Why Buffer pH Is Close to pKa When Acid and Base Amounts Are Similar

Imagine titrating ethanoic acid with sodium hydroxide. Halfway to the equivalence point, the solution is still acidic, but its pH is approximately 4.74 – the \(pK_a\) of ethanoic acid. Why does that happen?

The reason is the relative amounts of a weak acid and its conjugate base. At half-equivalence, half the original acid has been converted into its conjugate base, leaving approximately equal amounts of each. When the ratio of conjugate base to weak acid is 1, its logarithm is zero, so the pH equals the \(pK_a\).

If the amounts are similar rather than exactly equal, the logarithmic term is small. The pH therefore stays close to the \(pK_a\).

Before looking at the mathematics, make one prediction. If we doubled the amounts of both the weak acid and its conjugate base, would the pH double?

No. Their ratio would remain unchanged, so the pH would remain approximately the same. However, the buffer would have a larger supply of acid and base available to resist changes in pH. This distinction between the ratio of the components and their actual amounts is central to understanding buffers.

01Why similar acid and base amounts give a pH close to pKa

Consider a buffer containing a weak acid, \(\ce{HA}\), and its conjugate base, \(\mathrm{A}^{-}\).

The weak acid establishes an equilibrium with water:

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

The acid dissociation constant is:

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

Rearranging to make hydronium concentration the subject:

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

Now take the negative logarithm of both sides. Using \(pK_a=-\log_{10}K_a\), we obtain:

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

This is the Henderson-Hasselbalch equation.

The important feature is the ratio:

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

It compares the concentration of the conjugate base with the concentration of the weak acid.

When these concentrations are equal:

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

Therefore:

\[
\begin{aligned}
\mathrm{pH}
&=pK_a+\log_{10}(1)\\
&=pK_a+0\\
&=\boxed{pK_a}
\end{aligned}
\]

That is the mathematical reason equal acid and conjugate base concentrations produce a pH approximately equal to the \(pK_a\).

Notice that this does not mean the solution is neutral. It simply means the two buffer components have similar concentrations.

For ethanoic acid, with \(pK_a=4.74\), an equal-concentration buffer has a pH close to 4.74, not 7.

How close is close?

The logarithm changes slowly when the ratio is near 1.

Consider an ethanoic acid buffer with a \(pK_a\) of 4.74.

Notice what happens when the ratio changes from 0.80 to 1.25. The pH changes by only about 0.20 units across those two cases.

This is why a buffer can contain slightly more weak acid than conjugate base, or slightly more conjugate base than weak acid, while still having a pH close to its \(pK_a\).

A useful way to remember the relationship is:

• More weak acid than conjugate base: pH is below \(pK_a\).
• Equal concentrations: pH is approximately equal to \(pK_a\).
• More conjugate base than weak acid: pH is above \(pK_a\).

Why you can often use mole amounts instead of concentrations

You might notice that buffer calculations often use moles directly in the Henderson-Hasselbalch equation.

Why is that allowed?

For two substances in the same final solution:

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

and

\[
[\ce{HA}]=\frac{n(\ce{HA})}{V}
\]

Dividing these expressions gives:

\[
\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}
=
\frac{n(\mathrm{A}^{-})}{n(\ce{HA})}
\]

The common solution volume cancels.

This means you can use the mole ratio directly, provided you have correctly accounted for any reactions that occurred before calculating the pH.

There is one important qualification. The Henderson-Hasselbalch equation uses equilibrium concentrations. In ordinary HSC buffer calculations, we approximate these using the mole amounts remaining after neutralisation. This works when the equilibrium changes are small compared with the amounts of buffer components present.

The distinction becomes important when a buffer component is almost exhausted or when the solution is extremely dilute.

02Why buffer components have finite amounts

A buffer does not make added acid or base disappear. Instead, its two components react with the added substance and convert it into the other buffer component.

Think of the weak acid and conjugate base as two chemical reserves.

What happens when strong acid is added?

Suppose hydrochloric acid is added to an ethanoic acid/ethanoate buffer.

Hydrochloric acid ionises essentially completely in water, producing hydronium ions.

The added hydronium ions react with the conjugate base:

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

Two changes occur:

• The amount of ethanoate ions decreases.
• The amount of ethanoic acid increases.

The ratio of conjugate base to weak acid therefore decreases.

According to the Henderson-Hasselbalch equation, this causes the pH to decrease.

The weak acid equilibrium also shifts to the left, favouring the formation of more undissociated ethanoic acid.

The buffer resists the pH change, but it does not prevent it entirely.

What happens when strong base is added?

Now consider adding sodium hydroxide.

Sodium hydroxide dissociates to produce hydroxide ions, which react with the weak acid:

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

This time:

• The amount of ethanoic acid decreases.
• The amount of ethanoate ions increases.

The conjugate base-to-acid ratio increases, so the pH rises.

The equilibrium shifts to the right as hydronium ions are removed through neutralisation.

The buffer resists the increase in pH, but again, its ability to do so depends on how much weak acid remains.

The key point is that the two buffer components are finite. If enough strong acid is added, all the conjugate base can be consumed. If enough strong base is added, all the weak acid can be consumed.

Once one component is exhausted, the solution can no longer behave as an effective buffer in that direction.

For a step-by-step method of calculating these changes, see How to Calculate Buffer Composition After Adding Strong Acid or Base.

03An experiment that shows why pH equals pKa at half-equivalence

There is a useful experimental way to investigate this relationship: titrate a weak acid with a strong base and measure the pH as the titration progresses.

The important question is:

At what point in the titration should the pH equal the \(pK_a\)?

The prediction is that it should happen at half-equivalence, when equal amounts of weak acid and conjugate base remain.

Setting up the experiment

Consider a titration using ethanoic acid and sodium hydroxide.

A practical arrangement consists of:

• A burette containing \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.
• A conical flask containing \(25.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) ethanoic acid.
• A magnetic stirrer beneath the flask to mix the solution.
• A calibrated pH probe immersed in the solution and connected to a pH meter.

As sodium hydroxide is released from the burette, it enters the acid solution. The stirrer distributes it throughout the flask, while the pH probe measures the changing pH.

The burette tells us how much sodium hydroxide has been added. Its known concentration allows us to calculate the number of moles delivered.

The pH meter tells us the pH at each stage.

A first attempt might use universal indicator or indicator paper. However, colour changes provide only approximate pH values and make it difficult to identify the precise half-equivalence pH.

Using a calibrated pH probe gives a much more useful numerical record. The probe should be calibrated with appropriate standard buffer solutions, and the temperature should remain reasonably constant.

For more detail about pH probes, calibration, and indicator methods, see Measuring pH of Acids and Bases: HSC Chemistry Guide.

What should happen during the titration?

First, calculate the initial amount of ethanoic acid:

\[
\begin{aligned}
n&=cV\\
&=0.100\times0.0250\\
&=0.00250\ \mathrm{mol}
\end{aligned}
\]

Therefore, the flask initially contains 2.50 mmol of ethanoic acid.

The neutralisation reaction is:

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

The mole ratio is 1:1.

At the equivalence point, all 2.50 mmol of ethanoic acid have reacted. Therefore, 2.50 mmol of hydroxide ions must have been added.

At a sodium hydroxide concentration of \(0.100\ \mathrm{mol\,L^{-1}}\):

\[
V=\frac{n}{c}
=\frac{0.00250}{0.100}
=0.0250\ \mathrm{L}
\]

The equivalence volume is 25.0 mL.

Half-equivalence therefore occurs when:

\[
V_{\mathrm{NaOH}}=\frac{25.0}{2}
=12.5\ \mathrm{mL}
\]

At this point, only 1.25 mmol of hydroxide ions have been added.

The reaction has converted 1.25 mmol of ethanoic acid into 1.25 mmol of ethanoate ions.

The remaining amounts are:

\[
n(\ce{CH3COOH})=1.25\ \mathrm{mmol}
\]

\[
n(\mathrm{CH_{3}COO}^{-})=1.25\ \mathrm{mmol}
\]

The ratio is 1, so:

\[
\boxed{\mathrm{pH}\approx pK_a=4.74}
\]

This is the experimental prediction.

What should the measurements look like?

The following table contains illustrative values calculated using the stoichiometric mole changes and the Henderson-Hasselbalch approximation. They are predictions, not measurements from an actual laboratory experiment.

The pH should rise gradually through the buffer region. At half-equivalence, it should be close to the \(pK_a\).

The experiment supports the relationship between pH and the relative amounts of weak acid and conjugate base.

However, the measured pH might not be exactly 4.74. Temperature, electrode calibration, solution activities, and experimental uncertainties can all affect the result.

There is also a more important limitation: the Henderson-Hasselbalch approximation becomes unreliable as the titration approaches equivalence and the weak acid is almost completely consumed.

04Where the buffer approximation fails

The most important boundary is the point where one buffer component becomes too small to provide meaningful buffering.

Consider what happens at the equivalence point of our ethanoic acid titration.

All the original ethanoic acid has reacted with sodium hydroxide:

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

The solution contains ethanoate ions, but essentially no remaining ethanoic acid from the original buffer.

The Henderson-Hasselbalch equation would require:

\[
\mathrm{pH}
=
pK_a+\log_{10}
\left(\frac{[\mathrm{CH_{3}COO}^{-}]}{[\ce{CH3COOH}]}\right)
\]

The denominator approaches zero.

This does not mean that the pH becomes infinite. It means the buffer approximation is no longer appropriate.

The solution is now governed by the reaction of ethanoate ions with water:

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

This reaction produces hydroxide ions, making the solution basic.

A useful decision rule: Should you use Henderson-Hasselbalch?

Before substituting into the equation, check the composition of the solution.

A common practical guideline is that a buffer works most effectively when:

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

This corresponds approximately to:

\[
pK_a-1\leq\mathrm{pH}\leq pK_a+1
\]

It is a useful working range, not an absolute boundary at which the Henderson-Hasselbalch equation suddenly stops working.

The distinction is that a buffer becomes less effective as one component becomes much smaller than the other. Eventually, one reserve is exhausted and the solution can no longer resist further additions in that direction.

05Questions and solutions

Question 1

An ethanoic acid buffer contains 10.0 mmol of ethanoic acid and 9.00 mmol of ethanoate ions.

Given that \(pK_a=4.20\), calculate the approximate pH of the buffer.

Explain why the pH is close to the \(pK_a\).

Solution 1

The approximate pH is 4.15. The ethanoic acid amount is slightly greater than the ethanoate amount, so the pH is slightly below the \(pK_a\).

Step 1: Identify the mole ratio.

\[
\frac{n(\mathrm{CH_{3}COO}^{-})}{n(\ce{CH3COOH})}
=
\frac{9.00}{10.0}
=0.900
\]

Step 2: Apply the Henderson-Hasselbalch equation.

\[
\begin{aligned}
\mathrm{pH}
&=4.20+\log_{10}(0.900)\\
&=4.20-0.0458\\
&=\boxed{4.15}
\end{aligned}
\]

The ratio is close to 1, so the logarithmic term is small.

The pH is therefore close to the \(pK_a\), as expected for a buffer with similar amounts of weak acid and conjugate base.

Question 2

A \(30.0\ \mathrm{mL}\) sample of \(0.0800\ \mathrm{mol\,L^{-1}}\) monoprotic weak acid, HA, is titrated with \(0.100\ \mathrm{mol\,L^{-1}}\) sodium hydroxide.

The acid has a \(pK_a\) of 4.76.

Calculate:

a) The volume of sodium hydroxide required to reach equivalence.

b) The volume required to reach half-equivalence.

c) The approximate pH after \(18.0\ \mathrm{mL}\) of sodium hydroxide has been added.

Solution 2

The equivalence volume is 24.0 mL, the half-equivalence volume is 12.0 mL, and the pH after 18.0 mL of NaOH has been added is approximately 5.24.

Step 1: Calculate the initial moles of weak acid.

\[
\begin{aligned}
n(\ce{HA})
&=cV\\
&=0.0800\times0.0300\\
&=0.00240\ \mathrm{mol}\\
&=2.40\ \mathrm{mmol}
\end{aligned}
\]

Step 2: Calculate the equivalence volume.

The reaction is:

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

The mole ratio is 1:1, so 2.40 mmol of hydroxide ions are required.

\[
\begin{aligned}
V
&=\frac{n}{c}\\
&=\frac{0.00240}{0.100}\\
&=0.0240\ \mathrm{L}\\
&=\boxed{24.0\ \mathrm{mL}}
\end{aligned}
\]

Step 3: Calculate the half-equivalence volume.

\[
V_{\mathrm{half}}
=\frac{24.0}{2}
=\boxed{12.0\ \mathrm{mL}}
\]

At this point, half the original acid has reacted, leaving equal amounts of HA and A-.

Therefore:

\[
\mathrm{pH}\approx pK_a=4.76
\]

Step 4: Calculate the mole amounts after 18.0 mL of NaOH has been added.

The moles of hydroxide added are:

\[
\begin{aligned}
n(\mathrm{OH}^{-})
&=0.100\times0.0180\\
&=0.00180\ \mathrm{mol}\\
&=1.80\ \mathrm{mmol}
\end{aligned}
\]

The neutralisation reaction consumes 1.80 mmol of HA and produces 1.80 mmol of A-.

Therefore:

\[
n(\ce{HA})=2.40-1.80=0.60\ \mathrm{mmol}
\]

\[
n(\mathrm{A}^{-})=1.80\ \mathrm{mmol}
\]

Step 5: Calculate the pH.

\[
\begin{aligned}
\mathrm{pH}
&=4.76+\log_{10}\left(\frac{1.80}{0.60}\right)\\
&=4.76+\log_{10}(3)\\
&=4.76+0.477\\
&=\boxed{5.24}
\end{aligned}
\]

The pH is higher than the \(pK_a\) because there is now three times as much conjugate base as weak acid.

Question 3

A buffer initially contains:

• 15.0 mmol of a weak acid, HA.
• 5.00 mmol of its conjugate base, A-.

The \(pK_a\) of the acid is 4.74.

A student adds 4.00 mmol of sodium hydroxide.

Calculate:

a) The initial pH of the buffer.

b) The new pH after the sodium hydroxide has reacted completely.

c) Explain why the pH does not increase by the same amount that would be expected if the sodium hydroxide were added to water alone.

Solution 3

The initial pH is 4.26, and the final pH is approximately 4.65. The sodium hydroxide converts some weak acid into conjugate base, changing the ratio without leaving a large excess of hydroxide ions.

Step 1: Calculate the initial pH.

\[
\begin{aligned}
\mathrm{pH}
&=4.74+\log_{10}\left(\frac{5.00}{15.0}\right)\\
&=4.74+\log_{10}(0.333)\\
&=4.74-0.477\\
&=\boxed{4.26}
\end{aligned}
\]

Step 2: Identify the neutralisation reaction.

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

One mole of hydroxide ions reacts with one mole of weak acid.

The added 4.00 mmol of hydroxide therefore consumes 4.00 mmol of HA.

Step 3: Calculate the new buffer amounts.

The remaining weak acid is:

\[
n(\ce{HA})=15.0-4.00=11.0\ \mathrm{mmol}
\]

The new conjugate base amount is:

\[
n(\mathrm{A}^{-})=5.00+4.00=9.00\ \mathrm{mmol}
\]

Step 4: Calculate the new pH.

\[
\begin{aligned}
\mathrm{pH}
&=4.74+\log_{10}\left(\frac{9.00}{11.0}\right)\\
&=4.74+\log_{10}(0.818)\\
&=4.74-0.087\\
&=\boxed{4.65}
\end{aligned}
\]

Step 5: Explain the buffering effect.

The added hydroxide ions do not remain in solution as a large excess. They react with the weak acid to form water and its conjugate base.

The pH rises because the ratio of conjugate base to weak acid increases, but the change is limited because both buffer components remain present.

The buffer has resisted the pH increase rather than preventing it entirely.

Question 4

An ethanoic acid/ethanoate buffer has a measured pH of 5.04 at \(25^\circ\mathrm{C}\).

Given:

\[
pK_a=4.74
\]

The buffer initially contains 10.0 mmol of ethanoic acid.

a) Calculate the initial amount of ethanoate ions.

b) A student adds 5.00 mmol of hydrochloric acid. Calculate the new amounts of ethanoic acid and ethanoate ions.

c) Calculate the new approximate pH.

Explain why the final pH is close to the \(pK_a\).

Solution 4

The initial amount of ethanoate ions is approximately 20.0 mmol. After adding hydrochloric acid, both buffer components contain 15.0 mmol, giving a final pH of approximately 4.74.

Step 1: Rearrange the Henderson-Hasselbalch equation to find the initial ratio.

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

Substituting:

\[
5.04=4.74+
\log_{10}\left(\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}\right)
\]

Therefore:

\[
0.30=
\log_{10}\left(\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}\right)
\]

Taking the antilogarithm:

\[
\frac{[\mathrm{A}^{-}]}{[\ce{HA}]}
=10^{0.30}
\approx2.00
\]

Step 2: Calculate the initial amount of ethanoate.

The conjugate base amount is twice the weak acid amount.

\[
n(\mathrm{CH_{3}COO}^{-})
=2.00\times10.0
\]

\[
=\boxed{20.0\ \mathrm{mmol}}
\]

Step 3: Calculate the reaction with hydrochloric acid.

Hydrochloric acid produces hydronium ions, which react with ethanoate:

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

The reaction consumes 5.00 mmol of ethanoate and produces 5.00 mmol of ethanoic acid.

The new amounts are:

\[
n(\mathrm{CH_{3}COO}^{-})=20.0-5.00=15.0\ \mathrm{mmol}
\]

\[
n(\ce{CH3COOH})=10.0+5.00=15.0\ \mathrm{mmol}
\]

Step 4: Calculate the final pH.

\[
\begin{aligned}
\mathrm{pH}
&=4.74+\log_{10}\left(\frac{15.0}{15.0}\right)\\
&=4.74+\log_{10}(1)\\
&=\boxed{4.74}
\end{aligned}
\]

The final pH is approximately equal to the \(pK_a\) because the amounts of weak acid and conjugate base are now equal.

Notice that adding strong acid has moved the buffer towards an equal ratio. The pH has therefore decreased from 5.04 to 4.74.

Question 5

An ethanoic acid/ethanoate buffer is prepared by mixing:

• \(50.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) ethanoic acid.
• \(50.0\ \mathrm{mL}\) of \(0.100\ \mathrm{mol\,L^{-1}}\) sodium ethanoate.

Use \(pK_a=4.74\) at \(25^\circ\mathrm{C}\).

A student argues that because the buffer starts with equal amounts of acid and conjugate base, removing most of the solution should not affect its ability to resist added acid.

Test this claim.

a) Calculate the initial pH.

b) The student removes 90.0 mL of the well-mixed 100.0 mL buffer, leaving 10.0 mL. They then add 12.0 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid to the remaining solution. Calculate the final pH.

c) Suppose instead that the same 12.0 mL of hydrochloric acid had been added to the original 100.0 mL buffer without removing any solution. Calculate the resulting pH.

d) Explain why the two final pH values are so different, even though both solutions initially had the same pH.

Assume volumes are additive, hydrochloric acid ionises completely, and the excess strong acid determines the pH in part (b). Use the usual dilute-solution buffer approximation in part (c).

Solution 5

The remaining 10.0 mL aliquot has a final pH of approximately 1.50, whereas the original full buffer would have a final pH of approximately 4.53. Removing most of the buffer preserves its initial pH but greatly reduces the total amounts available to neutralise added acid.

This is the important distinction between buffer pH and buffer capacity.

Step 1: Calculate the initial moles of both buffer components.

For ethanoic acid:

\[
\begin{aligned}
n(\ce{CH3COOH})
&=cV\\
&=0.100\times0.0500\\
&=0.00500\ \mathrm{mol}\\
&=5.00\ \mathrm{mmol}
\end{aligned}
\]

The sodium ethanoate solution has the same concentration and volume, so it also supplies:

\[
n(\mathrm{CH_{3}COO}^{-})=5.00\ \mathrm{mmol}
\]

Step 2: Calculate the initial pH.

The conjugate base-to-acid ratio is:

\[
\frac{5.00}{5.00}=1
\]

Therefore:

\[
\begin{aligned}
\mathrm{pH}
&=4.74+\log_{10}(1)\\
&=\boxed{4.74}
\end{aligned}
\]

Step 3: Calculate the amounts remaining after removing 90.0 mL.

The student removes 90% of the original solution.

Because the buffer is well mixed, 90% of each component is removed.

Only 10% of each original amount remains:

\[
n(\ce{CH3COOH})=0.10\times5.00
=0.500\ \mathrm{mmol}
\]

\[
n(\mathrm{CH_{3}COO}^{-})=0.10\times5.00
=0.500\ \mathrm{mmol}
\]

The ratio remains 1.

Therefore, removing the solution has not significantly changed its initial pH.

However, the amounts of both buffer components have fallen to one-tenth of their original values.

Step 4: Calculate the amount of hydrochloric acid added.

\[
\begin{aligned}
n(\ce{HCl})
&=cV\\
&=0.100\times0.0120\\
&=0.00120\ \mathrm{mol}\\
&=1.20\ \mathrm{mmol}
\end{aligned}
\]

Step 5: Compare the added acid with the available conjugate base.

The remaining ethanoate amount is only 0.500 mmol.

The hydrochloric acid therefore consumes all 0.500 mmol of ethanoate and leaves:

\[
n(\text{excess }\ce{H3O+})
=1.20-0.500
=0.700\ \mathrm{mmol}
\]

The reaction has also produced 0.500 mmol of additional ethanoic acid.

The buffer has lost its conjugate base reserve. There is now excess strong acid in the solution.

Step 6: Calculate the final hydronium concentration.

The final volume is:

\[
V_{\mathrm{final}}=10.0+12.0=22.0\ \mathrm{mL}
\]

Convert to litres:

\[
V_{\mathrm{final}}=0.0220\ \mathrm{L}
\]

The excess hydronium concentration is approximately:

\[
\begin{aligned}
[\ce{H3O+}]
&=\frac{0.000700}{0.0220}\\
&=0.0318\ \mathrm{mol\,L^{-1}}
\end{aligned}
\]

Step 7: Calculate the final pH.

\[
\begin{aligned}
\mathrm{pH}
&=-\log_{10}(0.0318)\\
&=\boxed{1.50}
\end{aligned}
\]

The excess strong acid determines the pH. The remaining ethanoic acid contributes negligibly to the hydronium concentration under these conditions.

Step 8: Calculate what would happen if the full buffer received the same acid addition.

The original full buffer contains:

\[
n(\ce{CH3COOH})=5.00\ \mathrm{mmol}
\]

\[
n(\mathrm{CH_{3}COO}^{-})=5.00\ \mathrm{mmol}
\]

Adding 1.20 mmol of hydrochloric acid consumes 1.20 mmol of ethanoate and produces 1.20 mmol of ethanoic acid.

The new amounts are:

\[
n(\mathrm{CH_{3}COO}^{-})=5.00-1.20
=3.80\ \mathrm{mmol}
\]

\[
n(\ce{CH3COOH})=5.00+1.20
=6.20\ \mathrm{mmol}
\]

Both buffer components are still present in appreciable amounts.

Therefore, Henderson-Hasselbalch is appropriate:

\[
\begin{aligned}
\mathrm{pH}
&=4.74+\log_{10}\left(\frac{3.80}{6.20}\right)\\
&=4.74+\log_{10}(0.613)\\
&=4.74-0.213\\
&=\boxed{4.53}
\end{aligned}
\]

Step 9: Explain the difference.

Both samples initially had the same pH because they had the same ratio of weak acid to conjugate base.

However, removing 90% of the original buffer also removed 90% of its acid and base reserves.

The remaining aliquot contained only 0.500 mmol of conjugate base, which was insufficient to neutralise the 1.20 mmol of added hydrochloric acid.

The full buffer contained 5.00 mmol of conjugate base, so it could absorb the same addition without exhausting its reserve.

This demonstrates why knowing the pH alone is not enough to predict how a buffer will respond to added acid or base.

A buffer’s pH depends mainly on the ratio of its components, while its capacity to resist changes depends on how much of each component is actually present.

06What to take into your HSC calculations

When you see a question asking why a buffer has a pH close to its \(pK_a\), start with the Henderson-Hasselbalch equation:

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

Then think about three things:

1. If the weak acid and conjugate base have similar amounts, their ratio is close to 1. The logarithmic term is therefore close to zero, giving a pH close to the \(pK_a\).
2. If strong acid or base has been added, calculate the neutralisation first. The new mole ratio determines the new approximate pH.
3. If one buffer component has been exhausted, the solution is no longer functioning as a buffer in that direction. You must use the remaining chemical species and their relevant equilibria to calculate the pH.

The next useful step is learning to follow that changing ratio along an entire titration curve. It explains why the half-equivalence point identifies the \(pK_a\), why the buffer region is relatively flat, and why the pH changes rapidly as the solution approaches equivalence.