Why Buffers Need a Weak Acid and Its Conjugate Base

Two beakers each contain 0.100 mol of ethanoic acid in 1.00 L of solution. One also contains 0.100 mol of sodium ethanoate. You add 0.0100 mol of hydrochloric acid to each beaker. Which solution experiences the smaller pH change?

The solution containing sodium ethanoate. It already has a substantial supply of ethanoate ions, which can react with the added acid. The ethanoic acid solution on its own contains some ethanoate ions too, but only a small amount produced by the acid’s partial ionisation.

This is the reason a weak-acid buffer needs both a weak acid and its conjugate base. They have different jobs: the weak acid reacts with added hydroxide ions, while the conjugate base reacts with added hydronium ions. Together, they resist changes in pH in both directions.

The important detail is that both substances must be present in useful amounts. An equilibrium can shift, but it cannot provide an unlimited supply of either substance.

01The two substances have different jobs

Start with a general weak acid, which we’ll call \(\mathrm{HA}\). Its conjugate base is \(\mathrm{A^-}\).

A weak acid only partially ionises in water. It establishes a reversible equilibrium:

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

The forward reaction produces hydronium ions and conjugate base ions. The reverse reaction allows those ions to recombine and form the weak acid.

A buffer takes advantage of both directions of this chemistry.

What happens when you add an acid?

Suppose you add hydrochloric acid to the buffer. Hydrochloric acid is a strong acid, so it dissociates essentially completely, increasing the concentration of hydronium ions.

The conjugate base reacts with those added hydronium ions:

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

The added acid is converted into more weak acid.

This reduces the increase in hydronium ion concentration, so the pH falls by less than it would in a solution without the conjugate base.

The equilibrium also shifts to the left, favouring the formation of more \(\mathrm{HA}\).

What happens when you add a base?

Now suppose you add sodium hydroxide. The hydroxide ions react with hydronium ions, removing them from the solution.

The weak acid can replace those hydronium ions by reacting with hydroxide:

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

The added base is converted into more conjugate base.

The weak acid equilibrium shifts to the right as the solution responds to the removal of hydronium ions.

This limits the increase in pH.

The two components therefore protect against different disturbances.

A buffer doesn’t prevent the pH from changing. It makes the change smaller.

02Why does the acid need to be weak?

A weak acid is only partially ionised. This means a substantial proportion of its particles remain as undissociated acid molecules.

That matters because these molecules provide a reserve of acid that can react with added hydroxide ions.

Consider ethanoic acid:

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

Its acid dissociation constant at \(25^\circ\mathrm{C}\) is approximately:

\[
K_\mathrm{a}=1.8\times10^{-5}
\]

The small value tells us that ethanoic acid only partially ionises in water.

Most of the acid remains as \(\ce{CH3COOH}\), ready to react with added hydroxide ions.

A strong acid behaves differently. Hydrochloric acid, for example, is essentially completely ionised in water. There is no substantial reserve of undissociated HCl molecules.

Its conjugate base, \(\mathrm{Cl}^{-}\), is also an extremely weak base. It cannot effectively accept added protons.

A strong acid and its conjugate base therefore do not make a useful buffer pair.

This doesn’t mean a strong acid solution cannot neutralise added hydroxide. Its hydronium ions certainly can. The problem is that it cannot provide the two-way response needed to resist both added acid and added base.

Why a weak acid alone is not enough

Here’s a subtle point that often causes confusion.

A weak acid solution already contains some conjugate base because partial ionisation produces it. So why can’t we simply use a weak acid on its own as a buffer?

The problem is the amount of conjugate base available.

Consider a \(0.100\ \mathrm{mol\,L^{-1}}\) ethanoic acid solution.

Let \(x\) represent the concentration of ethanoic acid that ionises. At equilibrium:

• \([\ce{H3O+}]=x\)
• \([\ce{CH3COO-}]=x\)
• \([\ce{CH3COOH}]\approx0.100-x\)

Using the acid dissociation expression:

\[
K_\mathrm{a}
=
\frac{[\ce{H3O+}][\ce{CH3COO-}]}
{[\ce{CH3COOH}]}
\]

Substitute the equilibrium concentrations:

\[
1.8\times10^{-5}
=
\frac{x^2}{0.100-x}
\]

Since the acid is only slightly ionised, \(x\) is much smaller than \(0.100\). We can approximate:

\[
\begin{aligned}
1.8\times10^{-5} &\approx \frac{x^2}{0.100}\\
x &\approx \sqrt{1.8\times10^{-6}}\\
x &\approx 1.34\times10^{-3}\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

So the equilibrium concentration of ethanoate ions is only approximately \(0.00134\ \mathrm{mol\,L^{-1}}\).

Compare that with the original acid concentration of \(0.100\ \mathrm{mol\,L^{-1}}\).

There is plenty of weak acid available to react with added base, but a much smaller amount of conjugate base available to react with added acid.

Adding sodium ethanoate supplies the missing conjugate base:

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

The sodium ions are spectator ions. The important addition is the ethanoate ion.

Now the solution contains both:

• A substantial amount of \(\ce{CH3COOH}\), which can react with added hydroxide.
• A substantial amount of \(\ce{CH3COO-}\), which can react with added hydronium.

That’s the balanced arrangement required for an effective buffer.

The same principle works in reverse for a weak-base buffer: a weak base is paired with an appreciable amount of its conjugate acid.

For more on how these species are related through proton transfer, see Conjugate Acid-Base Pairs and Proton Transfer | HSC Chemistry.

03Equilibrium shifts, but the amounts are finite

Le Chatelier’s principle explains the direction in which a buffer responds. However, it doesn’t mean the buffer can absorb unlimited acid or base.

Each reaction consumes one component and produces the other.

For example, when hydrochloric acid is added:

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

The amount of conjugate base decreases, while the amount of weak acid increases.

When sodium hydroxide is added:

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

The amount of weak acid decreases, while the amount of conjugate base increases.

This gives us a useful way to track the buffer using moles.

Provided neither component is exhausted, the total number of moles of the weak acid and its conjugate base remains approximately constant.

The ratio between them changes, and that ratio determines the pH.

Buffer capacity

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

For a fixed volume of solution:

• The moles of \(\mathrm{A^-}\) determine the available reserve for added acid.
• The moles of \(\mathrm{HA}\) determine the available reserve for added base.

If you add more acid than the available conjugate base can consume, excess acid remains in the solution.

If you add more base than the available weak acid can consume, excess hydroxide remains.

Either way, the buffer approximation stops being useful.

There is also an important distinction between pH and buffer capacity.

The ratio of conjugate base to weak acid largely determines the pH. The total amount of these substances determines how much disturbance the buffer can handle.

A dilute buffer and a concentrated buffer can therefore have the same pH but very different capacities.

06When does the buffer approximation fail?

For an effective weak-acid buffer, we often use the Henderson-Hasselbalch equation:

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

Here:

• \(pH\) is the solution’s acidity.
• \(pK_\mathrm{a}=-\log_{10}K_\mathrm{a}\).
• \([\mathrm{A}^{-}]\) is the conjugate base concentration.
• \([\ce{HA}]\) is the weak acid concentration.

This relationship comes directly from the weak acid equilibrium expression.

It works well for a buffer when both components remain in appreciable amounts and the equilibrium changes are small compared with their existing concentrations.

For calculations involving added strong acids or bases, there is an important order of operations:

1. Calculate the moles of strong acid or strong base added.
2. Allow the strong acid or base to react completely with the appropriate buffer component.
3. Calculate the remaining moles of weak acid and conjugate base.
4. Check that both components are still present in useful amounts.
5. Only then use the buffer equation to estimate the pH.

The concentrations used in the equation should represent the solution after the neutralisation reaction, not the original buffer.

Three situations where you should stop using the shortcut

1. The conjugate base is exhausted.

If added strong acid consumes all the \(\mathrm{A^-}\), the solution no longer has a useful reserve for incoming acid.

If there is no excess strong acid, calculate the pH using the remaining weak acid equilibrium.

If strong acid remains in excess, calculate the pH from the excess strong acid, accounting for any relevant weak acid equilibrium if necessary.

2. The weak acid is exhausted.

If added strong base consumes all the \(\mathrm{HA}\), the solution is no longer a weak-acid buffer.

If no hydroxide remains in excess, the conjugate base may react with water:

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

You must now consider the base dissociation equilibrium, rather than treating the solution as a buffer.

If hydroxide remains in excess, it will generally dominate the pH.

3. One component becomes very small compared with the other.

A buffer doesn’t suddenly stop working only when a component reaches exactly zero.

Its effectiveness decreases as one component is depleted. A solution with a tiny amount of conjugate base may still technically contain both members of the pair, but it has very little capacity to absorb additional acid.

There is another limitation worth remembering: if the solution is extremely dilute, the equilibrium contribution from water may become significant. The simple buffer approximation may then need to be replaced by a fuller equilibrium calculation.

The key distinction is that the buffer equation is a useful model for a solution with two available components. It is not a formula that can be applied regardless of what remains in the beaker.

07Questions and solutions

The following questions build from identifying the two buffer components to tracking their finite reserves. Use the following illustrative data unless a question states otherwise:

• Ethanoic acid: \(K_\mathrm{a}=1.8\times10^{-5}\) at \(25^\circ\mathrm{C}\).
• Water: \(K_\mathrm{w}=1.0\times10^{-14}\) at \(25^\circ\mathrm{C}\).
• \(pK_\mathrm{a}=4.74\).
• Assume ideal solution behaviour and complete neutralisation by strong acids and bases.

Question 1

A \(1.00\ \mathrm{L}\) buffer contains \(0.0400\) mol of ethanoic acid and \(0.0600\) mol of sodium ethanoate.

A student adds \(0.0100\) mol of sodium hydroxide. Assume the final volume remains \(1.00\ \mathrm{L}\).

Calculate the final pH of the buffer. Explain why the pH increases.

Solution 1

The final pH is approximately \(5.11\).

The hydroxide ions react with ethanoic acid, converting it into ethanoate ions.

Step 1: Write the neutralisation reaction.

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

Step 2: Calculate the new mole amounts.

Initially:

\[
n(\ce{CH3COOH})=0.0400\ \mathrm{mol}
\]

\[
n(\ce{CH3COO-})=0.0600\ \mathrm{mol}
\]

The added \(0.0100\) mol of hydroxide consumes \(0.0100\) mol of ethanoic acid.

Therefore:

\[
n(\ce{CH3COOH})=0.0400-0.0100
=0.0300\ \mathrm{mol}
\]

\[
n(\ce{CH3COO-})=0.0600+0.0100
=0.0700\ \mathrm{mol}
\]

Step 3: Calculate the final concentrations.

The final volume is \(1.00\ \mathrm{L}\).

\[
[\ce{CH3COOH}]=0.0300\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{CH3COO-}]=0.0700\ \mathrm{mol\,L^{-1}}
\]

Step 4: Calculate the pH.

Using the buffer equation:

\[
\begin{aligned}
pH&=pK_\mathrm{a}
+\log_{10}\left(
\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}
\right)\\
&=4.74+\log_{10}\left(\frac{0.0700}{0.0300}\right)\\
&=4.74+0.368\\
&=5.11
\end{aligned}
\]

The pH increases because the hydroxide ions consume ethanoic acid and increase the proportion of conjugate base.

The buffer still works because \(0.0300\) mol of ethanoic acid remains available to neutralise further added base.

Question 2

A \(250.0\ \mathrm{mL}\) buffer contains:

• \(0.0200\) mol of ethanoic acid.
• \(0.0300\) mol of sodium ethanoate.

A student adds \(10.0\ \mathrm{mL}\) of \(0.500\ \mathrm{mol\,L^{-1}}\) hydrochloric acid. They then dilute the resulting solution to a final volume of \(500.0\ \mathrm{mL}\).

Calculate the final pH.

The student claims that dilution must reduce the buffer’s pH because both buffer components become less concentrated. Explain why this prediction is incorrect. Also compare the acid-neutralising reserve in a \(100.0\ \mathrm{mL}\) sample before and after dilution.

Solution 2

The final pH is approximately \(4.74\). Dilution does not significantly change the pH because it reduces both buffer concentrations by the same factor, leaving their ratio unchanged.

However, a \(100.0\ \mathrm{mL}\) portion of the diluted solution has a smaller acid-neutralising reserve.

Step 1: Calculate the moles of hydrochloric acid added.

Convert the volume into litres:

\[
10.0\ \mathrm{mL}=0.0100\ \mathrm{L}
\]

Then:

\[
\begin{aligned}
n&=cV\\
&=0.500\times0.0100\\
&=0.00500\ \mathrm{mol}
\end{aligned}
\]

Step 2: Calculate the new buffer composition.

Hydrochloric acid reacts with ethanoate ions:

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

Therefore:

\[
n(\ce{CH3COO-})=0.0300-0.00500
=0.0250\ \mathrm{mol}
\]

\[
n(\ce{CH3COOH})=0.0200+0.00500
=0.0250\ \mathrm{mol}
\]

Both components remain in equal amounts.

Step 3: Account for the dilution.

The final volume is \(0.5000\ \mathrm{L}\).

Therefore:

\[
[\ce{CH3COOH}]
=
\frac{0.0250}{0.5000}
=0.0500\ \mathrm{mol\,L^{-1}}
\]

\[
[\ce{CH3COO-}]
=
\frac{0.0250}{0.5000}
=0.0500\ \mathrm{mol\,L^{-1}}
\]

Step 4: Calculate the pH.

\[
\begin{aligned}
pH&=4.74+\log_{10}\left(\frac{0.0500}{0.0500}\right)\\
&=4.74+\log_{10}(1)\\
&=4.74
\end{aligned}
\]

The pH remains approximately \(4.74\).

Both concentrations have decreased, but their ratio is still one.

Step 5: Compare the acid-neutralising reserves.

After neutralisation but before dilution, the solution volume is:

\[
250.0+10.0=260.0\ \mathrm{mL}
\]

The ethanoate concentration at this point is:

\[
[\ce{CH3COO-}]
=
\frac{0.0250}{0.2600}
=0.0962\ \mathrm{mol\,L^{-1}}
\]

A \(100.0\ \mathrm{mL}\) sample would contain:

\[
n=0.0962\times0.1000
=0.00962\ \mathrm{mol}
\]

After dilution, a \(100.0\ \mathrm{mL}\) sample contains:

\[
n=0.0500\times0.1000
=0.00500\ \mathrm{mol}
\]

The diluted sample therefore contains only about half as many moles of ethanoate ions.

The distinction is important: the entire flask still contains \(0.0250\) mol of ethanoate ions, so its total stoichiometric acid-neutralising reserve has not changed. But a fixed \(100.0\ \mathrm{mL}\) portion has less reserve after dilution.

This is why pH and buffer capacity must be considered separately.

Question 3

A student prepares \(1.00\ \mathrm{L}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) ethanoic acid but does not add any sodium ethanoate.

They claim that the solution is already an effective buffer because the weak acid equilibrium produces ethanoate ions, and any added hydrochloric acid will simply shift the equilibrium to the left to replace them.

Given:

\[
K_\mathrm{a}=1.8\times10^{-5}
\]

Calculate the approximate initial concentration and number of moles of ethanoate ions.

The student then adds \(0.0100\) mol of hydrochloric acid. Assess their reasoning.

Solution 3

The initial solution contains only approximately \(1.34\times10^{-3}\) mol of ethanoate ions. It is therefore not an effective buffer against the proposed acid addition.

The student’s mistake is assuming that shifting equilibrium to the left creates more conjugate base. In fact, the reverse reaction consumes conjugate base.

Step 1: Write the weak acid equilibrium.

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

Step 2: Calculate the initial ethanoate concentration.

Let \(x\) represent the concentration of acid that ionises.

Using the weak acid approximation:

\[
K_\mathrm{a}\approx\frac{x^2}{0.100}
\]

Therefore:

\[
\begin{aligned}
x&\approx\sqrt{1.8\times10^{-5}\times0.100}\\
&=1.34\times10^{-3}\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Since the ethanoate concentration equals the amount of acid ionised:

\[
[\ce{CH3COO-}]
\approx1.34\times10^{-3}\ \mathrm{mol\,L^{-1}}
\]

Step 3: Calculate the initial moles of ethanoate.

\[
\begin{aligned}
n&=cV\\
&=1.34\times10^{-3}\times1.00\\
&=1.34\times10^{-3}\ \mathrm{mol}
\end{aligned}
\]

The solution initially contains approximately \(0.00134\) mol of ethanoate ions.

Step 4: Compare this with the hydrochloric acid addition.

The student adds \(0.0100\) mol of hydrochloric acid.

This is approximately:

\[
\frac{0.0100}{0.00134}=7.46
\]

So the added acid represents about 7.5 times the initial amount of conjugate base.

The reverse reaction is:

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

It consumes ethanoate ions. It does not produce more of them.

The equilibrium shifts to the left, reducing the concentration of ethanoate ions further.

The solution has a substantial reserve of ethanoic acid, so it can react with added hydroxide. However, it does not have enough conjugate base to provide effective resistance against the much larger acid addition.

The student’s prediction is therefore incorrect. The presence of some conjugate base is not enough. An effective buffer needs an appreciable amount of both components.

Question 4

A \(1.00\ \mathrm{L}\) buffer contains:

• \(0.0500\) mol of ethanoic acid.
• \(0.100\) mol of sodium ethanoate.

A student adds \(50.0\ \mathrm{mL}\) of \(1.00\ \mathrm{mol\,L^{-1}}\) sodium hydroxide. The final solution volume is \(1.05\ \mathrm{L}\).

The student attempts to use the Henderson-Hasselbalch equation to calculate the final pH.

Explain why this is no longer appropriate and calculate the actual approximate pH of the resulting solution.

Use:

\[
K_\mathrm{w}=1.0\times10^{-14}
\]

at \(25^\circ\mathrm{C}\).

Solution 4

The resulting solution has a pH of approximately \(8.95\), and it is no longer a buffer.

The added hydroxide consumes all the ethanoic acid. The remaining ethanoate ions behave as a weak base and establish a new equilibrium with water.

Step 1: Calculate the moles of sodium hydroxide added.

\[
\begin{aligned}
n&=cV\\
&=1.00\times0.0500\\
&=0.0500\ \mathrm{mol}
\end{aligned}
\]

Step 2: Determine the final buffer component amounts.

The neutralisation reaction is:

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

Initially, there are \(0.0500\) mol of ethanoic acid.

The added hydroxide also amounts to \(0.0500\) mol.

Therefore, all the ethanoic acid is consumed:

\[
n(\ce{CH3COOH})=0
\]

The final amount of ethanoate is:

\[
n(\ce{CH3COO-})=0.100+0.0500
=0.150\ \mathrm{mol}
\]

Only the conjugate base remains in appreciable concentration.

Step 3: Calculate the ethanoate concentration.

The final volume is \(1.05\ \mathrm{L}\).

\[
\begin{aligned}
[\ce{CH3COO-}]
&=\frac{0.150}{1.05}\\
&=0.1429\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Step 4: Calculate the base dissociation constant.

Ethanoate ions react with water:

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

The relationship between the acid and base dissociation constants is:

\[
K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}
\]

Therefore:

\[
\begin{aligned}
K_\mathrm{b}
&=\frac{K_\mathrm{w}}{K_\mathrm{a}}\\
&=\frac{1.0\times10^{-14}}{1.8\times10^{-5}}\\
&=5.56\times10^{-10}
\end{aligned}
\]

Step 5: Calculate the hydroxide concentration.

Let \(x\) represent the concentration of ethanoate ions that react with water.

Using the weak base approximation:

\[
K_\mathrm{b}\approx\frac{x^2}{0.1429}
\]

Therefore:

\[
\begin{aligned}
x&\approx\sqrt{5.56\times10^{-10}\times0.1429}\\
&=8.91\times10^{-6}\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Thus:

\[
[\ce{OH-}]=8.91\times10^{-6}\ \mathrm{mol\,L^{-1}}
\]

Step 6: Calculate the pH.

First, calculate the pOH:

\[
\begin{aligned}
pOH&=-\log_{10}(8.91\times10^{-6})\\
&=5.05
\end{aligned}
\]

At \(25^\circ\mathrm{C}\):

\[
pH+pOH=14.00
\]

Therefore:

\[
pH=14.00-5.05=8.95
\]

The final pH is approximately \(8.95\).

The important point is that the buffer equation cannot be used when one component has been completely consumed. The ethanoate ions now determine the pH through their reaction with water.

Question 5

A \(1.00\ \mathrm{L}\) buffer initially contains:

• \(0.100\) mol of a weak acid, \(\mathrm{HA}\).
• \(0.0600\) mol of its conjugate base, \(\mathrm{A^-}\).

Two students add the same amounts of strong acid and strong base, but in different orders.

Student X adds \(0.0800\) mol of HCl first, followed by \(0.0500\) mol of NaOH.

Student Y adds \(0.0500\) mol of NaOH first, followed by \(0.0800\) mol of HCl.

Assume:

• Both additions have negligible volume.
• The final volume remains \(1.00\ \mathrm{L}\).
• Both strong reagents react completely.
• No substances are removed from the solution.
• There are no other reactions.

Answer the following:

(a) Track the moles of \(\mathrm{HA}\) and \(\mathrm{A^-}\) after each addition for both students.

(b) Identify whether the solution remains a buffer at every stage for each student.

(c) Calculate the final pH for both students using \(K_\mathrm{a}=1.8\times10^{-5}\).

(d) Explain why the two students obtain the same final pH despite one of them temporarily destroying the buffer.

Solution 5

Both students finish with \(0.130\) mol of weak acid and \(0.0300\) mol of conjugate base. Their final pH is approximately \(4.11\).

However, Student X temporarily destroys the buffer, while Student Y maintains a buffer throughout both additions.

The key is to distinguish the final chemical composition from the path taken to reach it.

Part (a): Track the mole amounts.

Student X: HCl first, then NaOH

Initially:

\[
n(\mathrm{HA})=0.100\ \mathrm{mol}
\]

\[
n(\mathrm{A^-})=0.0600\ \mathrm{mol}
\]

First, \(0.0800\) mol of HCl is added.

The reaction is:

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

Only \(0.0600\) mol of conjugate base is available.

Therefore, all the conjugate base is consumed, producing \(0.0600\) mol of additional weak acid.

The amounts become:

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

\[
n(\mathrm{HA})=0.100+0.0600
=0.160\ \mathrm{mol}
\]

There is also excess hydrochloric acid:

\[
n(\text{excess HCl})=0.0800-0.0600
=0.0200\ \mathrm{mol}
\]

At this stage, the solution is no longer a buffer.

Now add \(0.0500\) mol of NaOH.

The hydroxide first neutralises the excess strong acid:

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

This consumes \(0.0200\) mol of hydroxide.

The remaining hydroxide is:

\[
0.0500-0.0200=0.0300\ \mathrm{mol}
\]

This remaining hydroxide reacts with the ethanoic acid:

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

The final amounts are:

\[
n(\mathrm{HA})=0.160-0.0300
=0.130\ \mathrm{mol}
\]

\[
n(\mathrm{A^-})=0+0.0300
=0.0300\ \mathrm{mol}
\]

Student Y: NaOH first, then HCl

Initially, the same amounts are present:

\[
n(\mathrm{HA})=0.100\ \mathrm{mol}
\]

\[
n(\mathrm{A^-})=0.0600\ \mathrm{mol}
\]

First, \(0.0500\) mol of NaOH is added.

The hydroxide reacts with the weak acid:

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

The new amounts are:

\[
n(\mathrm{HA})=0.100-0.0500
=0.0500\ \mathrm{mol}
\]

\[
n(\mathrm{A^-})=0.0600+0.0500
=0.110\ \mathrm{mol}
\]

Both components remain, so the solution is still a buffer.

Next, \(0.0800\) mol of HCl is added.

The hydronium ions react with the conjugate base:

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

The final amounts become:

\[
n(\mathrm{A^-})=0.110-0.0800
=0.0300\ \mathrm{mol}
\]

\[
n(\mathrm{HA})=0.0500+0.0800
=0.130\ \mathrm{mol}
\]

Both components remain, so the solution is still a buffer.

Part (b): Does the buffer survive throughout?

Student X temporarily loses the buffer when all \(0.0600\) mol of conjugate base is consumed. The remaining \(0.0200\) mol of hydrochloric acid creates an excess of strong acid.

Student Y never exhausts either component. The buffer remains functional after the first addition and after the second.

This difference matters because the intermediate pH values are very different.

Part (c): Calculate the final pH.

Both students finish with:

\[
n(\mathrm{HA})=0.130\ \mathrm{mol}
\]

\[
n(\mathrm{A^-})=0.0300\ \mathrm{mol}
\]

The final volume is \(1.00\ \mathrm{L}\), so:

\[
[\mathrm{HA}]=0.130\ \mathrm{mol\,L^{-1}}
\]

\[
[\mathrm{A^-}]=0.0300\ \mathrm{mol\,L^{-1}}
\]

Using the buffer equation:

\[
\begin{aligned}
pH&=pK_\mathrm{a}
+\log_{10}\left(
\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}
\right)\\
&=4.74+\log_{10}\left(\frac{0.0300}{0.130}\right)\\
&=4.74-0.636\\
&=4.10
\end{aligned}
\]

Using the unrounded value \(pK_\mathrm{a}=4.7447\), the final pH is approximately \(4.11\).

Both students therefore finish at the same pH.

Part (d): Explain why the order doesn’t change the final result.

The tempting prediction is that Student X must finish with a different pH because the first addition destroyed the buffer.

But this confuses an intermediate state with the final state.

In Student X’s solution, the second addition of sodium hydroxide first neutralises the excess hydrochloric acid. Only the remaining hydroxide reacts with the weak acid.

In Student Y’s solution, the hydroxide converts some weak acid into conjugate base first. The subsequent hydrochloric acid then converts some of that conjugate base back into weak acid.

In both cases, the overall difference between the strong acid and strong base additions is:

\[
0.0800-0.0500=0.0300\ \mathrm{mol}
\]

There is a net addition of \(0.0300\) mol of strong acid equivalents.

This converts \(0.0300\) mol of conjugate base into weak acid, giving the same final composition:

\[
0.100+0.0300=0.130\ \mathrm{mol\ HA}
\]

\[
0.0600-0.0300=0.0300\ \mathrm{mol\ A^-}
\]

No material has been lost, and the same final volume is assumed. Therefore, the final equilibrium composition and pH are the same.

The surprising part is that Student X’s solution can temporarily become strongly acidic, even though it eventually returns to a functioning buffer.

A buffer’s behaviour depends on the amounts of its components at a particular moment, not simply on what was present when the solution was first prepared.

08Where to go next

Once you understand why both components are necessary, the next step is learning to track their mole amounts accurately after an addition of strong acid or base.

That is the foundation of almost every quantitative buffer question.

Continue with How to Calculate Buffer Composition After Adding Strong Acid or Base.

The central rule to carry into those calculations is simple: a buffer only works while it has enough of the appropriate conjugate partner to consume the disturbance.