Why Diluting a Buffer Barely Changes pH but Reduces Its Capacity

Imagine you have an ethanoic acid/ethanoate buffer with a pH of 4.76. You add enough water to make its volume ten times larger. What happens to its pH?

Almost nothing. Now take 25.0 mL of the diluted buffer and add a small amount of hydrochloric acid. Its pH drops much more than the same acid addition would have caused in 25.0 mL of the original buffer.

There is no contradiction. Buffer pH depends mainly on the ratio of weak acid to conjugate base, whereas buffer capacity depends on how much of each component is available to react.

Adding water reduces both concentrations by the same factor, so their ratio stays approximately constant. However, a fixed volume of the diluted solution contains fewer moles of each component. It therefore has less capacity to neutralise added acid or base.

One important distinction: diluting an entire flask does not remove its buffer components. The total amount of buffer material remains, but its capacity per litre decreases.

01A practical that tests the difference

A useful way to investigate this question is to separate two measurements:

• The pH before and after dilution.
• The pH change when a known amount of strong acid is added.

These measurements test different properties.

Why measuring pH alone is insufficient

Suppose you prepare a buffer containing:

• \(0.100\ \mathrm{mol\,L^{-1}}\) ethanoic acid, \(\ce{CH3COOH}\).
• \(0.100\ \mathrm{mol\,L^{-1}}\) sodium ethanoate, \(\ce{CH3COONa}\).

You measure its pH, dilute it tenfold, and measure the pH again.

You would expect very similar readings.

However, this experiment alone cannot establish whether the buffer has retained its original capacity. You have measured its initial pH, but you have not challenged its ability to resist a change.

A better experimental design includes an identical acid addition to equal volumes of both solutions.

Improving the experimental design

Prepare two samples:

• Sample A: 25.0 mL of the original buffer.
• Sample B: 25.0 mL of the buffer after tenfold dilution with deionised water.

Place each sample in a labelled beaker. Use a calibrated pH probe to measure the initial pH of each solution.

Next, use a pipette to add exactly 1.00 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid to each sample. Stir both solutions and measure their final pH.

The apparatus should keep the sample volumes, acid concentration, acid volume, and temperature consistent.

What should happen?

The acid addition supplies:

\[ n(\ce{HCl})=cV \]

\[ =0.100\times 0.00100 \]

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

Therefore, each sample receives \(0.100\ \mathrm{mmol}\) of strong acid.

The acid reacts with ethanoate ions:

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

One mole of \(\mathrm{H}^{+}\) consumes one mole of \(\mathrm{CH_{3}COO}^{-}\).

The following table contains illustrative calculated results, rather than experimental measurements.

The concentrated buffer experiences a pH decrease of approximately 0.035 units.

The diluted buffer experiences a decrease of approximately 0.37 units.

Both started at almost the same pH, but their responses to the same acid addition are very different.

This is the experimental distinction we need to explain.

02Why dilution barely changes buffer pH

Let’s look at what actually controls the pH of a buffer.

A weak acid and its conjugate base exist in equilibrium:

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

Here:

• \(\ce{HA}\) represents the weak acid.
• \(\mathrm{A}^{-}\) represents its conjugate base.
• \(\mathrm{H}^{+}\) represents hydrogen ions.

For an ordinary buffer, the Henderson-Hasselbalch equation gives:

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

The \(pK_a\) is a measure of the weak acid’s strength. At a fixed temperature, it remains constant during dilution.

Notice what determines the other part of the equation: the ratio of conjugate base concentration to weak acid concentration.

It is not the total concentration of buffer particles.

What happens when you add water?

Suppose a buffer initially contains:

\[ [\ce{HA}]=0.100\ \mathrm{M} \]

\[ [\mathrm{A}^{-}]=0.100\ \mathrm{M} \]

Its concentration ratio is:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]}= \frac{0.100}{0.100}=1 \]

Therefore:

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

Since \(\log_{10}(1)=0\):

\[ pH=pK_a \]

Now dilute the buffer tenfold.

Both concentrations decrease by a factor of ten:

\[ [\ce{HA}]=0.0100\ \mathrm{M} \]

\[ [\mathrm{A}^{-}]=0.0100\ \mathrm{M} \]

The new ratio is:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]}= \frac{0.0100}{0.0100}=1 \]

Consequently:

\[ pH=pK_a \]

The calculated pH has not changed.

The same reasoning works for any dilution factor, provided both buffer components are diluted equally and the usual buffer approximation remains valid.

Why the equilibrium shift does not contradict this result

There is an important detail here.

You might remember from Le Chatelier’s principle that dilution can shift an equilibrium towards the side with more dissolved particles.

For a weak acid:

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

The forward reaction produces more dissolved particles, so dilution favours dissociation.

Doesn’t this mean the pH must change?

Not necessarily by much.

The buffer already contains a substantial amount of \(\mathrm{A}^{-}\). This common ion suppresses the dissociation of the weak acid. Under ordinary buffer conditions, the equilibrium correction caused by dilution is relatively small.

The Henderson-Hasselbalch calculation predicts that the pH stays constant because it uses the original acid-to-base ratio. The actual equilibrium can shift slightly, so the measured pH is not guaranteed to remain mathematically identical.

At ordinary buffer concentrations, however, the ratio approximation is usually very useful.

At extreme dilution, the equilibrium correction becomes more important. We will examine that limit shortly.

03Why buffer capacity decreases when you dilute

Now consider the other property: buffer capacity.

A buffer resists pH changes because its two components react with added strong acid or strong base.

The conjugate base neutralises added acid:

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

The weak acid neutralises added base:

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

These reactions consume the buffer components.

Every mole of added \(\mathrm{H}^{+}\) consumes one mole of \(\mathrm{A}^{-}\). Every mole of added \(\mathrm{OH}^{-}\) consumes one mole of \(\ce{HA}\).

The buffer therefore has a finite supply of reactants.

Imagine two identical-sized containers. One contains 2.50 mmol of each buffer component, while the other contains 0.250 mmol of each.

Both have the same initial acid-to-base ratio.

However, the first container has ten times as much material available to neutralise an added acid or base.

That is why its buffer capacity per unit volume is greater.

What do we mean by buffer capacity?

Buffer capacity describes how much strong acid or base must be added to produce a particular pH change.

A useful finite-change expression is:

\[ \beta\approx \frac{\Delta n}{V|\Delta pH|} \]

where:

• \(\beta\) is buffer capacity per unit volume.
• \(\Delta n\) is the amount of strong acid or base added, in moles.
• \(V\) is the buffer volume, in litres.
• \(\Delta pH\) is the resulting pH change.

The units are \(\mathrm{mol\,L^{-1}}\) per pH unit.

For buffers with the same initial composition ratio, diluting by a factor of ten reduces their capacity per litre by approximately a factor of ten.

The distinction is easiest to see by comparing what happens to different quantities.

An important distinction: the whole flask versus an aliquot

This is where students often make an incorrect generalisation.

Suppose you have 100.0 mL of buffer containing:

• 10.0 mmol of \(\ce{HA}\).
• 10.0 mmol of \(\mathrm{A}^{-}\).

You dilute the entire solution to 1.00 L.

Have you removed any of the weak acid or conjugate base?

No. You have added water, not removed buffer material.

The diluted 1.00 L solution still contains the original 10.0 mmol of each component.

However, a 100.0 mL aliquot of that diluted solution contains only:

• 1.00 mmol of \(\ce{HA}\).
• 1.00 mmol of \(\mathrm{A}^{-}\).

This aliquot has only one-tenth the buffer material of the original 100.0 mL sample.

So, when we say dilution decreases buffer capacity, we usually mean capacity per litre or the capacity of a fixed-volume sample.

For the entire flask, the total stoichiometric reserve is approximately unchanged under normal buffer conditions. The distinction becomes less exact at extreme dilution because the equilibrium distribution of the acid and conjugate base can change.

04When does the buffer approximation fail?

The Henderson-Hasselbalch equation is extremely useful, but it has limits.

There are two important situations where you need to be more careful.

1. The buffer becomes too dilute

The usual shortcut assumes that the equilibrium concentrations of the weak acid and conjugate base are approximately represented by their formal starting concentrations.

This works well when the weak acid dissociation is small compared with the amounts of buffer components present.

As you dilute the solution, that assumption becomes less reliable.

The weak acid can dissociate by a significant fraction of its total amount. The actual equilibrium ratio of \(\mathrm{A}^{-}\) to \(\ce{HA}\) can then differ substantially from the original salt-to-acid ratio.

At even greater dilution, the contribution of water’s own ionisation becomes important.

The pH then moves towards neutral rather than remaining fixed at the original \(pK_a\).

We will calculate an example of this in Question 5.

2. One buffer component becomes depleted

The second limitation occurs when you add too much strong acid or base.

Suppose a buffer contains:

• 2.00 mmol of \(\ce{HA}\).
• 0.500 mmol of \(\mathrm{A}^{-}\).

You add 0.500 mmol of HCl.

The strong acid consumes all the available conjugate base:

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

After the reaction, the solution no longer contains a meaningful reserve of \(\mathrm{A}^{-}\) to neutralise further acid.

You cannot continue using the ordinary buffer approximation as though both components are still present in substantial amounts.

Instead, you must consider the equilibrium of the remaining weak acid, or account for any excess strong acid if more was added.

This is why calculating the mole changes before using Henderson-Hasselbalch is so important.

A practical decision guide

For more practice with the neutralisation calculations, see How to Calculate Buffer Composition After Adding Strong Acid or Base.

You can also explore why two buffers with identical pH values can have very different capacities in How to Compare Buffers with the Same pH but Different Concentrations.

05Worked examples

06Questions and solutions

Question 1

A buffer contains 0.0200 mol of a weak acid, \(\ce{HA}\), and 0.0100 mol of its conjugate base, \(\mathrm{A}^{-}\), in 100.0 mL of solution.

The buffer is diluted to 400.0 mL.

Given \(pK_a=4.80\), calculate the pH before and after dilution. Assume the Henderson-Hasselbalch approximation is valid.

Solution 1

The pH is approximately 4.50 both before and after dilution.

The initial ratio is:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} =\frac{0.0100}{0.0200} =0.500 \]

Using Henderson-Hasselbalch:

\[ pH=4.80+\log_{10}(0.500) \]

\[ pH=4.80-0.301 \]

\[ \boxed{pH=4.50} \]

After dilution, the new concentrations are:

\[ [\ce{HA}]=\frac{0.0200}{0.400} =0.0500\ \mathrm{M} \]

\[ [\mathrm{A}^{-}]=\frac{0.0100}{0.400} =0.0250\ \mathrm{M} \]

The new ratio is:

\[ \frac{0.0250}{0.0500}=0.500 \]

Therefore:

\[ pH=4.80+\log_{10}(0.500) \]

\[ \boxed{pH=4.50} \]

The pH remains approximately constant because both concentrations decrease by the same factor.

However, the concentration of each buffer component has fallen to one-quarter of its original value. The capacity per litre is therefore substantially lower.

Question 2

Two 25.0 mL buffer samples have the following compositions:

• Sample A: \(0.0400\ \mathrm{M}\) \(\ce{HA}\) and \(0.0400\ \mathrm{M}\) \(\mathrm{A}^{-}\).
• Sample B: \(0.00800\ \mathrm{M}\) \(\ce{HA}\) and \(0.00800\ \mathrm{M}\) \(\mathrm{A}^{-}\).

Both samples initially have the same pH.

Each receives 0.100 mmol of hydrochloric acid.

Given \(pK_a=4.76\), calculate the final pH of each sample and compare their pH changes.

Solution 2

Sample A finishes at approximately pH 4.67, while Sample B finishes at approximately pH 4.28.

Both samples initially have equal concentrations of weak acid and conjugate base, so:

\[ pH=pK_a=4.76 \]

For Sample A:

The initial moles of each component are:

\[ n=0.0400\times0.0250 \]

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

After adding 0.100 mmol of HCl:

\[ n(\mathrm{A}^{-})=1.00-0.100=0.900\ \mathrm{mmol} \]

\[ n(\ce{HA})=1.00+0.100=1.10\ \mathrm{mmol} \]

Therefore:

\[ pH=4.76+\log_{10} \left(\frac{0.900}{1.10}\right) \]

\[ \boxed{pH=4.67} \]

The pH decreases by approximately 0.09 units.

For Sample B:

The initial moles of each component are:

\[ n=0.00800\times0.0250 \]

\[ =0.000200\ \mathrm{mol} =0.200\ \mathrm{mmol} \]

After adding the same 0.100 mmol of HCl:

\[ n(\mathrm{A}^{-})=0.200-0.100 =0.100\ \mathrm{mmol} \]

\[ n(\ce{HA})=0.200+0.100 =0.300\ \mathrm{mmol} \]

Therefore:

\[ pH=4.76+\log_{10} \left(\frac{0.100}{0.300}\right) \]

\[ pH=4.76-0.477 \]

\[ \boxed{pH=4.28} \]

The pH decreases by approximately 0.48 units.

Sample B experiences a much larger pH change because the same amount of acid consumes half its original conjugate base, compared with only one-tenth of Sample A’s conjugate base.

The initial pH values were identical, but the samples did not have identical capacities.

Question 3

A 200.0 mL buffer contains:

• 5.00 mmol of weak acid, \(\ce{HA}\).
• 5.00 mmol of conjugate base, \(\mathrm{A}^{-}\).

The entire solution is diluted to 1.00 L.

A laboratory technician performs two separate tests:

Test A: Adds 1.00 mmol of HCl to the entire 1.00 L solution.

Test B: Takes a 200.0 mL aliquot from the diluted solution and adds the same 1.00 mmol of HCl.

Assume the volume of added titrant is negligible.

Given \(pK_a=4.76\) and \(K_a=1.8\times10^{-5}\):

a) Calculate the final pH in Test A.

b) Explain why the Henderson-Hasselbalch approximation is no longer appropriate in Test B.

c) Calculate the approximate final pH in Test B.

Solution 3

Test A finishes at approximately pH 4.58, whereas Test B finishes at approximately pH 3.38.

The difference occurs because the two tests expose different amounts of buffer material to the same acid addition.

Part a: Test A

The entire diluted solution still contains:

• 5.00 mmol of \(\ce{HA}\).
• 5.00 mmol of \(\mathrm{A}^{-}\).

Adding 1.00 mmol of HCl consumes 1.00 mmol of conjugate base.

The final amounts are:

\[ n(\mathrm{A}^{-})=5.00-1.00 =4.00\ \mathrm{mmol} \]

\[ n(\ce{HA})=5.00+1.00 =6.00\ \mathrm{mmol} \]

The Henderson-Hasselbalch equation gives:

\[ pH=4.76+\log_{10} \left(\frac{4.00}{6.00}\right) \]

\[ \boxed{pH=4.58} \]

The buffer still contains substantial amounts of both components, so the approximation remains appropriate.

Part b: Why Test B is different

The 200.0 mL aliquot represents one-fifth of the entire diluted solution.

It therefore contains:

\[ n(\ce{HA})=1.00\ \mathrm{mmol} \]

\[ n(\mathrm{A}^{-})=1.00\ \mathrm{mmol} \]

The addition of 1.00 mmol of HCl consumes all the conjugate base.

The resulting solution contains 2.00 mmol of weak acid and essentially no remaining conjugate base from the original buffer pair.

The ordinary buffer approximation requires both components to be present in meaningful amounts.

That condition is no longer satisfied.

Part c: Calculate the final pH

After neutralisation, the solution contains:

\[ n(\ce{HA})=2.00\times10^{-3}\ \mathrm{mol} \]

The volume is 0.200 L, so:

\[ [\ce{HA}]=\frac{0.00200}{0.200} \]

\[ =0.0100\ \mathrm{M} \]

We now treat the solution as a weak acid solution.

The equilibrium is:

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

Let \(x\) represent the equilibrium concentration of \(\mathrm{H}^{+}\).

Using the acid dissociation expression:

\[ K_a=\frac{x^2}{0.0100-x} \]

Substitute:

\[ 1.8\times10^{-5} =\frac{x^2}{0.0100-x} \]

Rearranging:

\[ x^2+1.8\times10^{-5}x -1.8\times10^{-7}=0 \]

Solving the quadratic gives:

\[ x\approx4.15\times10^{-4}\ \mathrm{M} \]

Therefore:

\[ pH=-\log_{10}(4.15\times10^{-4}) \]

\[ \boxed{pH\approx3.38} \]

The key difference is that Test A retained enough conjugate base to function as a buffer. Test B exhausted its conjugate base completely.

The tempting mistake is to apply Henderson-Hasselbalch to both tests simply because they started with the same pH. The correct calculation depends on the actual amounts remaining after neutralisation.

Question 4

A buffer contains 8.00 mmol of weak acid, \(\ce{HA}\), and 2.00 mmol of its conjugate base, \(\mathrm{A}^{-}\), in 100.0 mL.

The buffer is diluted to 500.0 mL. A 100.0 mL aliquot is removed.

The aliquot receives:

• 0.300 mmol of HCl.
• 0.500 mmol of NaOH.

Given \(pK_a=4.76\):

a) Calculate the final amounts of \(\ce{HA}\) and \(\mathrm{A}^{-}\).

b) Calculate the final pH.

c) Would the final pH change if the NaOH were added before the HCl instead? Explain.

Solution 4

The final pH is approximately 4.39, regardless of the order of addition.

The important detail is that both additions can be completely neutralised by the available buffer components in either order.

Part a: Calculate the initial amounts in the aliquot

The original buffer is diluted fivefold.

The 100.0 mL aliquot therefore contains one-fifth of the original amounts:

\[ n(\ce{HA})=\frac{8.00}{5} =1.60\ \mathrm{mmol} \]

\[ n(\mathrm{A}^{-})=\frac{2.00}{5} =0.400\ \mathrm{mmol} \]

First consider adding HCl.

The acid reacts with the conjugate base:

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

Therefore:

\[ n(\mathrm{A}^{-})=0.400-0.300 =0.100\ \mathrm{mmol} \]

\[ n(\ce{HA})=1.60+0.300 =1.90\ \mathrm{mmol} \]

Now add NaOH.

The hydroxide reacts with the weak acid:

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

Therefore:

\[ n(\ce{HA})=1.90-0.500 =1.40\ \mathrm{mmol} \]

\[ n(\mathrm{A}^{-})=0.100+0.500 =0.600\ \mathrm{mmol} \]

Part b: Calculate the final pH

Apply Henderson-Hasselbalch:

\[ pH=4.76+\log_{10} \left(\frac{0.600}{1.40}\right) \]

\[ pH=4.76+\log_{10}(0.429) \]

\[ \boxed{pH=4.39} \]

Part c: Does the order matter?

No. The final composition is the same if NaOH is added first.

Starting with:

\[ n(\ce{HA})=1.60\ \mathrm{mmol} \]

\[ n(\mathrm{A}^{-})=0.400\ \mathrm{mmol} \]

Add 0.500 mmol of NaOH first:

\[ n(\ce{HA})=1.60-0.500 =1.10\ \mathrm{mmol} \]

\[ n(\mathrm{A}^{-})=0.400+0.500 =0.900\ \mathrm{mmol} \]

Then add 0.300 mmol of HCl:

\[ n(\mathrm{A}^{-})=0.900-0.300 =0.600\ \mathrm{mmol} \]

\[ n(\ce{HA})=1.10+0.300 =1.40\ \mathrm{mmol} \]

These are exactly the same final amounts as before.

The net addition is equivalent to 0.200 mmol of strong base, which converts 0.200 mmol of \(\ce{HA}\) into \(\mathrm{A}^{-}\).

The order does not matter here because neither reagent exceeds the available amount of the buffer component it reacts with at the time of addition.

This is a useful reminder that buffer calculations are fundamentally mole calculations before they become pH calculations.

Question 5

An ethanoic acid/ethanoate buffer is initially prepared with the following formal concentrations at 25°C:

\[ [\ce{CH3COOH}]=0.100\ \mathrm{M} \]

\[ [\mathrm{CH_{3}COO}^{-}]=0.100\ \mathrm{M} \]

The buffer is diluted until the formal concentration of each component is:

\[ 1.00\times10^{-5}\ \mathrm{M} \]

This represents a dilution by a factor of 10,000.

A student argues that the pH must remain equal to the original \(pK_a\), because the acid-to-conjugate-base ratio is still 1:1.

Another student argues that dilution shifts the weak acid equilibrium towards dissociation, so the pH must change.

Given:

\[ K_a=1.74\times10^{-5} \]

\[ K_w=1.00\times10^{-14} \]

Determine the actual pH after dilution, using equilibrium, mass balance, and charge balance.

Then explain:

a) Why the simple Henderson-Hasselbalch calculation using the formal concentrations is no longer reliable.

b) What has happened to the buffer capacity per litre.

Assume ideal solution behaviour and that no buffer material is lost during dilution.

Solution 5

The actual pH is approximately 5.27, rather than the original value of 4.76.

The second student has identified a real equilibrium shift, but the complete explanation requires considering the actual equilibrium concentrations rather than simply comparing the original formal concentrations.

This is the important boundary of the usual buffer approximation.

Step 1: Calculate the pH predicted by the simple buffer approximation.

The original acid-to-base ratio is:

\[ \frac{[\mathrm{CH_{3}COO}^{-}]}{[\ce{CH3COOH}]} =\frac{0.100}{0.100}=1 \]

Therefore:

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

\[ pH=pK_a \]

Since:

\[ pK_a=-\log_{10}(1.74\times10^{-5}) \]

\[ pK_a\approx4.76 \]

The simple calculation predicts that the diluted buffer also has a pH of approximately 4.76.

However, this assumes that the formal concentrations still accurately represent the equilibrium concentrations of the acid and conjugate base.

At this extreme dilution, that assumption needs checking.

Step 2: Define the formal concentration after dilution.

Let:

\[ c=1.00\times10^{-5}\ \mathrm{M} \]

This is the formal concentration of each component.

The total analytical acetate concentration is:

\[ [\ce{CH3COOH}]+[\mathrm{CH_{3}COO}^{-}]=2c \]

The sodium ion concentration is:

\[ [\mathrm{Na}^{+}]=c \]

Sodium ions come from the sodium ethanoate originally added.

Let:

\[ h=[\mathrm{H}^{+}] \]

represent the equilibrium hydrogen ion concentration.

Step 3: Apply charge balance.

The positive and negative charges in the solution must balance.

Therefore:

\[ [\mathrm{Na}^{+}]+[\mathrm{H}^{+}] = [\mathrm{CH_{3}COO}^{-}]+[\mathrm{OH}^{-}] \]

At the calculated acidic pH, the hydroxide concentration will be very small compared with the hydrogen ion concentration.

We can initially neglect \([\mathrm{OH}^{-}]\), giving:

\[ c+h\approx[\mathrm{CH_{3}COO}^{-}] \]

Therefore:

\[ [\mathrm{CH_{3}COO}^{-}]\approx c+h \]

Step 4: Apply the mass balance.

The total analytical acetate concentration remains \(2c\):

\[ [\ce{CH3COOH}]+[\mathrm{CH_{3}COO}^{-}]=2c \]

Substitute the charge balance result:

\[ [\ce{CH3COOH}]+c+h=2c \]

Therefore:

\[ [\ce{CH3COOH}]\approx c-h \]

Step 5: Substitute into the acid dissociation expression.

For ethanoic acid:

\[ K_a= \frac{[\mathrm{H}^{+}][\mathrm{CH_{3}COO}^{-}]} {[\ce{CH3COOH}]} \]

Substituting the expressions above:

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

Rearrange:

\[ K_a(c-h)=h(c+h) \]

Expanding:

\[ K_ac-K_ah=ch+h^2 \]

Collecting terms:

\[ h^2+(c+K_a)h-K_ac=0 \]

Step 6: Substitute the known values and solve.

\[ h^2+(1.00\times10^{-5} +1.74\times10^{-5})h \]

\[ -(1.74\times10^{-5} \times1.00\times10^{-5})=0 \]

Solving the quadratic gives:

\[ h\approx5.32\times10^{-6}\ \mathrm{M} \]

Therefore:

\[ pH=-\log_{10}(5.32\times10^{-6}) \]

\[ \boxed{pH\approx5.27} \]

Step 7: Check the water ionisation assumption.

Calculate the hydroxide concentration:

\[ [\mathrm{OH}^{-}]=\frac{K_w}{[\mathrm{H}^{+}]} \]

\[ =\frac{1.00\times10^{-14}} {5.32\times10^{-6}} \]

\[ =1.88\times10^{-9}\ \mathrm{M} \]

This is much smaller than the hydrogen ion concentration.

Therefore, neglecting hydroxide ions in the earlier charge balance was justified.

Step 8: Explain why the simple approximation failed.

The actual equilibrium concentrations are:

\[ [\mathrm{CH_{3}COO}^{-}] \approx c+h \]

\[ =1.53\times10^{-5}\ \mathrm{M} \]

And:

\[ [\ce{CH3COOH}] \approx c-h \]

\[ =4.68\times10^{-6}\ \mathrm{M} \]

Their actual ratio is approximately:

\[ \frac{[\mathrm{CH_{3}COO}^{-}]} {[\ce{CH3COOH}]} = \frac{1.53\times10^{-5}} {4.68\times10^{-6}} \]

\[ \approx3.27 \]

The equilibrium ratio is no longer 1:1.

The weak acid has dissociated to a significant extent relative to the very small formal concentrations present.

This is why using the original formal concentration ratio in the Henderson-Hasselbalch equation gives an incorrect result.

Importantly, the Henderson-Hasselbalch relationship itself can still describe the equilibrium using the actual equilibrium concentrations. It is the shortcut of substituting the formal starting concentrations that has failed.

Step 9: Explain the effect on buffer capacity.

Before dilution, the total formal buffer concentration was:

\[ 0.100+0.100=0.200\ \mathrm{M} \]

After dilution, it is:

\[ 1.00\times10^{-5} +1.00\times10^{-5} \]

\[ =2.00\times10^{-5}\ \mathrm{M} \]

The analytical concentration of buffer material has decreased by a factor of 10,000.

The buffer therefore has dramatically less capacity per litre.

The precise capacity change is not exactly 10,000-fold in this extreme case because the equilibrium distribution of the acid and conjugate base has also changed.

There is another important detail: the total analytical amount of acetate material in the entire flask has not disappeared. Dilution has not removed it. However, its equilibrium distribution between \(\ce{CH3COOH}\) and \(\mathrm{CH_{3}COO}^{-}\) is no longer the original 1:1 distribution.

This is the limit of the simple statement that dilution leaves buffer pH unchanged.

The key decision in buffer questions is to identify what has actually changed. If only water is added, the acid-to-conjugate-base ratio usually stays approximately constant. If strong acid or base is added, calculate the mole changes before calculating pH. If the solution becomes extremely dilute or one component is exhausted, return to the full equilibrium rather than relying on the usual buffer shortcut.