How to Recognise a Buffer from Mixed Solutions

Four beakers sit on a laboratory bench. One contains ethanoic acid and sodium ethanoate. Another contains ethanoic acid and sodium hydroxide. A third contains ethanoic acid and exactly enough sodium hydroxide to neutralise it. The fourth contains hydrochloric acid and sodium chloride.

Which ones are buffers?

The answer is the first two. But there’s a catch: the second mixture doesn’t initially contain a conjugate acid-base pair. It creates one when the chemicals react.

This is the key to recognising a buffer from a list of mixed solutions: work out what remains after all strong acid-base reactions have occurred. Don’t simply scan the ingredients and look for a weak acid and a salt.

01The recognition rule

A buffer solution contains appreciable amounts of either:

• A weak acid and its conjugate base.
• A weak base and its conjugate acid.

The two components must be present in the same solution. They must also be available in sufficient quantities to react with added acid and base.

For example, ethanoic acid and the ethanoate ion form a conjugate acid-base pair:

\[
\ce{CH3COOH <=> H+ + CH3COO-}
\]

Ethanoic acid, \(\ce{CH3COOH}\), is the weak acid. The ethanoate ion, \(\ce{CH3COO-}\), is its conjugate base.

They differ by one hydrogen ion, \(\mathrm{H}^{+}\).

A solution containing both in appreciable amounts can resist changes in pH when small quantities of strong acid or strong base are added.

The same principle applies to weak bases. For example, ammonia and ammonium ions form another conjugate pair:

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

Here, \(\ce{NH3}\) is the weak base and \(\ce{NH4+}\) is its conjugate acid.

Four mixtures: which are buffers?

Consider these four mixtures. All concentrations and volumes are illustrative.

Mixture A is straightforward. It directly combines a weak acid with a salt that supplies its conjugate base.

Mixture B is more interesting. The sodium hydroxide reacts with some of the ethanoic acid, producing ethanoate ions. Both members of the conjugate pair remain.

Mixture C is different. The sodium hydroxide completely neutralises the ethanoic acid, leaving ethanoate ions but no appreciable ethanoic acid.

Mixture D contains a strong acid and a salt. Chloride ions are not a useful conjugate base because hydrochloric acid is a strong acid.

The ingredients alone don’t tell you the answer. The final composition does.

02Why you must consider the reaction first

Suppose you mix a weak acid, \(\ce{HA}\), with a strong base, \(\ce{OH-}\).

The strong base reacts with the weak acid:

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

Every mole of hydroxide ions consumes one mole of weak acid and produces one mole of conjugate base.

This reaction is essentially complete. You should therefore account for it before considering the weak acid equilibrium.

There are three possible outcomes.

Case 1: The strong base is present in a smaller amount than the weak acid.

Some weak acid remains, and some conjugate base has been produced.

You have both components of a conjugate pair. Provided they are present in appreciable amounts, the mixture is a buffer.

Case 2: The strong base and weak acid are present in equal amounts.

The weak acid is completely neutralised.

Only the conjugate base remains. The mixture is not a buffer.

Case 3: The strong base is present in a greater amount than the weak acid.

The weak acid is completely neutralised, and excess hydroxide ions remain.

The mixture is not a buffer. Its pH is strongly influenced by the excess strong base.

The same reasoning works in reverse when a strong acid is mixed with a conjugate base.

The two reactions you should recognise immediately

For a weak acid buffer, \(\ce{HA/A-}\):

For a weak base buffer, \(\ce{B/BH+}\):

These reactions explain why buffers work in both directions.

03The four-step method for identifying a buffer

When you are given a list of mixed solutions in an HSC Chemistry question, use this method.

Step 1: Identify possible conjugate acid-base pairs

Look for a weak acid and its conjugate base, or a weak base and its conjugate acid.

Common examples include:

Remember that a salt can supply the conjugate ion.

For example, sodium ethanoate dissociates in water:

\[
\ce{CH3COONa -> Na+ + CH3COO-}
\]

The sodium ions are spectator ions. The ethanoate ions are the important part for buffer recognition.

Step 2: Calculate the initial number of moles

Never compare solution volumes alone.

A smaller volume of a concentrated solution can contain more moles than a larger volume of a dilute solution.

Use:

\[
n=cV
\]

where:

• \(n\) is the amount of substance in moles (mol).
• \(c\) is the molar concentration in \(\mathrm{mol\,L^{-1}}\).
• \(V\) is the solution volume in litres (L).

For quick calculations, you can use concentration in \(\mathrm{mol\,L^{-1}}\) multiplied by volume in mL to obtain the amount directly in mmol.

For example:

\[
0.100\times 25.0=2.50\text{ mmol}
\]

This shortcut works because \(1\text{ mol}=1000\text{ mmol}\) and \(1\text{ L}=1000\text{ mL}\).

Step 3: Complete any strong acid-base reactions

Write the balanced neutralisation equation.

Calculate which reactant is completely consumed and how much of each product forms.

For a monoprotic weak acid and hydroxide ions, the mole ratio is 1:1.

For a conjugate base and hydronium ions, it is also 1:1.

If a question involves several strong acids or bases, account for their reactions carefully. If both are added, they can also neutralise each other.

Step 4: Inspect what remains

Ask yourself:

• Is the weak acid still present?
• Is its conjugate base still present?
• Are both present in appreciable amounts?
• Is there any excess strong acid or strong base?

If both members of a conjugate pair remain in appreciable amounts, with no excess strong acid or base, you have a buffer.

If one member has been completely consumed, you don’t.

If one member is nearly exhausted, the solution may still show some buffering behaviour, but it has very little capacity in that direction.

06How could you test whether a solution is actually a buffer?

So far, we’ve identified buffers by calculating their chemical composition. What if you were given an unknown solution and had to investigate its behaviour experimentally?

A useful test is to measure how much its pH changes when small, controlled quantities of strong acid and strong base are added.

Setting up the experiment

A suitable apparatus arrangement includes:

• A beaker containing the solution being tested.
• A calibrated pH probe connected to a digital pH meter.
• A pipette or burette containing a known concentration of hydrochloric acid.
• A separate pipette or burette containing a known concentration of sodium hydroxide.
• A stirring mechanism to distribute each addition evenly.

First, place a measured volume of the unknown solution in the beaker. Immerse the pH probe and record the initial pH.

Add a small, measured amount of hydrochloric acid, stir, and record the new pH.

Repeat the experiment using a fresh sample of the unknown solution, but add sodium hydroxide instead.

Using fresh samples matters. If you add hydrochloric acid and then sodium hydroxide to the same beaker, the two strong reagents will neutralise each other. You won’t be testing the solution’s independent response to acid and base.

What should you predict?

A buffer should experience a relatively small pH change following a modest addition of either strong acid or strong base.

An unbuffered solution will generally experience a much larger change under the same conditions.

Consider an illustrative experiment using two 20.0 mL samples:

• Sample A is an acetate buffer containing 0.100 mol L⁻¹ ethanoic acid and 0.100 mol L⁻¹ sodium ethanoate.
• Sample B is a 0.00100 mol L⁻¹ sodium chloride solution prepared using carbon-dioxide-free water.

Each sample receives 1.00 mL of either 0.100 mol L⁻¹ hydrochloric acid or 0.100 mol L⁻¹ sodium hydroxide.

Before looking at the results, calculate how much strong acid or base has been added.

\[
n=cV
\]

\[
n=0.100\times0.00100
\]

\[
n=1.00\times10^{-4}\text{ mol}
\]

Therefore:

\[
n=0.100\text{ mmol}
\]

Sample A initially contains:

\[
n(\ce{CH3COOH})=2.00\text{ mmol}
\]

\[
n(\ce{CH3COO-})=2.00\text{ mmol}
\]

The added 0.100 mmol of hydrochloric acid consumes only 5% of the available ethanoate ions.

The buffer has enough conjugate base to absorb this addition without a large change in pH.

Illustrative results

The following values are calculated estimates, not experimental measurements. They assume a temperature of 25°C, negligible dissolved carbon dioxide, and ideal concentration behaviour. The acetate buffer calculation uses \(pK_a=4.74\).

The difference is clear.

The acetate buffer experiences only a small pH change. The unbuffered control experiences a much larger change.

What could make this experiment misleading?

A single small addition isn’t enough to establish the full buffering capacity of a solution.

Imagine testing a highly concentrated solution with an extremely small quantity of acid. Its pH might barely change simply because the disturbance is too small relative to the amount of material present.

A better experimental design uses a series of measured additions.

Add successive amounts of strong acid to one fresh sample, recording the pH after each addition. Repeat with strong base using another fresh sample.

Plot pH against the cumulative amount of strong acid or base added.

The resulting graph shows how long the solution can resist pH changes before its buffering capacity is exceeded.

The experiment supports the chemical prediction, but it doesn’t replace composition calculations. If you know the concentrations and volumes of the original solutions, calculating the remaining conjugate pair is the more direct way to identify a buffer.

07When does the buffer approximation fail?

Recognising a conjugate pair is only part of the problem. You also need to know when the usual buffer calculations are valid.

For a weak acid buffer, the acid dissociation constant is:

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

Rearranging and taking negative logarithms gives the Henderson-Hasselbalch equation:

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

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

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

\[
\mathrm{pH}
\approx
\mathrm{p}K_a+
\log_{10}\left(
\frac{n_{\mathrm{A}^{-}}}{n_{\ce{HA}}}
\right)
\]

This is a useful shortcut, but it has assumptions.

The usual approach assumes that:

• Strong acid-base neutralisation has been accounted for first.
• Both members of the conjugate pair remain in appreciable amounts.
• Further ionisation of the weak acid is small relative to the amounts of both buffer components.
• The solution is sufficiently dilute for concentrations to be a reasonable approximation to activities.

There are three important situations where you should be careful.

1. One buffer component has been completely consumed

This is the clearest failure.

Suppose a mixture initially contains 2.00 mmol of ethanoate ions and receives 2.50 mmol of hydrochloric acid.

The ethanoate ions can consume only 2.00 mmol of hydronium ions.

The remaining 0.50 mmol of hydronium ions is in excess.

There is no longer a conjugate acid-base pair available to provide normal buffer action.

Do not use the Henderson-Hasselbalch equation with a zero amount of conjugate base.

2. One component is almost exhausted

A buffer doesn’t suddenly become equally effective in both directions just because both components are technically still present.

Suppose a mixture contains:

• 4.00 mmol of ethanoic acid.
• 0.050 mmol of ethanoate ions.

Both species are present, so some buffer action remains.

However, the mixture can absorb only 0.050 mmol of additional strong acid before all its ethanoate ions are consumed.

It has a much larger reserve for absorbing strong base because 4.00 mmol of ethanoic acid is available.

This distinction is important in exam questions. The presence of both species tells you about the composition. Their relative amounts tell you how much disturbance the solution can withstand in each direction.

3. The solution is extremely dilute

The Henderson-Hasselbalch equation assumes that the weak acid’s additional ionisation is small compared with the amounts of both buffer components.

That assumption can fail if the solution is very dilute.

For example, imagine a 1.00 L solution containing:

• \(1.00\times10^{-5}\) mol of ethanoic acid.
• \(1.00\times10^{-5}\) mol of sodium ethanoate.

The initial mole ratio is 1:1.

A student might immediately predict:

\[
\mathrm{pH}=pK_a
\]

Using \(K_a=1.8\times10^{-5}\), this gives:

\[
\mathrm{pH}\approx4.74
\]

But check whether the approximation is physically consistent.

Rearranging the equilibrium expression:

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

Since the initial concentrations are equal:

\[
[\ce{H3O+}]
\approx1.8\times10^{-5}\text{ mol L}^{-1}
\]

There is only \(1.00\times10^{-5}\text{ mol L}^{-1}\) of ethanoic acid available.

The predicted hydronium concentration is greater than the total initial concentration of the weak acid. The calculation is therefore inconsistent with the assumption that only a negligible amount of the weak acid ionises.

The usual buffer approximation is not valid here.

The mixture contains the correct conjugate pair, but it is too dilute to behave as a useful buffer under the usual approximation. A full equilibrium treatment is needed to calculate its pH reliably.

The lesson is simple: don’t apply the buffer equation automatically. First check whether the mixture actually contains a suitable buffer and whether the equilibrium assumptions are reasonable.

08Questions and solutions

Try each question before reading its solution. The questions move from straightforward buffer recognition to situations where the reaction history and equilibrium assumptions become important.

Question 1

A student mixes 25.0 mL of 0.200 mol L⁻¹ ethanoic acid with 10.0 mL of 0.200 mol L⁻¹ sodium hydroxide.

Determine whether the final solution is a buffer. Calculate the final amounts of ethanoic acid and ethanoate ions.

Solution 1

Yes, the final solution is a buffer. It contains both ethanoic acid and ethanoate ions after the neutralisation reaction.

First, calculate the initial amounts:

\[
n(\ce{CH3COOH})
=0.200\times0.0250
=0.00500\text{ mol}
\]

\[
n(\ce{OH-})
=0.200\times0.0100
=0.00200\text{ mol}
\]

The reaction is:

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

The hydroxide ions consume 2.00 mmol of ethanoic acid.

Therefore:

\[
n(\ce{CH3COOH})=5.00-2.00
\]

\[
=3.00\text{ mmol}
\]

The amount of ethanoate ions produced is:

\[
n(\ce{CH3COO-})=2.00\text{ mmol}
\]

The final composition is:

Both members of the conjugate pair remain, and no strong base is left over.

The solution is therefore a buffer.

Question 2

A student prepares a solution by mixing:

• 40.0 mL of 0.125 mol L⁻¹ ammonia.
• 30.0 mL of 0.100 mol L⁻¹ ammonium chloride.
• 20.0 mL of 0.100 mol L⁻¹ sodium hydroxide.

The final mixture is then diluted to a total volume of 100.0 mL.

Determine whether the final solution is a buffer. Calculate the final concentrations of ammonia and ammonium ions.

Solution 2

Yes, the final solution is a buffer. The hydroxide ions consume some ammonium ions, producing additional ammonia, but both members of the conjugate pair remain.

Start by calculating the initial amounts.

For ammonia:

\[
n(\ce{NH3})
=0.125\times0.0400
=0.00500\text{ mol}
\]

\[
=5.00\text{ mmol}
\]

For ammonium ions:

\[
n(\ce{NH4+})
=0.100\times0.0300
=0.00300\text{ mol}
\]

\[
=3.00\text{ mmol}
\]

For hydroxide ions:

\[
n(\ce{OH-})
=0.100\times0.0200
=0.00200\text{ mol}
\]

The relevant reaction is:

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

One mole of hydroxide ions reacts with one mole of ammonium ions.

The hydroxide ions therefore consume 2.00 mmol of ammonium ions and produce 2.00 mmol of ammonia.

The final amounts are:

\[
n(\ce{NH4+})=3.00-2.00
=1.00\text{ mmol}
\]

\[
n(\ce{NH3})=5.00+2.00
=7.00\text{ mmol}
\]

Both species remain, and there is no excess hydroxide.

Now calculate their final concentrations using the total volume of 100.0 mL, or 0.1000 L.

For ammonia:

\[
[\ce{NH3}]
=\frac{0.00700}{0.1000}
=0.0700\text{ mol L}^{-1}
\]

For ammonium ions:

\[
[\ce{NH4+}]
=\frac{0.00100}{0.1000}
=0.0100\text{ mol L}^{-1}
\]

The final solution contains 0.0700 mol L⁻¹ ammonia and 0.0100 mol L⁻¹ ammonium ions.

These form a weak base/conjugate acid pair, so the solution is a buffer.

Notice that dilution changes the concentrations but does not change the mole ratio of the two components.

Question 3

An acetate buffer initially contains:

• 4.00 mmol of ethanoic acid.
• 0.80 mmol of ethanoate ions.

The buffer receives 0.75 mmol of hydrochloric acid.

(a) Calculate the final amounts of ethanoic acid and ethanoate ions.

(b) Is the resulting solution still a buffer?

(c) Two separate portions of this resulting solution are tested. The first receives a further 0.020 mmol of hydrochloric acid. The second receives 1.00 mmol of sodium hydroxide.

Which portion has the greater remaining stoichiometric reserve for the added disturbance? Explain.

Solution 3

The solution remains a buffer after the initial hydrochloric acid addition, but its ability to absorb further acid is extremely limited. It has a much larger reserve for absorbing strong base.

Part (a): Calculate the final composition.

Hydrochloric acid produces hydronium ions, which react with ethanoate ions:

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

The reaction consumes 0.75 mmol of ethanoate ions and produces 0.75 mmol of ethanoic acid.

Therefore:

\[
n(\ce{CH3COO-})=0.80-0.75
\]

\[
=0.05\text{ mmol}
\]

And:

\[
n(\ce{CH3COOH})=4.00+0.75
\]

\[
=4.75\text{ mmol}
\]

Part (b): Determine whether it is still a buffer.

Both species remain:

• 4.75 mmol of ethanoic acid.
• 0.05 mmol of ethanoate ions.

There is no excess strong acid.

The solution is still technically a buffer, although it is poorly equipped to absorb additional acid.

The distinction is that having both components doesn’t mean they are present in equally useful amounts.

Part (c): Compare the two disturbances.

For the hydrochloric acid addition:

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

The solution has only 0.05 mmol of ethanoate ions available.

Adding 0.020 mmol consumes:

\[
\frac{0.020}{0.050}\times100
=40\%
\]

of its remaining acid-absorbing reserve.

For the sodium hydroxide addition:

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

The solution has 4.75 mmol of ethanoic acid available.

Adding 1.00 mmol consumes:

\[
\frac{1.00}{4.75}\times100
\approx21.1\%
\]

of its base-absorbing reserve.

Therefore, the solution has the greater remaining stoichiometric capacity towards the addition of sodium hydroxide.

Both additions leave some of the relevant conjugate pair components, but the acid addition uses a much larger fraction of its available reserve.

This is why buffer capacity must be considered separately from simply identifying the presence of a conjugate pair.

Question 4

A 1.00 L solution contains \(1.00\times10^{-5}\) mol of ethanoic acid and \(1.00\times10^{-5}\) mol of sodium ethanoate.

The acid dissociation constant of ethanoic acid is:

\[
K_a=1.8\times10^{-5}
\]

A student claims that the solution is a buffer and that its pH must equal the \(pK_a\), because the two components are present in equal amounts.

Assess the student’s reasoning. Explain whether the usual buffer approximation is valid.

Solution 4

The student has identified the correct conjugate pair, but the predicted pH is not reliable because the usual buffer approximation breaks down at this concentration.

First, calculate the concentrations.

For ethanoic acid:

\[
[\ce{CH3COOH}]
=\frac{1.00\times10^{-5}}{1.00}
\]

\[
=1.00\times10^{-5}\text{ mol L}^{-1}
\]

The ethanoate concentration is identical:

\[
[\ce{CH3COO-}]
=1.00\times10^{-5}\text{ mol L}^{-1}
\]

The student uses the Henderson-Hasselbalch equation:

\[
\mathrm{pH}
\approx
\mathrm{p}K_a+
\log_{10}
\left(
\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}
\right)
\]

Since the concentrations are equal:

\[
\mathrm{pH}\approx\mathrm{p}K_a
\]

And:

\[
\mathrm{p}K_a=-\log_{10}(1.8\times10^{-5})
\]

\[
\mathrm{p}K_a\approx4.74
\]

However, we must check the underlying equilibrium assumption.

Using:

\[
K_a=
\frac{[\ce{H3O+}][\ce{CH3COO-}]}
{[\ce{CH3COOH}]}
\]

Rearranging:

\[
[\ce{H3O+}]
=K_a
\frac{[\ce{CH3COOH}]}{[\ce{CH3COO-}]}
\]

The equal concentration ratio predicts:

\[
[\ce{H3O+}]
=1.8\times10^{-5}\text{ mol L}^{-1}
\]

But the total initial ethanoic acid concentration is only:

\[
1.00\times10^{-5}\text{ mol L}^{-1}
\]

The predicted hydronium concentration would require more ethanoic acid to ionise than is actually available.

That contradicts the assumption that the additional ionisation of the weak acid is negligible.

The usual buffer approximation is therefore invalid.

The mixture contains the correct conjugate pair, but it is too dilute to be treated as a useful buffer using the standard shortcut.

A full equilibrium calculation is needed for a reliable pH value.

The important lesson is that the Henderson-Hasselbalch equation is not automatically valid just because the ingredients look like a buffer.

Question 5

A student prepares an acetate solution by mixing:

• 100.0 mL of 0.0500 mol L⁻¹ ethanoic acid.
• 20.0 mL of 0.100 mol L⁻¹ sodium ethanoate.

The student then performs two additions in sequence:

1. Adds 30.0 mL of 0.100 mol L⁻¹ hydrochloric acid.
2. Adds 40.0 mL of 0.100 mol L⁻¹ sodium hydroxide.

Assume that:

• All strong acid-base reactions occur completely.
• The solution volumes are additive.
• No other reactions occur.

A student argues:

Once the hydrochloric acid destroys the buffer, the subsequent sodium hydroxide cannot restore it.

Evaluate this claim.

Determine whether the solution is a buffer:

(a) Initially.

(b) Immediately after the hydrochloric acid addition.

(c) After the sodium hydroxide addition.

Calculate the final amounts of ethanoic acid and ethanoate ions. Explain whether calculating the net amount of strong acid or base added gives the same final composition as following the reactions in sequence.

Solution 5

The initial solution is a buffer, the solution immediately after the hydrochloric acid addition is not a buffer, and the final solution is a buffer again. The student’s claim is incorrect.

The key is to distinguish the intermediate composition from the final composition.

Part (a): Determine the initial composition.

Calculate the initial amount of ethanoic acid:

\[
n(\ce{CH3COOH})
=0.0500\times0.1000
\]

\[
=0.00500\text{ mol}
=5.00\text{ mmol}
\]

Calculate the initial amount of ethanoate ions:

\[
n(\ce{CH3COO-})
=0.100\times0.0200
\]

\[
=0.00200\text{ mol}
=2.00\text{ mmol}
\]

The initial mixture contains:

• 5.00 mmol of ethanoic acid.
• 2.00 mmol of ethanoate ions.

Both members of the conjugate pair are present.

The initial solution is therefore a buffer.

Part (b): Calculate the composition after adding hydrochloric acid.

First, calculate the amount of hydrochloric acid added:

\[
n(\ce{HCl})
=0.100\times0.0300
\]

\[
=0.00300\text{ mol}
=3.00\text{ mmol}
\]

The hydronium ions react with the ethanoate ions:

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

Initially, only 2.00 mmol of ethanoate ions are available.

Therefore, the reaction consumes all 2.00 mmol of ethanoate ions and produces 2.00 mmol of ethanoic acid.

The remaining 1.00 mmol of hydronium ions is in excess.

The intermediate composition is:

There is no longer a conjugate pair in appreciable amounts.

The solution is not a buffer immediately after the hydrochloric acid addition.

Part (c): Calculate the composition after adding sodium hydroxide.

First, calculate the amount of sodium hydroxide added:

\[
n(\ce{NaOH})
=0.100\times0.0400
\]

\[
=0.00400\text{ mol}
=4.00\text{ mmol}
\]

There are two reactions to consider.

First, the excess hydronium ions react with hydroxide ions:

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

There are 1.00 mmol of excess hydronium ions, so the first 1.00 mmol of hydroxide ions neutralises them.

This leaves:

\[
4.00-1.00=3.00\text{ mmol OH-}
\]

The remaining hydroxide ions react with ethanoic acid:

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

The reaction consumes 3.00 mmol of ethanoic acid and produces 3.00 mmol of ethanoate ions.

Therefore, the final ethanoic acid amount is:

\[
n(\ce{CH3COOH})=7.00-3.00
\]

\[
=4.00\text{ mmol}
\]

The final ethanoate amount is:

\[
n(\ce{CH3COO-})=0+3.00
\]

\[
=3.00\text{ mmol}
\]

The final composition is:

Both members of the conjugate pair are present again, with no excess strong acid or base.

The final solution is a buffer.

Check the net strong acid-base addition.

There is another way to verify the final result.

The student added:

• 3.00 mmol of strong acid.
• 4.00 mmol of strong base.

The net addition is equivalent to 1.00 mmol of hydroxide ions.

If we apply this net amount directly to the original buffer:

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

One mmol of hydroxide ions consumes one mmol of the original ethanoic acid.

Therefore:

\[
n(\ce{CH3COOH})=5.00-1.00
=4.00\text{ mmol}
\]

\[
n(\ce{CH3COO-})=2.00+1.00
=3.00\text{ mmol}
\]

These are exactly the same final amounts obtained by following the reactions in sequence.

The net calculation works because the strong acid and strong base neutralise each other quantitatively, and there are no competing reactions.

However, it does not describe the intermediate stage.

Immediately after the hydrochloric acid addition, the mixture contained excess strong acid and no ethanoate ions. It was not a buffer at that point.

The later sodium hydroxide addition neutralised the excess acid and then converted some ethanoic acid back into ethanoate ions.

This restored the conjugate pair.

The deeper principle is that buffer classification depends on the chemical composition at the time you assess it, not on whether the solution was previously a buffer.

09What to learn next

Recognising a buffer is the first step. The next step is calculating how its composition and pH change when a known amount of strong acid or base is added.

For that, see How to Calculate Buffer Composition After Adding Strong Acid or Base.

For a broader explanation of how conjugate acid-base pairs stabilise pH in natural systems, read Buffers in Natural Systems: HSC Chemistry Explained.