How to Find pKa at Half-Equivalence in Weak-Acid Titrations

A weak acid is titrated with sodium hydroxide. The pH rises gradually, then jumps sharply when 17.60 mL of NaOH has been added. At that point, the pH is about 8.5. Is 8.5 the acid’s \(pK_a\)?

No. To find \(pK_a\), you need the pH at half-equivalence, not at equivalence. For a monoprotic weak acid initially containing no significant amount of its conjugate base, half-equivalence occurs when half the acid has been neutralised.

If equivalence is at 17.60 mL, look at 8.80 mL of added NaOH. If the pH there is 4.34, then \(pK_a\approx4.34\).

The calculation is short. The difficult part is knowing which volume to halve, why the method works, and when it can give the wrong answer.

01What the titration experiment actually measures

Imagine trying to measure the strength of an unknown weak acid, \(\ce{HA}\).

You could place the acid in a conical flask, add an indicator, and deliver sodium hydroxide from a burette until the indicator changes colour.

That tells you approximately how much base is needed to neutralise the acid. But it doesn’t tell you the pH throughout the titration, so you cannot reliably read \(pK_a\) from that single observation.

The colour change is the indicator’s endpoint. It may be close to the equivalence point, but the two are not necessarily identical.

For a \(pK_a\) measurement, a better arrangement uses a pH probe.

1. Measure a known volume of weak acid into a beaker.
2. Place a calibrated pH electrode in the solution and stir gently.
3. Add standardised NaOH from a burette, recording the total volume delivered.
4. Allow the pH reading to stabilise after each addition.
5. Record smaller additions near the sharp pH rise to locate equivalence more precisely.

The probe measures pH, while the burette measures how much NaOH has entered the flask. Plotting these two measurements gives the titration curve.

Suppose the probe records a gradual pH increase followed by a steep rise.

What should that steep rise tell you?

It identifies the region where the added base has almost completely neutralised the original acid. The equivalence point lies near the inflection point of that rise, where the curve changes its direction of bending.

The steep rise is useful for finding equivalence. The flatter region before it is where we’ll find \(pK_a\).

02Step 1: Identify the reaction and the equivalence point

Before looking at any pH values, write the neutralisation equation.

For a monoprotic weak acid reacting with NaOH:

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

Here:

• \(\ce{HA}\) is the weak acid.
• \(\mathrm{OH}^{-}\) is the hydroxide ion supplied by NaOH.
• \(\mathrm{A}^{-}\) is the conjugate base formed when \(\ce{HA}\) loses a proton.

The mole ratio is 1:1. One mole of hydroxide neutralises one mole of \(\ce{HA}\).

This is important because a titration curve’s horizontal axis gives volume, but neutralisation is determined by moles.

The amount of hydroxide delivered is

\[ n=cV \]

where \(n\) is the amount in moles, \(c\) is the titrant concentration in \(\mathrm{mol\,L^{-1}}\), and \(V\) is its volume in litres.

For our illustrative experiment, suppose:

• The flask initially contains 25.0 mL of a monoprotic weak acid.
• The burette contains \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH.
• The equivalence point occurs at 17.60 mL of added NaOH.

At equivalence:

\[ \begin{aligned} n(\mathrm{OH}^{-}) &=0.100\times0.01760\\ &=1.760\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Because the reaction is 1:1, the original sample contained \(1.760\times10^{-3}\) mol of \(\ce{HA}\).

Notice that we haven’t used the pH yet. We’ve established the chemical amount first.

03Step 2: Find when half the acid has reacted

At half-equivalence, half the original \(\ce{HA}\) has reacted with hydroxide.

For our example:

\[ \begin{aligned} V_{\frac12} &=\frac{17.60}{2}\\ &=8.80\ \mathrm{mL} \end{aligned} \]

Here, \(V_{\frac12}\) is the volume of NaOH delivered at half-equivalence, and \(V_e\) is the equivalence volume.

The general rule for this setup is

\[ V_{\frac12}=\frac{V_e}{2} \]

Why is that volume so useful?

Imagine following the acid molecules through the titration. Every hydroxide ion that neutralises an \(\ce{HA}\) molecule produces one \(\mathrm{A}^{-}\) ion.

At half-equivalence, the acid has been divided into two approximately equal groups: molecules that remain as \(\ce{HA}\), and molecules that have been converted into \(\mathrm{A}^{-}\).

For the illustrative experiment:

These are stoichiometric amounts, ignoring the small adjustments caused by acid-base equilibrium. One millimole (mmol) is \(10^{-3}\) mol.

At half-equivalence, both species share the same total solution volume. Therefore, equal amounts of \(\ce{HA}\) and \(\mathrm{A}^{-}\) also mean approximately equal concentrations.

That is the chemical reason the pH at this point reveals \(pK_a\).

04Step 3: Explain why pH equals pKa at half-equivalence

The weak acid establishes an equilibrium with water:

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

Its acid dissociation constant is

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

The square brackets represent equilibrium concentrations.

A larger \(K_a\) means the acid has a greater tendency to donate a proton. We usually express acid strength on a logarithmic scale:

\[ pK_a=-\log_{10}(K_a) \]

A smaller \(pK_a\) therefore indicates a stronger acid.

Rearranging the equilibrium expression gives the Henderson-Hasselbalch equation:

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

This connects pH to the ratio of conjugate base to weak acid.

At half-equivalence:

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

Their ratio is approximately 1. Since \(\log_{10}(1)=0\),

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

So the result is

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

This relationship is sometimes described as the half-neutralisation principle.

It is an approximation in the usual HSC concentration-based treatment. The derivation assumes an initially unbuffered monoprotic weak acid, a strong-base titrant, and sufficiently dilute, well-behaved solutions. Small equilibrium shifts and ion activity effects can affect highly precise measurements.

The useful intuition is that the pH depends on the balance between \(\ce{HA}\) and \(\mathrm{A}^{-}\). At half-equivalence, neither dominates.

05Step 4: Read the correct pH from experimental data

Here is an illustrative data set for our 25.0 mL weak-acid sample titrated with \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH.

These are modelled teaching values, not measurements from a real experiment. They represent a monoprotic weak acid with \(pK_a\) near 4.34 at approximately \(25^\circ\mathrm{C}\).

The steepest part of the rise is centred near 17.60 mL.

Once that volume is identified, halve it and read the pH at 8.80 mL.

The result is \(pK_a\approx4.34\).

06When halfway along the graph is not half-equivalence

The instruction to halve \(V_e\) assumes that the graph’s volume axis starts at zero NaOH delivered.

That sounds obvious, until a data logger starts recording halfway through an experiment.

07A quick way to diagnose the curve before calculating

Use this order when an HSC question supplies a titration curve or a pH data table.

Two other distinctions are important.

First, half-equivalence does not mean half the equivalence-point pH. pH is logarithmic, and the titration curve is not linear.

Second, equivalence does not mean pH 7.

For a weak acid titrated with a strong base, the conjugate base remains at equivalence and can react with water:

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

That produces hydroxide ions. At \(25^\circ\mathrm{C}\), the equivalence-point pH is generally above 7.

So a curve with an equivalence pH of 8.5 can still give a perfectly reasonable \(pK_a\) of 4.3.

The broader method for interpreting the entire graph, including uncertainty when locating its inflection, is covered in How to Find pKa from a Weak-Acid Titration Curve.

08Questions and solutions

Assume aqueous solutions at approximately \(25^\circ\mathrm{C}\), a calibrated pH probe, and ideal dilute-solution behaviour unless otherwise stated. Use complete neutralisation stoichiometry and the usual buffer approximations. All experimental readings below are original illustrative data.

Question 1

A monoprotic weak acid is titrated with NaOH.

The steepest rise of the titration curve is centred at 26.0 mL, and the pH at 13.0 mL is 4.54.

Calculate \(pK_a\) and \(K_a\). Explain why the equivalence-point pH should not be used.

Solution 1

The acid has \(pK_a\approx4.54\) and \(K_a\approx2.9\times10^{-5}\).

The half-equivalence volume is

\[ V_{\frac12}=\frac{26.0}{2} =13.0\ \mathrm{mL} \]

At this volume, approximately equal concentrations of the weak acid and its conjugate base are present.

Therefore:

\[ pK_a\approx\mathrm{pH}=4.54 \]

Convert to \(K_a\):

\[ \begin{aligned} K_a&=10^{-pK_a}\\ &=10^{-4.54}\\ &=2.88\times10^{-5} \end{aligned} \]

The equivalence-point pH is not the acid’s \(pK_a\) because the original acid has been neutralised stoichiometrically. The conjugate base dominates the solution and reacts with water.

The tempting mistake is to read pH at the sharp jump. That jump locates equivalence; half of the equivalence volume locates the useful \(pK_a\) reading.

Question 2

Two identical 20.0 mL aliquots of a pure monoprotic weak acid each contain 2.00 mmol of \(\ce{HA}\).

In experiment A, the acid is titrated with \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH. The equivalence volume is 20.0 mL, and the pH at 10.0 mL is 4.55.

In experiment B, the second acid aliquot is diluted with 80.0 mL of distilled water before titration. This time, the titrant is \(0.0500\ \mathrm{mol\,L^{-1}}\) NaOH.

Determine:

• The equivalence and half-equivalence volumes in experiment B.
• The expected pH at half-equivalence in experiment B.
• The expected pH in experiment B after 10.0 mL of NaOH has been delivered.
• Whether adding water alone, without changing the titrant concentration, would alter the ideal half-equivalence pH.

Explain why comparing the same volume of NaOH in the two experiments can be misleading.

Solution 2

Experiment B reaches equivalence at 40.0 mL, half-equivalence at 20.0 mL, and has \(pK_a\approx4.55\). Its pH after just 10.0 mL of NaOH is approximately 4.07, despite the initial acid having been diluted.

First, calculate the NaOH volume required to neutralise the original 2.00 mmol of acid.

\[ \begin{aligned} V_e&=\frac{n}{c}\\ &=\frac{0.00200}{0.0500}\\ &=0.0400\ \mathrm{L}\\ &=40.0\ \mathrm{mL} \end{aligned} \]

Half-equivalence occurs at

\[ V_{\frac12}=\frac{40.0}{2} =20.0\ \mathrm{mL} \]

The acid is unchanged, so under the ideal buffer approximation:

\[ pK_a\approx4.55 \]

The half-equivalence pH is therefore approximately 4.55.

Now consider experiment B after only 10.0 mL of NaOH has been added.

The amount of hydroxide delivered is

\[ \begin{aligned} n(\mathrm{OH}^{-}) &=0.0500\times0.0100\\ &=0.000500\ \mathrm{mol}\\ &=0.500\ \mathrm{mmol} \end{aligned} \]

This forms 0.500 mmol of \(\mathrm{A}^{-}\) and leaves

\[ 2.00-0.500=1.50\ \mathrm{mmol} \]

of \(\ce{HA}\).

Both species share the same solution volume, so their concentration ratio equals their mole ratio:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} \approx\frac{0.500}{1.50} =\frac13 \]

Substitute into the Henderson-Hasselbalch equation:

\[ \begin{aligned} \mathrm{pH} &\approx4.55+\log_{10}\left(\frac13\right)\\ &\approx4.55-0.477\\ &\approx4.07 \end{aligned} \]

Experiment A has pH 4.55 at 10.0 mL because that volume delivers 1.00 mmol of hydroxide, exactly half of the original acid amount.

Experiment B receives only 0.500 mmol at the same volume.

This is the important interaction: experiment B is initially more dilute, which tends to increase the initial pH of the weak acid, but its lower NaOH concentration means less acid has been neutralised after a fixed titrant volume.

If experiment B had used \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH instead, adding distilled water alone would not change the neutralisation amount or the equivalence volume. In the ideal buffer model, it would also leave the half-equivalence pH approximately unchanged because both \(\ce{HA}\) and \(\mathrm{A}^{-}\) would be diluted by the same factor.

The acid’s \(pK_a\) is not determined by how much water is in the flask. It is a property of the acid equilibrium at the specified temperature and solution conditions.

Question 3

Two identical 25.0 mL samples of a pure monoprotic weak acid are titrated using different strong bases.

Experiment A uses \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH. Equivalence occurs at 24.0 mL, and the pH at 12.0 mL is 4.44.

Experiment B uses \(0.100\ \mathrm{mol\,L^{-1}}\) \(\ce{Ba(OH)2}\). Equivalence occurs at 12.0 mL, and the pH at 6.0 mL is also 4.44.

Assume that barium hydroxide dissociates completely and that no precipitation or other reactions occur.

Determine whether the two experiments are consistent with the same weak-acid concentration and \(pK_a\). Calculate both quantities and \(K_a\).

A student argues that 12.0 mL must represent half-equivalence in both experiments because the acid samples are identical. Explain the error.

Solution 3

The experiments are consistent. Both give an initial acid concentration of \(0.0960\ \mathrm{mol\,L^{-1}}\), \(pK_a\approx4.44\), and \(K_a\approx3.63\times10^{-5}\).

The different equivalence volumes arise because one mole of \(\ce{Ba(OH)2}\) supplies two moles of hydroxide ions.

For experiment A:

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

The amount of acid originally present is

\[ \begin{aligned} n(\ce{HA}) &=0.100\times0.0240\\ &=2.40\times10^{-3}\ \mathrm{mol} \end{aligned} \]

For experiment B, the overall reaction is

\[ \ce{2HA + Ba(OH)2 -> BaA2 + 2H2O} \]

The amount of barium hydroxide used is

\[ \begin{aligned} n(\ce{Ba(OH)2}) &=0.100\times0.0120\\ &=1.20\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Each mole of \(\ce{Ba(OH)2}\) provides two hydroxide ions, so this neutralises

\[ n(\ce{HA})=2(1.20\times10^{-3}) \]

or \(2.40\times10^{-3}\) mol of acid.

Both experiments therefore give the same original acid amount.

Using the 25.0 mL sample volume:

\[ \begin{aligned} c(\ce{HA}) &=\frac{0.00240}{0.0250}\\ &=0.0960\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Now calculate half-equivalence separately for each titration.

For experiment A:

\[ V_{\frac12}=\frac{24.0}{2} =12.0\ \mathrm{mL} \]

For experiment B:

\[ V_{\frac12}=\frac{12.0}{2} =6.0\ \mathrm{mL} \]

Both curves give pH 4.44 at their respective half-equivalence volumes.

Therefore:

\[ \begin{aligned} pK_a&\approx4.44\\ K_a&=10^{-4.44}\\ &=3.63\times10^{-5} \end{aligned} \]

The student’s error is using the same physical titrant volume without considering how many hydroxide ions that volume supplies.

At 12.0 mL, experiment A is at half-equivalence, but experiment B is at equivalence.

In experiment B, the weak acid has been neutralised stoichiometrically, and the solution contains its conjugate base. Its pH is expected to be basic, rather than 4.44.

The half-equivalence method still works with a strong base that supplies two hydroxide ions per formula unit. You must first use the correct stoichiometry to establish the equivalence volume.

Question 4

An extension of the half-equivalence method involves a diprotic weak acid, \(\ce{H2A}\), which can donate two protons.

Its titration with NaOH produces two clearly separated equivalence regions.

The first equivalence point occurs at 14.0 mL, and the second at 28.0 mL.

The measured pH values include:

Assume the two acid dissociation steps are well separated, allowing the usual sequential-neutralisation approximation.

Determine \(pK_{a1}\) and \(pK_{a2}\).

A student halves the final equivalence volume, obtains 14.0 mL, and reports \(pK_{a1}=5.90\).

Explain why this reasoning fails, even though the student correctly halved an equivalence volume.

Solution 4

The first dissociation has \(pK_{a1}\approx3.35\), and the second has \(pK_{a2}\approx8.40\). The pH of 5.90 at 14.0 mL is associated with the first equivalence point, not half-neutralisation of the first proton.

A diprotic acid loses its protons in two stages:

\[ \ce{H2A + H2O <=> H3O+ + HA^-} \]

and

\[ \ce{HA^- + H2O <=> H3O+ + A^2-} \]

The first neutralisation stage converts \(\ce{H2A}\) into \(\mathrm{HA}^{-}\).

It runs from 0 to 14.0 mL of added NaOH.

Halfway through that stage is

\[ V_{\frac12,1}=\frac{14.0}{2} =7.0\ \mathrm{mL} \]

At this point, \(\ce{H2A}\) and \(\mathrm{HA}^{-}\) are present in approximately equal concentrations.

Therefore:

\[ pK_{a1}\approx3.35 \]

The second neutralisation stage converts \(\mathrm{HA}^{-}\) into \(\mathrm{A}^{2-}\).

It runs from 14.0 to 28.0 mL.

Halfway through the second stage is

\[ \begin{aligned} V_{\frac12,2} &=\frac{14.0+28.0}{2}\\ &=21.0\ \mathrm{mL} \end{aligned} \]

At this point, \(\mathrm{HA}^{-}\) and \(\mathrm{A}^{2-}\) are present in approximately equal concentrations.

Therefore:

\[ pK_{a2}\approx8.40 \]

Why is the pH at 14.0 mL about 5.90?

At the first equivalence point, the dominant species is \(\mathrm{HA}^{-}\).

This species is amphiprotic: it can donate another proton to become \(\mathrm{A}^{2-}\), or accept a proton to become \(\ce{H2A}\).

Its pH is consequently controlled by both acid-base equilibria. With well-separated dissociation constants, the first-equivalence pH is often approximately midway between the two \(pK_a\) values.

Here:

\[ \frac{3.35+8.40}{2}=5.875 \]

This is consistent with the illustrative pH of about 5.90.

The student’s reasoning fails because half of the total two-proton neutralisation corresponds to complete neutralisation of the first proton.

For a multiprotic acid, each well-separated neutralisation stage has its own halfway point. You cannot identify \(pK_{a1}\) simply by halving the final equivalence volume.

Question 5

An unknown acidic solution contains the monoprotic weak acid \(\ce{HA}\). It may also contain hydrochloric acid, \(\ce{HCl}\).

Two models are proposed:

• Model A: The sample originally contained only \(\ce{HA}\).
• Model B: The sample originally contained \(\ce{HA}\) and some \(\ce{HCl}\).

No conjugate-base salt was initially added.

The solution is titrated with \(0.100\ \mathrm{mol\,L^{-1}}\) NaOH. Its final stoichiometric equivalence point occurs at 30.0 mL.

The only clearly resolved sharp rise is near this final equivalence point. Two accurate pH readings in the buffer region are:

Assume complete neutralisation, that strong-acid protons are consumed before substantial neutralisation of \(\ce{HA}\), and that the Henderson-Hasselbalch approximation is valid once excess strong acid has been consumed. Treat the pH values as rounded to two decimal places.

Using only these observations and the stated assumptions:

1. Determine which model is consistent with both readings.
2. Estimate the original amount of \(\ce{HCl}\) and \(\ce{HA}\).
3. Determine the weak acid’s \(pK_a\) and \(K_a\).
4. Find the volume at which \([\ce{HA}]\approx[\mathrm{A}^{-}]\).
5. Explain what would happen to the equivalence volume and buffer-region pH readings if an identical sample were diluted with distilled water before titration, using the same NaOH concentration.

A student argues that 15.0 mL must be the correct half-equivalence volume because it is half of 30.0 mL. Test that argument rather than accepting it.

Solution 5

Model B is consistent with the data. The solution initially contained approximately 1.01 mmol of \(\ce{HCl}\) and 1.99 mmol of \(\ce{HA}\). The weak acid has \(pK_a\approx4.79\) and \(K_a\approx1.6\times10^{-5}\). Equal acid and conjugate-base concentrations occur near 20.0 mL of NaOH added, not 15.0 mL.

The surprising point is that a final equivalence volume of 30.0 mL does not tell us how much of the NaOH reacted with the weak acid rather than with the strong acid.

Step 1: Test the pure weak-acid model.

If Model A is correct, all 30.0 mL of NaOH neutralises \(\ce{HA}\).

For a pure monoprotic weak acid, the buffer-region concentration ratio is approximately

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

where \(V\) is the NaOH volume delivered, and \(V_e=30.0\ \mathrm{mL}\).

At 15.0 mL:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} =\frac{15.0}{30.0-15.0}=1 \]

At 25.0 mL:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} =\frac{25.0}{30.0-25.0}=5 \]

If Model A is correct, the predicted pH difference is

\[ \begin{aligned} \Delta\mathrm{pH} &=\log_{10}(5)-\log_{10}(1)\\ &=\log_{10}(5)\\ &\approx0.70 \end{aligned} \]

But the observed difference is

\[ 5.26-4.30=0.96 \]

The difference is too large to explain by rounding the stated pH readings.

Crucially, the unknown \(pK_a\) cancels when we subtract the two buffer equations. Choosing a different \(pK_a\) cannot rescue Model A.

Step 2: Build the strong-acid mixture model.

In Model B, some of the added hydroxide first neutralises \(\ce{HCl}\):

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

Once the strong acid has been consumed, additional hydroxide converts \(\ce{HA}\) into \(\mathrm{A}^{-}\).

Let \(s\) be the volume of NaOH, in mL, required to neutralise the original \(\ce{HCl}\).

The remaining NaOH volume available to neutralise all the original \(\ce{HA}\) is

\[ 30.0-s \]

At any volume \(V\) beyond \(s\), the amount of conjugate base formed is proportional to \(V-s\).

The amount of weak acid remaining is proportional to \(30.0-V\).

Therefore:

\[ \frac{[\mathrm{A}^{-}]}{[\ce{HA}]} \approx\frac{V-s}{30.0-V} \]

This is different from the pure-acid model. The numerator has an offset because some NaOH was consumed by strong acid before conjugate base began accumulating.

Step 3: Use both pH readings to find the hidden strong-acid amount.

At 15.0 mL, the ratio is

\[ r_{15}=\frac{15.0-s}{15.0} \]

At 25.0 mL:

\[ r_{25}=\frac{25.0-s}{5.0} \]

The Henderson-Hasselbalch equation gives

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

Subtracting the two pH equations eliminates \(pK_a\):

\[ 0.96=\log_{10}\left(\frac{r_{25}}{r_{15}}\right) \]

Therefore:

\[ \begin{aligned} \frac{r_{25}}{r_{15}} &=10^{0.96}\\ &\approx9.12 \end{aligned} \]

Substitute the volume expressions:

\[ 9.12=\frac{3(25.0-s)}{15.0-s} \]

Rearrange:

\[ \begin{aligned} 9.12(15.0-s)&=3(25.0-s)\\ 136.8-9.12s&=75.0-3s\\ s&\approx10.1\ \mathrm{mL} \end{aligned} \]

So approximately 10.1 mL of NaOH was consumed by the strong acid.

The original amount of \(\ce{HCl}\) was

\[ \begin{aligned} n(\ce{HCl}) &=0.100\times0.0101\\ &\approx1.01\times10^{-3}\ \mathrm{mol} \end{aligned} \]

or 1.01 mmol.

Step 4: Determine the original weak-acid amount.

The total equivalence volume was 30.0 mL.

After accounting for the strong acid, the volume required to neutralise \(\ce{HA}\) was approximately

\[ 30.0-10.1=19.9\ \mathrm{mL} \]

Therefore:

\[ \begin{aligned} n(\ce{HA}) &\approx0.100\times0.0199\\ &=1.99\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The sample originally contained approximately 1.99 mmol of \(\ce{HA}\).

Together, the strong and weak acids account for approximately 3.00 mmol of neutralisable protons, consistent with the 30.0 mL equivalence volume.

Step 5: Calculate the weak acid’s \(pK_a\).

Use the 15.0 mL pH reading.

At this volume:

\[ \begin{aligned} r_{15} &=\frac{15.0-10.1}{15.0}\\ &\approx0.327 \end{aligned} \]

Rearrange the Henderson-Hasselbalch equation:

\[ \begin{aligned} pK_a &\approx\mathrm{pH}-\log_{10}(r_{15})\\ &\approx4.30-\log_{10}(0.327)\\ &\approx4.79 \end{aligned} \]

Then:

\[ \begin{aligned} K_a&=10^{-pK_a}\\ &\approx10^{-4.79}\\ &\approx1.6\times10^{-5} \end{aligned} \]

The estimate from the 25.0 mL reading is consistent with this result.

Step 6: Locate the actual equal-concentration point.

Equal concentrations of \(\ce{HA}\) and \(\mathrm{A}^{-}\) occur when half the original weak acid has been neutralised.

But the NaOH must first account for the strong acid.

The required volume is therefore

\[ V_{\text{equal}}=s+\frac{30.0-s}{2} \]

Substitute \(s\approx10.1\ \mathrm{mL}\):

\[ \begin{aligned} V_{\text{equal}} &\approx10.1+\frac{19.9}{2}\\ &\approx20.0\ \mathrm{mL} \end{aligned} \]

The pH near 20.0 mL should therefore be approximately 4.79.

Notice what happened. At 15.0 mL, the sample is halfway to its final equivalence volume, but it is not halfway through the weak-acid neutralisation.

About 10.1 mL was consumed by strong acid. Only the remaining approximately 4.9 mL has formed conjugate base.

The mole balance, not the visual midpoint of the graph, determines where pH equals \(pK_a\).

Step 7: Predict the effect of dilution before titration.

Adding distilled water does not change the initial numbers of moles of \(\ce{HCl}\) and \(\ce{HA}\).

If the NaOH concentration remains \(0.100\ \mathrm{mol\,L^{-1}}\), the stoichiometric equivalence volume remains 30.0 mL.

The volume needed to neutralise the strong acid also remains approximately 10.1 mL, and the equal-\(\ce{HA}\)/\(\mathrm{A}^{-}\) point remains near 20.0 mL.

In the ideal buffer model, the pH at 15.0, 20.0, and 25.0 mL also remains approximately unchanged. Dilution reduces both \([\ce{HA}]\) and \([\mathrm{A}^{-}]\), but their ratio stays the same.

However, the initial pH and the pH near equivalence can change because the absolute concentrations of the species controlling those regions have changed.

This illustrates a useful distinction: dilution can alter a titration curve without changing the stoichiometric volumes or the acid’s intrinsic \(pK_a\). In sufficiently dilute real solutions, the ideal approximations become less reliable.

The deeper lesson from this question is that even a seemingly ordinary titration curve can conceal a more complicated starting composition. Before assigning a pH reading to \(pK_a\), identify which chemical species the added base has actually neutralised.

09What to do with an unfamiliar titration curve

When you next see a weak-acid titration question, begin with the neutralisation reaction and the volume scale. Establish the equivalence amount, then ask what species remain halfway through the relevant neutralisation stage.

If the original sample contains only a monoprotic weak acid and the titrant is a strong base, the method is simply

\[ V_{\frac12}=\frac{V_e}{2}, \qquad pK_a\approx\mathrm{pH}_{\frac12} \]

If the starting solution contains other acids, conjugate-base salts, or multiple ionisable protons, work out the mole balance before using that shortcut.

Once you’ve located \(pK_a\) correctly, the next useful skill is predicting the pH elsewhere in the buffer region using the ratio of conjugate base to remaining acid. That same reasoning helps you recognise whether an experimental curve matches the chemical model you’ve assumed.

And if an experiment supplies an indicator colour change rather than a clear pH curve, first establish whether you have an endpoint or a true equivalence volume. The difference is explained in Endpoint vs Equivalence Point: The Difference in Titrations.