How to Compare Buffers with the Same pH but Different Concentrations

Two acetate buffer solutions both have a pH of 4.74. One contains four times the concentration of acetic acid and acetate ions as the other. If you add the same amount of hydrochloric acid to each, will their pH values change by the same amount?

No. The more concentrated buffer will experience a smaller pH change, provided both samples have the same volume and contain the same buffer system.

The reason is that pH depends on the ratio of the weak acid to its conjugate base, while buffer capacity depends on how much of each substance is available to react.

This is the key distinction when comparing buffers with the same pH but different concentrations. You need to consider both the equilibrium ratio and the actual number of moles present.

01Why two buffers can have the same pH

A buffer contains a weak acid and its conjugate base.

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

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

The two buffer components have different jobs:

• HA is the weak acid. It reacts with added hydroxide ions.
• A⁻ is the conjugate base. It reacts with added hydronium ions.

For an acetate buffer, these substances are acetic acid, \(\ce{CH3COOH}\), and acetate ions, \(\mathrm{CH_{3}COO}^{-}\).

The pH of a buffer depends on the relative concentrations of these two components.

Starting with the acid dissociation constant:

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

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

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

Here:

• \(K_a\) is the acid dissociation constant.
• \(pK_a=-\log_{10}(K_a)\).
• \([\mathrm{A}^{-}]\) is the conjugate base concentration in \(\mathrm{mol\,L^{-1}}\).
• \([\ce{HA}]\) is the weak acid concentration in \(\mathrm{mol\,L^{-1}}\).

The important feature is the concentration ratio.

If you multiply both concentrations by the same factor, their ratio stays unchanged. Therefore, the calculated pH also stays unchanged.

An example: Same ratio, different concentrations

Consider two acetate buffers at \(25^\circ\mathrm{C}\).

For acetic acid, use:

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

Therefore:

\[
pK_a=4.74
\]

For Buffer A:

\[
pH=4.74+\log_{10}\left(\frac{0.0500}{0.0500}\right)
\]

\[
pH=4.74+\log_{10}(1)
\]

\[
pH=4.74
\]

Buffer B has exactly the same concentration ratio:

\[
pH=4.74+\log_{10}\left(\frac{0.0125}{0.0125}\right)
\]

\[
pH=4.74
\]

Both buffers therefore have the same calculated pH, despite having different concentrations.

This is why measuring pH alone cannot tell you how concentrated a buffer is.

This relationship assumes the same weak acid/conjugate base pair, the same temperature, and the usual dilute-solution approximation used in HSC calculations. Different buffer systems can have the same pH with different concentration ratios because their \(pK_a\) values differ.

02Why concentration affects buffer capacity

Now consider what happens when you add hydrochloric acid.

Hydrochloric acid ionises essentially completely in water, producing hydronium ions. These react with the conjugate base in the buffer:

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

Two things happen:

1. The amount of conjugate base decreases.
2. The amount of weak acid increases.

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

The weak acid equilibrium shifts to the left as hydronium ions are introduced. This helps resist the decrease in pH.

However, the reaction consumes acetate ions. They are not replaced indefinitely.

This is where concentration matters.

A more concentrated buffer contains more moles of both components in the same volume. It therefore has a larger supply of conjugate base available to react with added acid.

The same reasoning applies when adding a strong base:

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

This time, the weak acid is consumed.

What exactly is buffer capacity?

Buffer capacity describes how much strong acid or strong base a buffer can absorb before its pH changes significantly.

For a useful comparison, consider the amount of acid or base required to change the pH by a specified amount.

For example, suppose two equal-volume buffers both start at pH 4.74.

• Buffer A requires 2.00 mmol of HCl to lower its pH to 4.24.
• Buffer B requires only 0.50 mmol of HCl to reach the same pH.

Buffer A has the greater capacity against added acid.

It takes four times as much acid to produce the same pH change.

This gives us an important distinction:

For equal sample volumes and the same buffer system at the same pH, increasing both buffer component concentrations increases buffer capacity.

However, if the sample volumes differ, concentration alone is not enough. You must calculate the total moles of each component.

03How to compare two buffers experimentally

Suppose you have two acetate buffers with the same initial pH but different concentrations.

How could you demonstrate experimentally that one has a greater buffer capacity?

Simply placing a pH probe into each solution is not enough. Both may give exactly the same initial reading.

You need to challenge the buffers with a known quantity of strong acid or base.

The experimental apparatus

A suitable arrangement contains:

• A burette containing hydrochloric acid of known concentration.
• A beaker containing a measured volume of buffer.
• A magnetic stirrer to mix the solution.
• A pH electrode connected to a digital pH meter.

The electrode sits in the buffer solution without touching the magnetic stir bar.

The experimental procedure

Step 1: Measure equal volumes of the two buffer solutions into separate beakers.

Both solutions should have the same initial pH and temperature.

Step 2: Record the initial pH of each buffer using a calibrated pH meter.

Step 3: Add a measured amount of hydrochloric acid to the first buffer.

Allow the solution to mix thoroughly before recording the new pH.

Step 4: Repeat with the second buffer, using exactly the same amount of hydrochloric acid at the same concentration.

Step 5: Continue adding acid in measured increments, recording the pH after each addition.

You can then compare the pH changes as more acid is introduced.

To investigate resistance to added base, repeat the experiment using fresh buffer samples and sodium hydroxide instead.

What should you observe?

Consider the following illustrative model predictions.

Both buffers have an initial volume of 100.0 mL and an initial pH of 4.74.

• Buffer A contains 0.0500 M acetic acid and 0.0500 M acetate ions.
• Buffer B contains 0.0125 M acetic acid and 0.0125 M acetate ions.

Both receive the same amount of hydrochloric acid at each stage.

The predicted pH values are:

These are illustrative Henderson-Hasselbalch calculations, not recorded experimental measurements.

Notice that Buffer B experiences a much larger pH decrease after receiving the same amount of acid.

Buffer A has a greater supply of acetate ions, so it can neutralise more of the added hydronium ions before its composition changes substantially.

Why experimental controls matter

A comparison is only meaningful if the important conditions are controlled.

One common experimental mistake is adding the same volume of acid without checking its concentration.

For example, 5.00 mL of 0.100 mol L⁻¹ HCl contains twice as many moles of acid as 5.00 mL of 0.0500 mol L⁻¹ HCl.

Equal volumes do not necessarily mean equal amounts.

The better approach is to calculate the moles of acid added and compare the resulting pH changes.

There is also an important measurement limitation: a pH meter measures the pH of the solution, not the individual concentrations of acetic acid and acetate ions. The stoichiometric calculations predict how those amounts change.

04Worked examples: Comparing buffer concentrations

05When does the buffer approximation stop working?

The Henderson-Hasselbalch equation is useful, but it relies on the buffer components remaining available.

Remember that the buffer has a finite supply of weak acid and conjugate base.

If you add too much strong acid, all the conjugate base can be consumed.

If you add too much strong base, all the weak acid can be consumed.

Once one component is exhausted, the solution is no longer functioning as an ordinary buffer.

The practical buffer range

A commonly used guide is:

\[
pH\approx pK_a\pm1
\]

This corresponds approximately to:

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

Within this range, both buffer components are present in reasonably useful proportions.

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

This is a practical guide, not an exact boundary where the Henderson-Hasselbalch equation suddenly stops working.

What happens when one component is completely consumed?

Consider an acetate buffer containing:

• 1.50 mmol of acetic acid.
• 0.40 mmol of acetate ions.

Now add 1.00 mmol of hydrochloric acid.

The acetate ions can only neutralise 0.40 mmol of the added acid.

Therefore:

\[
n(\ce{H3O+})_{\text{excess}}
=1.00-0.40
\]

\[
=0.60\text{ mmol}
\]

All the acetate ions have been consumed.

The solution now contains acetic acid and excess strong acid, rather than an ordinary functioning buffer.

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

Instead, calculate the concentration of excess hydronium ions using the final solution volume, then determine the pH.

The same principle applies if all the weak acid is consumed by added hydroxide ions.

For a complete method covering these calculations, see How to Calculate Buffer Composition After Adding Strong Acid or Base.

Other limitations to remember

There are two further reasons the simple comparison can become less accurate.

Very dilute buffers: If the buffer components become extremely dilute, water’s own ionisation can become significant relative to the buffer concentrations. You may need to consider the equilibrium of water as well.

Non-ideal solutions: At higher concentrations, ion interactions affect chemical activities. The simple concentration-based Henderson-Hasselbalch equation may then give a less accurate pH prediction.

For typical HSC calculations, the dilute-solution approximation is generally the intended starting model unless the question specifies otherwise.

06A reliable method for comparing two buffers

When a question asks you to compare two buffers with the same pH but different concentrations, use the following decision process.

One final distinction is worth remembering.

A buffer can have a high capacity against added acid but a different capacity against added base. The amount of conjugate base determines how much acid it can neutralise, while the amount of weak acid determines how much base it can neutralise.

For a fixed total buffer concentration, capacity against both directions is most balanced when the pH is close to the \(pK_a\).

If you need to revise the logarithmic relationship between pH and hydrogen ion concentration, see pH, pOH and Ion Concentrations for HSC Chemistry.

07Questions and solutions

Use \(pK_a=4.74\) for acetic acid unless otherwise stated. Assume a temperature of \(25^\circ\mathrm{C}\) and ideal dilute-solution behaviour.

Question 1

An acetate buffer contains:

• \([\ce{CH3COOH}]=0.0300\text{ mol L}^{-1}\)
• \([\mathrm{CH_{3}COO}^{-}]=0.0150\text{ mol L}^{-1}\)

Calculate its pH.

A second buffer contains twice the concentration of each component.

Explain whether its pH will be different.

Solution 1

Both buffers have a pH of approximately 4.44.

The second buffer has the same pH because doubling both concentrations does not change their ratio.

Step 1: Calculate the pH of the first buffer.

Using the Henderson-Hasselbalch equation:

\[
pH=4.74+\log_{10}
\left(\frac{0.0150}{0.0300}\right)
\]

\[
pH=4.74+\log_{10}(0.50)
\]

\[
pH=4.44
\]

Step 2: Calculate the concentration ratio of the second buffer.

Doubling both concentrations gives:

\[
[\ce{CH3COOH}]=0.0600\text{ M}
\]

\[
[\mathrm{CH_{3}COO}^{-}]=0.0300\text{ M}
\]

The new ratio is:

\[
\frac{0.0300}{0.0600}=0.50
\]

Therefore:

\[
pH=4.74+\log_{10}(0.50)
\]

\[
pH=4.44
\]

The pH remains unchanged because the ratio is unchanged.

However, if the two buffers have equal volumes, the second contains twice as many moles of both components and therefore has greater buffer capacity.

Question 2

Two 50.0 mL acetate buffers, X and Y, have the following initial compositions:

Both buffers receive 0.100 mmol of hydrochloric acid.

Calculate:

(a) The initial pH of both buffers.

(b) The final pH of each buffer.

(c) Which buffer experiences the greater pH change?

Solution 2

Buffer Y experiences the greater pH decrease. Its final pH is approximately 4.14, compared with 4.37 for Buffer X.

Although both buffers start at the same pH, Buffer Y contains fewer moles of each buffer component.

Step 1: Calculate the initial pH.

Both buffers have the same ratio:

\[
\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}=0.50
\]

Therefore:

\[
pH=4.74+\log_{10}(0.50)
\]

\[
pH=4.44
\]

Step 2: Calculate the initial moles in Buffer X.

For acetic acid:

\[
n=0.0400\times0.0500
\]

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

For acetate ions:

\[
n=0.0200\times0.0500
\]

\[
=0.00100\text{ mol}
=1.00\text{ mmol}
\]

Step 3: Calculate the new composition of Buffer X.

The reaction is:

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

The added acid consumes 0.100 mmol of acetate ions.

Therefore:

\[
n(\mathrm{CH_{3}COO}^{-})=1.00-0.100
\]

\[
=0.900\text{ mmol}
\]

The acetic acid increases:

\[
n(\ce{CH3COOH})=2.00+0.100
\]

\[
=2.10\text{ mmol}
\]

Step 4: Calculate the final pH of Buffer X.

\[
pH=4.74+\log_{10}
\left(\frac{0.900}{2.10}\right)
\]

\[
pH=4.74+\log_{10}(0.4286)
\]

\[
pH\approx4.37
\]

Step 5: Calculate the initial moles in Buffer Y.

Acetic acid:

\[
n=0.0100\times0.0500
\]

\[
=0.500\text{ mmol}
\]

Acetate ions:

\[
n=0.00500\times0.0500
\]

\[
=0.250\text{ mmol}
\]

Step 6: Calculate the new composition of Buffer Y.

Acetate ions:

\[
n(\mathrm{CH_{3}COO}^{-})=0.250-0.100
\]

\[
=0.150\text{ mmol}
\]

Acetic acid:

\[
n(\ce{CH3COOH})=0.500+0.100
\]

\[
=0.600\text{ mmol}
\]

Step 7: Calculate the final pH of Buffer Y.

\[
pH=4.74+\log_{10}
\left(\frac{0.150}{0.600}\right)
\]

\[
pH=4.74+\log_{10}(0.25)
\]

\[
pH\approx4.14
\]

Step 8: Compare the pH changes.

Buffer X:

\[
\Delta pH=4.37-4.44=-0.07
\]

Buffer Y:

\[
\Delta pH=4.14-4.44=-0.30
\]

Buffer Y experiences a much greater pH decrease.

The reason is that the same amount of acid consumes a much larger proportion of its available acetate ions.

Question 3

A student wants to investigate whether two acetate buffers with the same initial pH have different buffer capacities.

The student places a pH electrode into each solution and observes that both have a pH of 4.74.

The student concludes that the buffers have identical buffer capacities.

Evaluate this conclusion.

Describe an improved experimental method that would allow the student to compare their capacities reliably. Identify the variables that should be controlled and explain the expected observations.

Solution 3

The student’s conclusion is incorrect. Identical initial pH readings do not establish that two buffers have identical capacities.

The pH meter measures the hydrogen ion level of each solution. It does not measure the total amounts of weak acid and conjugate base available to react with added substances.

Two buffers can have the same pH because they have the same conjugate base-to-acid ratio, even when their absolute concentrations are very different.

Improved experimental method

The student should perform a controlled titration using hydrochloric acid.

First, measure equal volumes of the two buffer solutions into separate beakers.

Record their initial pH values, ensuring both are at the same temperature.

Next, use a burette containing hydrochloric acid of a known concentration. Add the same measured amount of acid to each buffer, allowing each solution to mix before recording its pH.

Repeat this process in small increments.

Variables to control

• Initial buffer volume: both samples should have equal volumes.
• Initial pH: both solutions should start at the same pH.
• Temperature: both should remain at the same temperature.
• Titrant concentration: use the same hydrochloric acid concentration.
• Titrant amount: compare equal numbers of moles of added acid.
• Mixing and measurement: use consistent stirring and the same calibrated pH meter.

Expected observations

The more concentrated buffer should experience a smaller pH change for the same amount of added acid.

Its titration curve should also be less steep over the effective buffering range.

Alternatively, the student could determine how much hydrochloric acid is needed to produce a fixed pH decrease, such as 0.50 units.

The buffer requiring more acid to produce that change has the greater capacity against added acid.

The experiment should also be repeated with fresh samples and sodium hydroxide to compare resistance against added base.

The important conclusion is that buffer capacity must be tested by challenging the buffer with a known amount of acid or base. An initial pH measurement alone cannot establish it.

Question 4

A 50.0 mL acetate buffer contains:

• 1.50 mmol of acetic acid.
• 0.40 mmol of acetate ions.

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

Assume volumes are additive.

(a) Calculate the amount of hydrochloric acid added.

(b) Determine the final amounts of acetic acid and acetate ions.

(c) Explain why the Henderson-Hasselbalch equation is no longer appropriate.

(d) Calculate the approximate final pH, assuming the excess strong acid controls the hydrogen ion concentration.

Solution 4

The buffer is exhausted because all 0.40 mmol of acetate ions are consumed. The excess hydrochloric acid determines the final pH, which is approximately 2.00.

Step 1: Calculate the amount of hydrochloric acid added.

\[
n=cV
\]

\[
n=0.100\times0.0100
\]

\[
n=0.00100\text{ mol}
\]

\[
n=1.00\text{ mmol}
\]

Step 2: Write the neutralisation reaction.

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

The reaction has a 1:1 mole ratio.

The initial acetate amount is only 0.40 mmol.

Therefore, all 0.40 mmol reacts.

Step 3: Calculate the final buffer component amounts.

Acetate ions:

\[
0.40-0.40=0\text{ mmol}
\]

Acetic acid:

\[
1.50+0.40=1.90\text{ mmol}
\]

Excess hydronium ions:

\[
1.00-0.40=0.60\text{ mmol}
\]

Step 4: Explain why the buffer approximation fails.

The Henderson-Hasselbalch equation requires both the weak acid and its conjugate base to be present.

Here, all acetate ions have been consumed.

The solution no longer has both components available to function as a buffer.

The excess strong acid must now be considered.

Step 5: Calculate the final hydronium concentration.

The final volume is:

\[
V=50.0+10.0=60.0\text{ mL}
\]

Therefore:

\[
[\ce{H3O+}]
=\frac{0.60\times10^{-3}}{0.0600}
\]

\[
=0.0100\text{ mol L}^{-1}
\]

Step 6: Calculate the pH.

\[
pH=-\log_{10}(0.0100)
\]

\[
pH=2.00
\]

The final pH is approximately 2.00.

The small additional contribution from acetic acid dissociation is negligible under the stated assumption that excess strong acid controls the pH.

The important point is that you must check whether both buffer components remain before applying the Henderson-Hasselbalch equation.

Question 5

Two acetate buffer samples, P and Q, are prepared at \(25^\circ\mathrm{C}\).

Their initial compositions are:

Each sample receives 15.0 mL of 0.100 mol L⁻¹ hydrochloric acid.

Assume that:

• Hydrochloric acid dissociates completely.
• The neutralisation reactions occur completely before equilibrium is considered.
• Solution volumes are additive.
• The usual dilute-solution approximations apply.
• Any excess strong acid dominates the pH when a buffer component is exhausted.

Answer the following:

(a) Verify that both buffers have the same initial pH.

(b) Calculate the initial moles of acetic acid and acetate ions in each sample.

(c) Determine the composition of each solution after adding hydrochloric acid.

(d) Identify which solution remains a functioning buffer.

(e) Calculate the final pH of both solutions.

(f) A student argues that Buffer P must have the greater total buffer capacity because its component concentrations are five times higher. Evaluate this statement, distinguishing between capacity per litre and capacity of the entire sample.

Solution 5

Buffer Q, despite being less concentrated, has the greater total capacity in this comparison. It contains twice as many initial moles of each buffer component as Buffer P.

After adding hydrochloric acid, Buffer P is exhausted and has an approximate pH of 1.70. Buffer Q retains both buffer components and has a final pH of approximately 3.89.

This question tests an important distinction: concentration per litre and total amount in the actual sample are not the same thing.

Step 1: Verify the initial pH values.

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

Therefore:

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

Using the Henderson-Hasselbalch equation:

\[
pH=4.74+\log_{10}(1)
\]

\[
pH=4.74
\]

Both initial pH values are correct.

Step 2: Calculate the initial moles in Buffer P.

For acetic acid:

\[
n=0.100\times0.0100
\]

\[
=0.00100\text{ mol}
=1.00\text{ mmol}
\]

The acetate amount is identical:

\[
n(\mathrm{CH_{3}COO}^{-})=1.00\text{ mmol}
\]

Step 3: Calculate the initial moles in Buffer Q.

For acetic acid:

\[
n=0.0200\times0.100
\]

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

The acetate amount is also:

\[
n(\mathrm{CH_{3}COO}^{-})=2.00\text{ mmol}
\]

Although Buffer Q is less concentrated, it contains twice as many moles of each buffer component because its sample volume is ten times greater.

Step 4: Calculate the amount of hydrochloric acid added.

Both samples receive:

\[
n=cV
\]

\[
n=0.100\times0.0150
\]

\[
=0.00150\text{ mol}
\]

\[
=1.50\text{ mmol}
\]

The same amount of acid is added to each buffer.

Step 5: Determine the composition of Buffer P.

Initially:

• Acetic acid = 1.00 mmol.
• Acetate ions = 1.00 mmol.

The 1.50 mmol of added hydrochloric acid can only react with the available 1.00 mmol of acetate ions.

Therefore, all acetate ions are consumed.

The final acetic acid amount is:

\[
n(\ce{CH3COOH})=1.00+1.00
\]

\[
=2.00\text{ mmol}
\]

The excess hydronium amount is:

\[
n(\ce{H3O+})_{\text{excess}}
=1.50-1.00
\]

\[
=0.50\text{ mmol}
\]

Buffer P no longer functions as a buffer.

Step 6: Calculate the final pH of Buffer P.

The final solution volume is:

\[
V=10.0+15.0=25.0\text{ mL}
\]

The excess hydronium concentration is:

\[
[\ce{H3O+}]
=\frac{0.50\times10^{-3}}{0.0250}
\]

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

Therefore:

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

\[
pH\approx1.70
\]

The excess strong acid controls the pH.

Step 7: Determine the composition of Buffer Q.

Initially:

• Acetic acid = 2.00 mmol.
• Acetate ions = 2.00 mmol.

The 1.50 mmol of added acid consumes 1.50 mmol of acetate ions.

The remaining acetate amount is:

\[
n(\mathrm{CH_{3}COO}^{-})=2.00-1.50
\]

\[
=0.50\text{ mmol}
\]

The acetic acid amount increases:

\[
n(\ce{CH3COOH})=2.00+1.50
\]

\[
=3.50\text{ mmol}
\]

Both components remain, so Buffer Q is still functioning as a buffer.

Step 8: Calculate the final pH of Buffer Q.

The final volume is:

\[
V=100.0+15.0=115.0\text{ mL}
\]

Using the Henderson-Hasselbalch equation:

\[
pH=4.74+\log_{10}
\left(\frac{0.50}{3.50}\right)
\]

\[
pH=4.74+\log_{10}(0.1429)
\]

\[
pH\approx3.89
\]

Buffer Q has a final pH of approximately 3.89.

Step 9: Evaluate the student’s statement.

The statement is only partly correct.

Buffer P has five times the concentration of each component per litre:

\[
\frac{0.100}{0.0200}=5
\]

Therefore, P has greater buffer capacity per litre under the same conditions.

However, the entire samples have different volumes.

Buffer P initially contains 1.00 mmol of each component, whereas Buffer Q contains 2.00 mmol of each.

Consequently, Buffer Q has twice the total amount of each buffer component and greater total capacity in the actual samples being compared.

This is why Buffer Q can neutralise more added acid before losing its buffering ability.

The tempting approach is to compare only the molar concentrations. The missing constraint is the sample volume.

Always use:

\[
n=cV
\]

when comparing the total amounts available in different samples.

The central idea is simple: the ratio of conjugate base to weak acid determines the initial pH, but the absolute amounts of those components determine how much added acid or base the buffer can resist.

Once you can distinguish those two quantities, you can move on to the next step: calculating how the buffer composition and pH change after a specified strong-acid or strong-base addition.