How to Calculate Buffer Composition After Adding Strong Base

Add 1.00 mmol of strong base to a buffer containing 10.0 mmol of a weak acid and 10.0 mmol of its conjugate base. Does the pH stay the same?

No. The hydroxide ions react with the weak acid, converting some of it into its conjugate base. The buffer now contains 9.00 mmol of weak acid and 11.0 mmol of conjugate base.

For an ethanoic acid/ethanoate buffer with a pKa of 4.76 at 25°C, the new pH is:

\[ \mathrm{pH}=4.76+\log_{10}\left(\frac{11.0}{9.00}\right) \]

\[ \boxed{\mathrm{pH}=4.85} \]

The pH has increased by approximately 0.09 units. The important step is to calculate the new amounts of acid and conjugate base before calculating pH.

This is the method you need for calculating buffer composition after adding a small amount of strong base. It also works in reverse when you add a strong acid.

01Why does adding strong base change the buffer composition?

A buffer contains two substances:

• A weak acid, represented by \(\ce{HA}\).
• Its conjugate base, represented by \(\mathrm{A}^{-}\).

The weak acid can donate hydrogen ions, while the conjugate base can accept them.

For example, an ethanoic acid/ethanoate buffer contains ethanoic acid and ethanoate ions.

The weak acid exists in equilibrium with its ions:

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

Now imagine adding sodium hydroxide to this solution.

Sodium hydroxide dissociates completely in water, releasing hydroxide ions. These hydroxide ions react with the weak acid:

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

Notice what happens:

• One mole of \(\mathrm{OH}^{-}\) reacts with one mole of \(\ce{HA}\).
• One mole of \(\ce{HA}\) disappears.
• One mole of \(\mathrm{A}^{-}\) is produced.

The strong base therefore changes the actual amounts of both buffer components.

What happens to the equilibrium?

There is an important distinction between the neutralisation reaction and the weak acid equilibrium.

When hydroxide ions are added, they remove hydrogen ions from the solution by forming water. This causes the weak acid equilibrium to shift to the right, producing more hydrogen ions and conjugate base.

However, those hydrogen ions are also neutralised while hydroxide remains.

The overall result is the net reaction:

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

The buffer doesn’t prevent the strong base from reacting. It uses up some of its weak acid to neutralise it.

Think of the weak acid as a reserve that deals with incoming hydroxide ions, while the conjugate base is a reserve that deals with incoming hydrogen ions. Both reserves are finite.

This is why the amount of each component matters. A buffer containing 10 mmol of weak acid cannot neutralise the same amount of strong base as one containing 100 mmol of weak acid.

02A simple experiment: does a buffer really resist pH change?

Consider a paired experiment designed to test the effect of adding the same amount of strong base to water and a buffer.

The apparatus consists of two 100.0 mL beakers:

• Beaker A contains 100.0 mL of distilled water.
• Beaker B contains 100.0 mL of a buffer made from 0.0100 mol of \(\ce{HA}\) and 0.0100 mol of \(\mathrm{A}^{-}\).

Place a calibrated pH probe in each beaker. Stir both solutions, then use a pipette to add 1.00 mL of 1.00 mol/L sodium hydroxide to each.

What should happen?

The prediction is that the water will experience a much larger pH change.

In distilled water, there is no significant reserve of weak acid to react with the added hydroxide. Most of the hydroxide remains in solution.

In the buffer, the hydroxide reacts with the available weak acid.

The amount of sodium hydroxide added is:

\[ n=cV \]

\[ =1.00\times 0.00100 \]

\[ =0.00100\text{ mol} \]

For the buffer, the updated composition is:

\[ n(\ce{HA})=0.0100-0.00100 \]

\[ =0.00900\text{ mol} \]

\[ n(\mathrm{A}^{-})=0.0100+0.00100 \]

\[ =0.0110\text{ mol} \]

Using the Henderson-Hasselbalch equation, the predicted final pH is 4.85.

For the water sample, the final volume is 101.0 mL. Its hydroxide concentration is:

\[ [\mathrm{OH}^{-}]=\frac{0.00100}{0.1010} \]

\[ =0.00990\text{ mol/L} \]

Therefore:

\[ \mathrm{pOH}=-\log_{10}(0.00990)=2.00 \]

At 25°C:

\[ \mathrm{pH}=14.00-2.00 \]

\[ \boxed{\mathrm{pH}=12.00} \]

The following table summarises the calculated predictions. These are illustrative model values, not recorded experimental measurements.

The buffer experiences a much smaller pH change because its weak acid reacts with the added hydroxide.

What does this experiment actually establish?

A single addition demonstrates that the buffer resists a particular dose of strong base. It does not tell us how much more base the buffer can neutralise.

There is also a limitation in using pH readings alone: a pH meter measures the equilibrium pH, not the number of moles of weak acid remaining.

A better experimental design would add measured increments of sodium hydroxide using a burette. After each addition, allow the solution to mix, record the stable pH, and calculate the cumulative moles of hydroxide added.

Repeating the procedure with a water control allows the two responses to be compared.

This revised investigation reveals something a single pH reading cannot: the buffer has a limited capacity, and its resistance to further base addition decreases as its weak acid is consumed.

03The calculation method: update the moles first

The most reliable way to calculate buffer composition is to separate the problem into two stages:

1. Calculate the neutralisation reaction using mole ratios.
2. Calculate the new pH using the updated buffer composition.

Do not jump straight into the pH equation.

Step 1: Write the neutralisation reaction

For a strong base:

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

The mole ratio is 1:1.

This means that every mole of hydroxide added consumes one mole of weak acid and produces one mole of conjugate base.

Step 2: Calculate the moles of strong base added

If the question gives a concentration and volume, use:

\[ n=cV \]

where:

• \(n\) is the amount of substance in mol.
• \(c\) is the concentration in mol/L.
• \(V\) is the volume in L.

Always convert millilitres into litres before substituting.

Step 3: Calculate the new buffer composition

Let:

• \(n_{\mathrm{HA},0}\) be the initial moles of weak acid.
• \(n_{\mathrm{A^-},0}\) be the initial moles of conjugate base.
• \(n_{\mathrm{OH^-}}\) be the moles of strong base added.

Provided that the added hydroxide is less than the initial amount of weak acid:

\[ n_{\mathrm{HA,final}} = n_{\mathrm{HA},0}-n_{\mathrm{OH^-}} \]

\[ n_{\mathrm{A^-,final}} = n_{\mathrm{A^-},0}+n_{\mathrm{OH^-}} \]

These equations are the heart of the calculation.

The weak acid decreases, while the conjugate base increases by exactly the same number of moles.

Step 4: Calculate the new pH

Once the composition has been updated, use the Henderson-Hasselbalch equation:

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

Here, \(\mathrm{p}K_a\) is the negative base-10 logarithm of the acid dissociation constant \(K_a\).

Because both buffer components occupy the same final solution volume, their concentration ratio is equal to their mole ratio:

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

Therefore, you can use:

\[ \boxed{ \mathrm{pH}=\mathrm{p}K_a+ \log_{10}\left( \frac{n_{\mathrm{A^-,final}}} {n_{\mathrm{HA,final}}} \right) } \]

This saves time because you do not need to calculate both final concentrations separately.

The final volume still matters if you need an actual concentration, but it cancels when calculating this mole ratio.

04What if you add a strong acid instead?

The same method applies, but the reaction runs in the opposite direction.

A strong acid supplies hydrogen ions, which react with the conjugate base:

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

Each mole of \(\ce{H3O+}\) consumes one mole of \(\mathrm{A}^{-}\) and produces one mole of \(\ce{HA}\).

If \(n_{\mathrm{H_3O^+}}\) is the number of moles of strong acid added:

\[ n_{\mathrm{A^-,final}} = n_{\mathrm{A^-,0}}-n_{\mathrm{H_3O^+}} \]

\[ n_{\mathrm{HA,final}} = n_{\mathrm{HA,0}}+n_{\mathrm{H_3O^+}} \]

The updated mole ratio is then substituted into the same Henderson-Hasselbalch equation.

If a question involves both a strong acid and a strong base, update the mole amounts after each reaction. If excess strong acid or base is present, account for its neutralisation before calculating the final composition.

05Worked examples

06When does the buffer approximation fail?

A buffer has a limited ability to neutralise added acid or base.

The important question is not simply how many moles of strong base are added. It is how many moles are added compared with the amount of weak acid available to react.

Consider two situations:

• Adding 1.00 mmol of hydroxide to a buffer containing 100 mmol of weak acid consumes only 1% of its acid reserve.
• Adding 9.00 mmol of hydroxide to a buffer containing 10.0 mmol of weak acid consumes 90% of its acid reserve.

Both additions are finite amounts, but their effects are very different.

The practical buffer range

A useful rule of thumb is that a buffer works most effectively when the ratio of conjugate base to weak acid is between 0.1 and 10.

In other words:

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

This corresponds approximately to:

\[ \mathrm{p}K_a-1 \leq\mathrm{pH}\leq \mathrm{p}K_a+1 \]

This is called the effective buffer range.

It is a practical guideline, not a mathematical boundary where the Henderson-Hasselbalch equation suddenly stops working.

A solution can still contain both buffer components outside this range. The equation may still give a reasonable pH estimate, provided its assumptions remain valid. However, the solution has much less ability to resist further additions in one direction.

For example, if the ratio of conjugate base to weak acid is 20, the weak acid is present in a much smaller amount. The solution has little reserve for neutralising additional strong base.

What happens when the strong base exceeds the available weak acid?

Consider a buffer containing:

• 0.0100 mol of weak acid.
• 0.00200 mol of conjugate base.
• An initial volume of 100.0 mL.

Now add 12.0 mL of 1.00 mol/L NaOH.

Step 1: Calculate the hydroxide added.

\[ n(\mathrm{OH}^{-})=1.00\times0.0120 \]

\[ =0.0120\text{ mol} \]

Step 2: Compare the hydroxide amount with the available weak acid.

Only 0.0100 mol of weak acid is available.

Therefore, all the weak acid is consumed, and excess hydroxide remains:

\[ n_{\mathrm{excess\ OH^-}} = 0.0120-0.0100 \]

\[ =0.00200\text{ mol} \]

The conjugate base amount becomes:

\[ n(\mathrm{A}^{-})=0.00200+0.0100 \]

\[ =0.0120\text{ mol} \]

There is now no remaining weak acid to act as the buffer’s base-neutralising reserve.

Step 3: Calculate the final hydroxide concentration.

The final volume is:

\[ V=100.0+12.0=112.0\text{ mL} \]

\[ =0.1120\text{ L} \]

Therefore:

\[ [\mathrm{OH}^{-}] = \frac{0.00200}{0.1120} \]

\[ =0.0179\text{ mol/L} \]

Step 4: Calculate the pH.

\[ \mathrm{pOH}=-\log_{10}(0.0179) \]

\[ =1.75 \]

At 25°C:

\[ \mathrm{pH}=14.00-1.75 \]

\[ \boxed{\mathrm{pH}=12.25} \]

The final pH is controlled by the excess strong base, not by the original buffer ratio.

Trying to substitute a negative amount of weak acid into the Henderson-Hasselbalch equation would be physically meaningless.

There is also an important boundary case: if the added hydroxide exactly equals the initial amount of weak acid, all the weak acid is consumed, but no excess hydroxide remains. The solution then contains the conjugate base, whose reaction with water determines the pH. You must consider base hydrolysis rather than using the buffer equation.

The same reasoning applies in reverse when a strong acid exceeds the available amount of conjugate base.

A useful decision process

One final distinction matters: the approximation that the pH changes very little is not the same as the Henderson-Hasselbalch equation itself.

You must always update the buffer composition after adding strong acid or base. The pH change can be substantial even while both buffer components remain.

07Questions and solutions

Use \(\mathrm{p}K_a=4.76\) for ethanoic acid at 25°C in all questions unless otherwise stated. Assume ideal behaviour and complete dissociation of strong electrolytes.

Question 1

An ethanoic acid/ethanoate buffer contains 0.0200 mol of ethanoic acid and 0.0100 mol of ethanoate ions.

A student adds 2.00 mL of 1.00 mol/L NaOH.

Calculate:

a) The final number of moles of ethanoic acid.

b) The final number of moles of ethanoate ions.

c) The final pH of the buffer.

Solution 1

The final buffer contains 0.0180 mol of ethanoic acid and 0.0120 mol of ethanoate ions, giving a pH of 4.58.

Step 1: Calculate the moles of hydroxide added.

\[ n(\mathrm{OH}^{-})=cV \]

\[ =1.00\times0.00200 \]

\[ =0.00200\text{ mol} \]

Step 2: Calculate the final amount of ethanoic acid.

Hydroxide reacts with ethanoic acid in a 1:1 ratio.

\[ n(\ce{CH3COOH}) =0.0200-0.00200 \]

\[ \boxed{0.0180\text{ mol}} \]

Step 3: Calculate the final amount of ethanoate ions.

The amount of ethanoate increases by the amount of ethanoic acid consumed.

\[ n(\mathrm{CH_{3}COO}^{-}) =0.0100+0.00200 \]

\[ \boxed{0.0120\text{ mol}} \]

Step 4: Calculate the final pH.

\[ \mathrm{pH} = 4.76+\log_{10} \left(\frac{0.0120}{0.0180}\right) \]

\[ =4.76+\log_{10}(0.6667) \]

\[ =4.76-0.176 \]

\[ \boxed{\mathrm{pH}=4.58} \]

The pH increases because the ratio of ethanoate ions to ethanoic acid increases.

The mistake to avoid is using the original mole ratio of 0.0100/0.0200. The buffer composition has changed, so the original ratio is no longer relevant.

Question 2

An ethanoic acid/ethanoate buffer contains 0.0150 mol of ethanoic acid and 0.0250 mol of ethanoate ions in 100.0 mL of solution.

A student adds 5.00 mL of 0.400 mol/L HCl.

Calculate the final pH and explain why the increase in solution volume does not require a separate dilution correction when using the mole-ratio form of the Henderson-Hasselbalch equation.

Solution 2

The final buffer contains 0.0170 mol of ethanoic acid and 0.0230 mol of ethanoate ions. Its final pH is 4.89.

Step 1: Calculate the moles of hydrochloric acid added.

\[ n(\mathrm{H_{3}O}^{+})=cV \]

\[ =0.400\times0.00500 \]

\[ =0.00200\text{ mol} \]

Step 2: Identify the neutralisation reaction.

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

The added acid consumes ethanoate ions and produces ethanoic acid.

Step 3: Calculate the updated composition.

Ethanoate ions:

\[ n(\mathrm{CH_{3}COO}^{-}) =0.0250-0.00200 \]

\[ =0.0230\text{ mol} \]

Ethanoic acid:

\[ n(\ce{CH3COOH}) =0.0150+0.00200 \]

\[ =0.0170\text{ mol} \]

Step 4: Calculate the final pH.

\[ \mathrm{pH} = 4.76+\log_{10} \left(\frac{0.0230}{0.0170}\right) \]

\[ =4.76+\log_{10}(1.3529) \]

\[ =4.76+0.131 \]

\[ \boxed{\mathrm{pH}=4.89} \]

The final volume is 105.0 mL.

However, the concentration ratio is:

\[ \frac{[\mathrm{CH_{3}COO}^{-}]}{[\ce{CH3COOH}]} = \frac{0.0230/0.1050}{0.0170/0.1050} \]

The common volume cancels:

\[ =\frac{0.0230}{0.0170} \]

This is why using the mole ratio is convenient.

The pH decreases compared with the initial value because strong acid converts some of the conjugate base into weak acid.

Question 3

A buffer contains 0.0200 mol of weak acid HA and 0.0100 mol of conjugate base A−.

A student adds 6.00 mL of 1.00 mol/L NaOH.

a) Calculate the final pH.

b) Calculate the maximum additional amount of NaOH that can be added before the pH reaches the upper boundary of the conventional buffer range, defined here as \(\mathrm{p}K_a+1\).

Assume the buffer components remain in the same solution and the mole-ratio approximation applies.

Solution 3

The pH after the first addition is 4.82. A further 0.0113 mol of NaOH can be added before the pH reaches the specified upper boundary.

Step 1: Calculate the initial moles of hydroxide added.

\[ n(\mathrm{OH}^{-})=1.00\times0.00600 \]

\[ =0.00600\text{ mol} \]

Step 2: Update the buffer composition.

Weak acid:

\[ n(\mathrm{HA})=0.0200-0.00600 \]

\[ =0.0140\text{ mol} \]

Conjugate base:

\[ n(\mathrm{A^-})=0.0100+0.00600 \]

\[ =0.0160\text{ mol} \]

Step 3: Calculate the new pH.

\[ \mathrm{pH} = 4.76+\log_{10}\left(\frac{0.0160}{0.0140}\right) \]

\[ =4.76+\log_{10}(1.1429) \]

\[ =4.76+0.058 \]

\[ \boxed{\mathrm{pH}=4.82} \]

Step 4: Determine the upper boundary of the buffer range.

The specified upper boundary is:

\[ \mathrm{pH}=\mathrm{p}K_a+1 \]

Therefore:

\[ \mathrm{pH}=5.76 \]

Rearrange the Henderson-Hasselbalch equation:

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

At the upper boundary:

\[ 1=\log_{10}\left(\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}\right) \]

Therefore:

\[ \frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}=10 \]

Step 5: Calculate the additional hydroxide required.

Let \(x\) represent the additional moles of hydroxide added.

The updated amounts will be:

\[ n_{\mathrm{A^-}}=0.0160+x \]

\[ n_{\mathrm{HA}}=0.0140-x \]

Set their ratio equal to 10:

\[ \frac{0.0160+x}{0.0140-x}=10 \]

Rearrange:

\[ 0.0160+x=0.140-10x \]

\[ 11x=0.124 \]

\[ \boxed{x=0.0113\text{ mol}} \]

The buffer can therefore neutralise approximately another 0.0113 mol of hydroxide before reaching the specified upper boundary.

Notice that the calculation uses the updated composition after the first addition, not the original amounts.

Question 4

Two ethanoic acid/ethanoate buffers have the same initial pH and the same volume of 100.0 mL.

Their compositions are:

Both buffers receive 1.00 mL of 1.00 mol/L NaOH.

a) Calculate the final pH of each buffer.

b) Explain why the two buffers experience different pH changes despite having identical initial pH values.

Solution 4

Buffer A finishes at pH 4.85, while buffer B finishes at pH 4.79.

Buffer B experiences a smaller pH change because it contains larger absolute amounts of both buffer components.

Step 1: Calculate the hydroxide added to each buffer.

\[ n(\mathrm{OH}^{-})=1.00\times0.00100 \]

\[ =0.00100\text{ mol} \]

Step 2: Calculate the final composition of buffer A.

Ethanoic acid:

\[ 0.0100-0.00100=0.00900\text{ mol} \]

Ethanoate ions:

\[ 0.0100+0.00100=0.0110\text{ mol} \]

Calculate the pH:

\[ \mathrm{pH}_A = 4.76+\log_{10}\left(\frac{0.0110}{0.00900}\right) \]

\[ \boxed{\mathrm{pH}_A=4.85} \]

Step 3: Calculate the final composition of buffer B.

Ethanoic acid:

\[ 0.0300-0.00100=0.0290\text{ mol} \]

Ethanoate ions:

\[ 0.0300+0.00100=0.0310\text{ mol} \]

Calculate the pH:

\[ \mathrm{pH}_B = 4.76+\log_{10}\left(\frac{0.0310}{0.0290}\right) \]

\[ \boxed{\mathrm{pH}_B=4.79} \]

Step 4: Explain the difference.

Both buffers initially have equal amounts of weak acid and conjugate base.

Therefore, both have an initial pH equal to their pKa:

\[ \mathrm{pH}=4.76 \]

However, buffer A loses 10% of its initial weak acid when the hydroxide is added.

Buffer B loses only approximately 3.33%.

The two buffers have the same initial pH because their ratios are identical. They do not have the same buffer capacity because their total amounts of buffer components are different.

This is an important distinction: pH depends on the ratio of buffer components, while buffer capacity depends on how much of those components is available.

Question 5

Two identical 1.00 L buffer samples each initially contain:

• 0.0100 mol of weak acid HA.
• 0.0100 mol of conjugate base A−.

Their pKa is 4.76.

Two different addition sequences are tested.

Sample X:

1. Add 0.0120 mol of NaOH.
2. After the reaction is complete, add 0.00500 mol of HCl.

Sample Y:

1. Add 0.00500 mol of HCl.
2. After the reaction is complete, add 0.0120 mol of NaOH.

Assume the final volume is 1.00 L, titrant volume changes are negligible for the mole calculations, all strong acid and base reactions proceed to completion, and no other reactions occur.

a) Determine the final moles of HA and A− in sample X.

b) Determine the final moles of HA and A− in sample Y.

c) Calculate the final pH of both samples.

d) A student argues that sample X must have a different final pH because its buffer was completely overwhelmed by the first NaOH addition. Evaluate this claim. Distinguish between the intermediate state and the final equilibrium state.

Solution 5

Both samples finish with 0.00300 mol of HA and 0.0170 mol of A−, giving the same final pH of 5.51. However, sample X temporarily loses its buffering capacity after the first addition, while sample Y remains a buffer throughout the sequence.

This distinction between the path taken and the final composition is the key to the problem.

Step 1: Track sample X after the first NaOH addition.

Initially:

\[ n(\mathrm{HA})=0.0100\text{ mol} \]

\[ n(\mathrm{A^-})=0.0100\text{ mol} \]

The added hydroxide is:

\[ n(\mathrm{OH}^{-})=0.0120\text{ mol} \]

There is more hydroxide than available weak acid.

All 0.0100 mol of HA is consumed, producing 0.0100 mol of additional A−.

Therefore:

\[ n(\mathrm{HA})=0 \]

\[ n(\mathrm{A^-})=0.0200\text{ mol} \]

The excess hydroxide is:

\[ 0.0120-0.0100=0.00200\text{ mol} \]

At this intermediate stage, sample X is no longer a buffer. It contains excess strong base.

Step 2: Add HCl to sample X.

The added strong acid is:

\[ n(\ce{H3O+})=0.00500\text{ mol} \]

The first 0.00200 mol neutralises the excess hydroxide:

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

This leaves:

\[ 0.00500-0.00200=0.00300\text{ mol} \]

of hydrogen ions.

These remaining hydrogen ions react with ethanoate ions:

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

Therefore, 0.00300 mol of A− is converted back into HA.

The final composition of sample X is:

\[ n(\mathrm{HA})=0+0.00300 \]

\[ \boxed{n(\mathrm{HA})=0.00300\text{ mol}} \]

And:

\[ n(\mathrm{A^-})=0.0200-0.00300 \]

\[ \boxed{n(\mathrm{A^-})=0.0170\text{ mol}} \]

Step 3: Track sample Y after the first HCl addition.

The initial conjugate base amount is 0.0100 mol.

The HCl consumes 0.00500 mol of A−:

\[ n(\mathrm{A^-})=0.0100-0.00500 \]

\[ =0.00500\text{ mol} \]

The weak acid amount increases:

\[ n(\mathrm{HA})=0.0100+0.00500 \]

\[ =0.0150\text{ mol} \]

Sample Y still contains both buffer components.

Step 4: Add NaOH to sample Y.

The 0.0120 mol of hydroxide reacts with the available weak acid.

Since 0.0150 mol of HA is available, all the hydroxide can be neutralised.

The final weak acid amount is:

\[ n(\mathrm{HA})=0.0150-0.0120 \]

\[ \boxed{n(\mathrm{HA})=0.00300\text{ mol}} \]

The conjugate base amount becomes:

\[ n(\mathrm{A^-})=0.00500+0.0120 \]

\[ \boxed{n(\mathrm{A^-})=0.0170\text{ mol}} \]

Both samples now have identical compositions.

Step 5: Calculate the final pH.

Since both samples contain the same buffer components, their final pH values must be equal under the stated assumptions.

Use the Henderson-Hasselbalch equation:

\[ \mathrm{pH} = 4.76+\log_{10}\left(\frac{0.0170}{0.00300}\right) \]

\[ =4.76+\log_{10}(5.667) \]

\[ =4.76+0.753 \]

\[ \boxed{\mathrm{pH}=5.51} \]

Step 6: Evaluate the student’s claim.

The student is correct that sample X temporarily loses its buffering capacity. The first addition of NaOH consumes all the available weak acid and leaves excess hydroxide.

However, this does not mean the original buffer components cannot be regenerated by a later acid addition.

The HCl first neutralises the excess hydroxide. Any remaining hydrogen ions then react with the conjugate base, producing weak acid again.

The final composition depends on the net amount of strong acid and strong base added:

\[ n_{\mathrm{net\ OH^-}}=0.0120-0.00500 \]

\[ =0.00700\text{ mol} \]

If 0.00700 mol of NaOH had been added directly to the original buffer, it would consume 0.00700 mol of HA and produce the same final composition:

\[ n(\mathrm{HA})=0.0100-0.00700 =0.00300\text{ mol} \]

\[ n(\mathrm{A^-})=0.0100+0.00700 =0.0170\text{ mol} \]

The final equilibrium state is therefore identical.

However, the intermediate conditions are not identical. Sample X experiences an excess of strong base, whereas sample Y never has excess strong acid or base.

This distinction matters if other substances in the solution undergo irreversible reactions at high or low pH. In that situation, the order of addition could affect the final outcome.

For the ideal buffer system in this question, though, the final composition is determined by the net acid-base equivalents.

08What to learn next

Once you can calculate the updated buffer composition, the next important skill is recognising when a solution has moved beyond the effective buffer range and entered the region where excess strong acid, excess strong base, or conjugate-ion hydrolysis controls its pH.

For more practice with the surrounding concepts, see:

• pH After Dilution and Acid-Base Mixing | HSC Chemistry – particularly useful for distinguishing changes in concentration from changes in buffer composition.
• Weak Acid-Strong Base Titrations: HSC Chemistry Guide – extends the same mole-accounting method to the full titration process, including the equivalence region.

The essential habit to carry into those problems is simple: calculate what reacts first, update the remaining moles, and only then calculate the pH.