3: Calculate the new mole amounts.

Acetate ions:

\[ 0.0150-0.00100=0.0140\text{ mol} \]

Acetic acid:

\[ 0.0200+0.00100=0.0210\text{ mol} \]

Both components remain, so the buffer equation is still valid.

Step 4: Calculate the final pH.

\[ \begin{aligned} pH&=4.74+\log_{10} \left(\frac{0.0140}{0.0210}\right)\\ &=4.74+\log_{10}(0.6667)\\ &=4.57 \end{aligned} \]

The pH decreases because the conjugate base-to-acid ratio decreases after adding hydrochloric acid.

Question 2

A student prepares an ammonia/ammonium buffer by mixing:

• 30.0 mL of 0.300 mol L⁻¹ ammonia.
• 50.0 mL of 0.200 mol L⁻¹ ammonium chloride.

The student then adds 20.0 mL of 0.150 mol L⁻¹ sodium hydroxide.

Calculate the final pH and the amount of ammonium ions remaining.

Solution 2

The final buffer contains 0.00700 mol of ammonium ions and 0.0120 mol of ammonia. Its final pH is approximately 9.49.

Step 1: Calculate the initial ammonia amount.

\[ \begin{aligned} n(\ce{NH3})&=0.300\times0.0300\\ &=0.00900\text{ mol} \end{aligned} \]

Step 2: Calculate the initial ammonium amount.

\[ \begin{aligned} n(\ce{NH4+})&=0.200\times0.0500\\ &=0.0100\text{ mol} \end{aligned} \]

Step 3: Calculate the moles of hydroxide ions added.

\[ \begin{aligned} n(\mathrm{OH}^{-})&=0.150\times0.0200\\ &=0.00300\text{ mol} \end{aligned} \]

Step 4: Write the neutralisation reaction.

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

Hydroxide ions consume ammonium ions and produce ammonia.

Step 5: Calculate the final composition.

Ammonium ions:

\[ 0.0100-0.00300=0.00700\text{ mol} \]

Ammonia:

\[ 0.00900+0.00300=0.0120\text{ mol} \]

Step 6: Calculate the final pH.

\[ \begin{aligned} pH&=9.25+\log_{10} \left(\frac{0.0120}{0.00700}\right)\\ &=9.25+\log_{10}(1.7143)\\ &=9.49 \end{aligned} \]

The amount of ammonium ions remaining is 0.00700 mol.

The pH increases because hydroxide ions convert the weak acid component into its conjugate base.

Question 3

Two acetate buffers are prepared at 25°C.

Their initial compositions are:

Both buffers receive 15.0 mL of 0.100 mol L⁻¹ hydrochloric acid.

(a) Calculate the final pH of each buffer.

(b) Determine which buffer experiences the greater pH change.

(c) Explain why two buffers with identical initial pH values can have different buffering capacities.

Solution 3

Buffer P finishes at approximately pH 4.66, while Buffer Q finishes at approximately pH 4.27. Buffer Q experiences the greater pH change because it contains smaller absolute amounts of both buffer components.

Step 1: Calculate the moles of hydrochloric acid added.

\[ \begin{aligned} n(\ce{HCl})&=0.100\times0.0150\\ &=0.00150\text{ mol} \end{aligned} \]

This is 0.00150 mol of hydronium ions.

Step 2: Calculate the composition of Buffer P.

The reaction consumes 0.00150 mol of acetate ions.

Acetic acid:

\[ 0.0150+0.00150=0.0165\text{ mol} \]

Acetate ions:

\[ 0.0150-0.00150=0.0135\text{ mol} \]

Both components remain.

Step 3: Calculate the pH of Buffer P.

\[ \begin{aligned} pH_P&=4.74+\log_{10} \left(\frac{0.0135}{0.0165}\right)\\ &=4.74+\log_{10}(0.8182)\\ &=4.66 \end{aligned} \]

Step 4: Calculate the composition of Buffer Q.

Acetic acid:

\[ 0.00300+0.00150=0.00450\text{ mol} \]

Acetate ions:

\[ 0.00300-0.00150=0.00150\text{ mol} \]

Both components remain, but half of the original acetate ions have been consumed.

Step 5: Calculate the pH of Buffer Q.

\[ \begin{aligned} pH_Q&=4.74+\log_{10} \left(\frac{0.00150}{0.00450}\right)\\ &=4.74+\log_{10}(0.3333)\\ &=4.27 \end{aligned} \]

Step 6: Compare the pH changes.

Both buffers initially have equal amounts of acetic acid and acetate ions.

Therefore:

\[ pH_{\mathrm{initial}}=pK_a=4.74 \]

The pH changes are approximately:

• Buffer P: \(4.74-4.66=0.08\).
• Buffer Q: \(4.74-4.27=0.47\).

Buffer Q experiences a much greater change.

The tempting assumption is that identical initial pH values mean identical buffering abilities. However, pH depends on the ratio of conjugate base to weak acid, whereas buffering capacity also depends on their absolute amounts.

Buffer P has a much larger reserve of acetate ions and can neutralise more added acid before its composition changes substantially.

Question 4

An acetate buffer initially contains:

• 0.0100 mol of acetic acid.
• 0.00600 mol of acetate ions.

After mixing with hydrochloric acid, the final solution volume is 100.0 mL.

Three separate trials are performed. The amounts of hydrochloric acid added are:

• Trial A: 0.00500 mol.
• Trial B: 0.00600 mol.
• Trial C: 0.00800 mol.

For each trial:

(a) Determine the amounts of acetic acid and acetate ions remaining after neutralisation.

(b) Identify whether the final solution should be treated as a buffer, a weak acid solution, or a solution containing excess strong acid.

(c) Calculate the approximate final pH.

Explain why the same calculation method cannot be used for all three trials.

Solution 4

The approximate final pH values are:

• Trial A: 3.57.
• Trial B: 2.77.
• Trial C: 1.70.

The calculation method changes because the amount of hydrochloric acid eventually equals or exceeds the available acetate ions.

Step 1: Identify the neutralisation reaction.

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

The initial amount of acetate ions is 0.00600 mol.

This means the buffer can consume a maximum of 0.00600 mol of added hydronium ions before all its acetate is used up.

Step 2: Determine the composition after neutralisation.

Now consider each trial separately.

Trial A: The buffer remains.

Step 3: Calculate the pH using the updated buffer composition.

\[ \begin{aligned} pH&=4.74+\log_{10} \left(\frac{0.00100}{0.0150}\right)\\ &=4.74+\log_{10}(0.06667)\\ &=3.57 \end{aligned} \]

Both buffer components remain, so the Henderson-Hasselbalch equation can still estimate the pH.

However, the conjugate base-to-acid ratio is now below 0.1. The buffer has little remaining capacity to neutralise additional acid.

Trial B: All conjugate base is consumed.

Step 4: Recognise that the buffer equation is no longer appropriate.

All 0.00600 mol of acetate ions react.

The remaining acetic acid amount is:

\[ 0.0100+0.00600=0.0160\text{ mol} \]

Its concentration is:

\[ [\ce{CH3COOH}]=\frac{0.0160}{0.100} =0.160\text{ mol L}^{-1} \]

We now have a weak acid equilibrium, not a buffer equilibrium.

Step 5: Calculate the hydronium ion concentration.

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

\[ K_a=\frac{x^2}{0.160-x} \]

Using the small-\(x\) approximation:

\[ \begin{aligned} x&\approx\sqrt{1.8\times10^{-5}\times0.160}\\ &=1.70\times10^{-3}\text{ mol L}^{-1} \end{aligned} \]

Step 6: Calculate the pH.

\[ \begin{aligned} pH&=-\log_{10}(1.70\times10^{-3})\\ &=2.77 \end{aligned} \]

The final solution contains weak acid only, so the weak acid equilibrium must be used.

Trial C: Strong acid remains in excess.

Step 7: Calculate the excess hydronium ions.

The original acetate ions consume 0.00600 mol of the added acid.

Therefore:

\[ \begin{aligned} n(\ce{H3O+})_{\mathrm{excess}} &=0.00800-0.00600\\ &=0.00200\text{ mol} \end{aligned} \]

Step 8: Calculate the excess hydronium concentration.

\[ \begin{aligned} [\ce{H3O+}]&=\frac{0.00200}{0.100}\\ &=0.0200\text{ mol L}^{-1} \end{aligned} \]

Step 9: Calculate the pH.

\[ \begin{aligned} pH&=-\log_{10}(0.0200)\\ &=1.70 \end{aligned} \]

The excess strong acid dominates the pH. The contribution from acetic acid dissociation is negligible under these conditions.

The key lesson is that the buffer calculation depends on whether both buffer components survive the initial neutralisation.

A negative mole amount for acetate or an attempt to use the logarithm of zero is a warning that you have chosen the wrong calculation method.

Question 5

Two acetate buffers, P and Q, are prepared at 25°C.

Their initial compositions are:

Both buffers initially have the same pH.

Each buffer undergoes the following sequence:

1. Add 8.00 mmol of hydrochloric acid.
2. After mixing, add 8.00 mmol of sodium hydroxide.

Assume both reagents are strong electrolytes that dissociate completely, the acid and base are monoprotic, and no other reactions occur.

(a) Calculate the initial pH of both buffers.

(b) Determine the composition of each buffer immediately after the hydrochloric acid addition.

(c) Identify which buffer still functions as a buffer at this intermediate stage.

(d) Calculate the final composition of both buffers after the sodium hydroxide addition.

(e) Determine the final pH of each buffer.

(f) Evaluate the statement:

“Because both buffers finish at their original pH, they must have resisted the addition of hydrochloric acid equally well.”

Solution 5

Both buffers finish with their original compositions and a pH of 4.74. However, Buffer P remains a buffer throughout the first addition, while Buffer Q completely loses its buffering ability temporarily because its acetate ions are exhausted.

This is an important distinction between the final state of a system and what happens while the system changes.

Step 1: Calculate the initial pH values.

Both buffers have equal amounts of acetic acid and acetate ions.

Therefore:

\[ \frac{n_{\mathrm{A}^{-}}}{n_{\ce{HA}}}=1 \]

Using the Henderson-Hasselbalch equation:

\[ \begin{aligned} pH&=4.74+\log_{10}(1)\\ &=4.74 \end{aligned} \]

Both buffers initially have pH 4.74.

Step 2: Calculate the composition of Buffer P after adding hydrochloric acid.

Buffer P initially contains:

• 20.0 mmol HA.
• 20.0 mmol A⁻.

The 8.00 mmol of added hydrochloric acid reacts with 8.00 mmol of acetate ions.

The new amounts are:

\[ n(\mathrm{A}^{-})=20.0-8.00=12.0\text{ mmol} \]

\[ n(\ce{HA})=20.0+8.00=28.0\text{ mmol} \]

Both components remain, so Buffer P is still a buffer.

Its intermediate pH is:

\[ \begin{aligned} pH&=4.74+\log_{10} \left(\frac{12.0}{28.0}\right)\\ &=4.37 \end{aligned} \]

Step 3: Calculate the composition of Buffer Q after adding hydrochloric acid.

Buffer Q initially contains:

• 6.00 mmol HA.
• 6.00 mmol A⁻.

The added 8.00 mmol of hydrochloric acid can only react with the 6.00 mmol of acetate ions available.

Therefore:

\[ n(\mathrm{A}^{-})=0 \]

The amount of acetic acid becomes:

\[ 6.00+6.00=12.0\text{ mmol} \]

There is also excess hydronium:

\[ 8.00-6.00=2.00\text{ mmol} \]

Buffer Q is no longer a buffer at this point.

It contains weak acid and excess strong acid, with no acetate ions remaining.

Step 4: Consider what happens when sodium hydroxide is added to Buffer P.

The 8.00 mmol of hydroxide ions react with acetic acid:

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

Acetic acid decreases:

\[ 28.0-8.00=20.0\text{ mmol} \]

Acetate ions increase:

\[ 12.0+8.00=20.0\text{ mmol} \]

Buffer P has returned to its original composition.

Step 5: Consider what happens when sodium hydroxide is added to Buffer Q.

This is the more interesting part.

Buffer Q contains 2.00 mmol of excess hydronium ions after the first addition.

The 8.00 mmol of hydroxide ions must react with this excess acid first:

\[ \ce{H3O+ + OH^- -> 2H2O} \]

The first 2.00 mmol of hydroxide ions neutralise the excess hydronium ions.

This leaves:

\[ 8.00-2.00=6.00\text{ mmol OH}^- \]

These remaining hydroxide ions now react with the 12.0 mmol of acetic acid:

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

The final acetic acid amount is:

\[ 12.0-6.00=6.00\text{ mmol} \]

The final acetate amount is:

\[ 0+6.00=6.00\text{ mmol} \]

Buffer Q has also returned to its original composition.

Step 6: Calculate the final pH values.

Both buffers now have equal amounts of acetic acid and acetate ions.

Therefore:

\[ \begin{aligned} pH&=4.74+\log_{10} \left(\frac{6.00}{6.00}\right)\\ &=4.74 \end{aligned} \]

The same result applies to Buffer P.

Step 7: Evaluate the statement.

The statement is incorrect.

Both buffers finish at their original pH, but their behaviour during the additions was very different.

Buffer P had sufficient acetate ions to consume the hydrochloric acid without exhausting either buffer component.

Buffer Q did not. The hydrochloric acid exceeded the available acetate ions, leaving excess hydronium ions and temporarily destroying the buffer system.

When sodium hydroxide was subsequently added, it first neutralised the excess hydronium ions. The remaining hydroxide ions then reacted with the acetic acid, regenerating the acetate ions.

The original buffer composition was restored.

There is a tempting shortcut here: because equal amounts of strong acid and strong base were added, you might simply cancel them and conclude that nothing happened.

That shortcut gives the correct final composition, but it hides the crucial intermediate state. It cannot tell you whether the buffer maintained its pH throughout the process.

This is why the order of additions matters when explaining the behaviour of a system, even when the final mole balance is unchanged.

08What to learn next

The central decision in every buffer calculation is whether both members of the conjugate pair remain after neutralisation.

• If both remain, calculate the new mole ratio and use the buffer equilibrium.
• If one component is exactly exhausted, switch to a weak acid or weak base equilibrium calculation.
• If strong acid or base remains in excess, calculate the pH from the excess reagent, accounting for any significant equilibrium contributions.

This connects directly to acid-base titrations, where the dominant chemical species change as more reagent is added.

For the broader pH changes during neutralisation, continue with Weak Acid-Strong Base Titrations: HSC Chemistry Guide and Strong Acid-Strong Base Titration Curves for HSC Chemistry.

NSW HSC syllabus context (November 2026): The Chemistry Stage 6 Syllabus (2017) remains the current syllabus for Year 12 students through 2027. Its buffer content includes preparing a buffer and demonstrating its properties. NESA’s 2021 HSC marking feedback also emphasises identifying buffer components, writing their equilibrium reactions, and explaining the response to added hydrochloric acid using Le Chatelier’s principle. The quantitative calculations in this resource extend those ideas through mole calculations and acid-base equilibrium; they should not be mistaken for a claim that every Henderson-Hasselbalch calculation is explicitly required. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029. See NESA’s current Chemistry syllabus information and 2021 HSC Chemistry marking feedback.