How to Read the Buffer Region on a Titration Curve

Adding sodium hydroxide to a weak acid produces a curious result. For a while, the pH rises quite slowly. Then, after only a little more sodium hydroxide has been added, the pH shoots upwards.

Why doesn't the pH change much at first? And why does that protection suddenly become much weaker?

The gently sloping section of the titration curve is called the **buffer region**. It appears because the solution contains significant amounts of a weak acid and its conjugate base. Together, these substances react with added acid or base, limiting the change in pH.

The important detail is that their amounts are finite. A buffer can only absorb so much added acid or base before one component becomes depleted. Reading the buffer region means recognising both when this protection develops and when it starts to fail.

## What is happening inside the buffer region?

Consider a titration in which sodium hydroxide is added to a solution of a weak monoprotic acid, \(\ce{HA}\).

A standard experimental arrangement has:
- A burette containing a known concentration of \(\ce{NaOH}\), positioned above a conical flask.
- A measured volume of weak acid in the flask.
- A magnetic stirrer to mix the solution.
- A calibrated pH probe immersed in the flask and connected to a meter or data logger.

The chemist adds measured volumes of sodium hydroxide, allows the solution to mix, and records the pH after each addition. Plotting pH against the cumulative volume of sodium hydroxide produces the titration curve.

[IMAGE: A clearly labelled weak-acid/strong-base titration apparatus viewed from the side. Show a burette containing NaOH above a conical flask containing weak acid HA, a magnetic stir bar and stirrer beneath the flask, and a pH electrode immersed in the solution and connected to a digital meter. Show the burette tip positioned over the flask and the electrode separate from the stir bar.]

Before looking at the chemistry, make a prediction.

If sodium hydroxide is added to the weak acid, what happens to the acid molecules? Do they simply disappear as hydroxide ions neutralise the hydrogen ions already present, or does something else happen?

The second explanation is closer to the full picture. The weak acid itself is converted into its conjugate base.

A weak acid partially ionises in water:

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

Here:
- \(\ce{HA}\) is the weak acid.
- \(\ce{A-}\) is its conjugate base.
- \(\ce{H3O+}\) is the hydronium ion.

When sodium hydroxide is added, its hydroxide ions react with the weak acid:

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

Two important changes occur:

1. Some of the weak acid, \(\ce{HA}\), is consumed.
2. Its conjugate base, \(\ce{A-}\), is produced.

The solution now contains appreciable amounts of both \(\ce{HA}\) and \(\ce{A-}\). This is the combination needed to form a buffer.

[IMAGE: A scientifically accurate particle-level illustration with two panels. In the first, show mostly HA molecules in water with relatively few A- ions. In the second, after a small addition of NaOH, show fewer HA molecules and more A- ions, with each added OH- consumed through HA + OH- -> A- + H2O. Include water molecules and clear labels. Do not depict OH- remaining in excess.]

### Why does the pH change slowly?

A buffer works in both directions.

If extra hydroxide ions are added, the weak acid reacts with them:

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

If extra hydronium ions are added, the conjugate base reacts with them:

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

In each case, the added acid or base is consumed by one of the buffer components.

This also makes sense through Le Chatelier's principle.

When hydroxide ions remove \(\ce{H3O+}\) from the solution, the weak acid equilibrium shifts to the right, replacing some of the hydronium ions. When additional hydronium ions are introduced, the equilibrium shifts to the left, consuming them and producing more \(\ce{HA}\).

The solution doesn't prevent the reactions from happening. It changes the relative amounts of its acid and conjugate base to oppose the pH change.

Think of the two components as limited reserves:
- \(\ce{HA}\) is the reserve that reacts with added base.
- \(\ce{A-}\) is the reserve that reacts with added acid.

Like two containers being used to absorb different kinds of disturbance, each reserve can be used up. The analogy stops there: the chemical components aren't physically storing the added substances. They react and establish a new equilibrium.

## How to identify the buffer region on a titration curve

The buffer region is the broad, gently rising section of a weak-acid/strong-base titration curve before the equivalence point.

It is not completely horizontal. The pH still increases as more sodium hydroxide is added. However, the change is much smaller than the dramatic increase near equivalence.

[IMAGE: An annotated pH-versus-volume titration curve for a monoprotic weak acid titrated with strong NaOH. Label the initial weak-acid region, the gently sloping buffer region, the half-equivalence point at half the equivalence volume, pH = pKa at that point, the steep rise, the equivalence point above pH 7, and the post-equivalence region where excess OH- controls pH. Use clear axes labelled pH and volume of NaOH added.]

There are three features worth identifying.

**1. The buffer region**

This is the gradual rise before equivalence. Both the weak acid and its conjugate base are present in significant amounts.

**2. The half-equivalence point**

This occurs when half of the original weak acid has been neutralised.

At this point, the amounts of weak acid and conjugate base are approximately equal:

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

Consequently:

\[
\boxed{\text{pH}\approx\text{p}K_a}
\]

This is one of the most useful features of a weak-acid titration curve. You can estimate the acid's \(\text{p}K_a\) directly from the pH at half-equivalence.

**3. The equivalence point**

This is where the original weak acid has been stoichiometrically neutralised.

The conjugate base, \(\ce{A-}\), remains in solution. It reacts with water to produce hydroxide ions:

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

The solution is therefore basic at equivalence, so the equivalence pH is greater than 7 at \(25^\circ\text{C}\).

The steep rise occurs around this point because very little weak acid remains to react with additional hydroxide ions.

### The difference between a buffer region and a flat line

A common misconception is that a buffer region should have a constant pH.

It doesn't.

The pH depends on the relative amounts of weak acid and conjugate base. As sodium hydroxide is added, more \(\ce{HA}\) is converted into \(\ce{A-}\). Their ratio changes, so the pH changes too.

The buffer simply makes that change more gradual.

A useful rule of thumb is that a buffer works most effectively when its pH is within approximately one pH unit of its \(\text{p}K_a\).

That corresponds to a conjugate-base-to-acid ratio between approximately \(0.1:1\) and \(10:1\).

These are practical guidelines, not exact boundaries. A solution does not suddenly stop being a buffer at a particular pH.

## Using the half-equivalence point to find pKa

The relationship between pH and the relative amounts of acid and conjugate base can be obtained from the acid dissociation constant.

For the weak acid equilibrium:

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

Rearranging:

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

Taking the negative logarithm gives:

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

This is the Henderson-Hasselbalch equation.

For a titration, it is often convenient to use moles instead of concentrations:

\[
\text{pH}\approx\text{p}K_a+
\log_{10}\left(
\frac{n(\ce{A-})}{n(\ce{HA})}
\right)
\]

Why is this allowed? Both substances are in the same solution, so their concentrations have the same total-volume denominator. That denominator cancels when the ratio is calculated.

The approximation assumes that the stoichiometric amounts produced by neutralisation are a good estimate of the equilibrium amounts. It is most useful when both buffer components are present in appreciable quantities.

For a weak monoprotic acid titrated with a strong base, before equivalence:

\[
n(\ce{A-})\approx n(\ce{OH-}\text{ added})
\]

\[
n(\ce{HA})\approx n(\ce{HA}\text{ initially})-
n(\ce{OH-}\text{ added})
\]

At half-equivalence, the two amounts are equal. Therefore:

\[
\text{pH}\approx\text{p}K_a+\log_{10}(1)
\]

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

\[
\boxed{\text{pH}\approx\text{p}K_a}
\]

Remember that half-equivalence means half the original acid has reacted. It does not mean halfway between the initial pH and the equivalence pH.

### Worked example: Find pKa from a titration curve

A \(25.00\ \text{mL}\) sample of a monoprotic weak acid is titrated with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide. The equivalence point occurs after \(20.00\ \text{mL}\) of sodium hydroxide has been added. At \(10.00\ \text{mL}\), the measured pH is 4.74.

Calculate the original concentration of the acid and its acid dissociation constant, \(K_a\).

**Step 1**: Find the amount of acid in the original sample.

At equivalence, the moles of hydroxide added equal the original moles of monoprotic acid.

Using \(n=cV\):

\[
\begin{aligned}
n(\ce{OH-})
&=0.1000\times0.02000\\
&=0.002000\ \text{mol}
\end{aligned}
\]

Therefore, the original sample contained \(0.002000\ \text{mol}\) of acid.

**Step 2**: Calculate the original acid concentration.

\[
\begin{aligned}
c(\ce{HA})
&=\frac{n}{V}\\
&=\frac{0.002000}{0.02500}\\
&=0.08000\ \text{mol L}^{-1}
\end{aligned}
\]

**Step 3**: Identify the half-equivalence point.

Half of the equivalence volume is:

\[
V_{\frac{1}{2}\text{eq}}=
\frac{20.00}{2}=10.00\ \text{mL}
\]

The measured pH of 4.74 is therefore approximately equal to the acid's \(\text{p}K_a\).

\[
\text{p}K_a=4.74
\]

**Step 4**: Calculate \(K_a\).

By definition:

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

Rearranging:

\[
\begin{aligned}
K_a&=10^{-4.74}\\
&=1.82\times10^{-5}
\end{aligned}
\]

The original acid concentration is \(0.08000\ \text{mol L}^{-1}\), and its acid dissociation constant is approximately \(1.82\times10^{-5}\).

The key observation is that the pH at half-equivalence gives the \(\text{p}K_a\), not the initial pH or the equivalence pH.

### Worked example: Calculate pH near the end of the buffer region

A \(25.00\ \text{mL}\) sample containing \(0.0800\ \text{mol L}^{-1}\) of a weak monoprotic acid is titrated with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide. Its \(\text{p}K_a\) is 4.74.

Calculate the approximate pH after \(19.00\ \text{mL}\) of sodium hydroxide has been added. Then determine the approximate pH at equivalence and explain why the buffer approximation cannot be used there.

**Step 1**: Calculate the initial amount of weak acid.

\[
\begin{aligned}
n(\ce{HA})
&=cV\\
&=0.0800\times0.02500\\
&=0.00200\ \text{mol}
\end{aligned}
\]

The original sample contains \(2.00\ \text{mmol}\) of acid.

**Step 2**: Calculate the amount of hydroxide added.

\[
\begin{aligned}
n(\ce{OH-})
&=0.1000\times0.01900\\
&=0.00190\ \text{mol}\\
&=1.90\ \text{mmol}
\end{aligned}
\]

**Step 3**: Determine the remaining amounts of acid and conjugate base.

The reaction is:

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

Therefore:

\[
n(\ce{HA})=2.00-1.90=0.10\ \text{mmol}
\]

\[
n(\ce{A-})=1.90\ \text{mmol}
\]

**Step 4**: Apply the Henderson-Hasselbalch approximation.

\[
\begin{aligned}
\text{pH}
&=\text{p}K_a+
\log_{10}\left(\frac{n(\ce{A-})}{n(\ce{HA})}\right)\\
&=4.74+\log_{10}\left(\frac{1.90}{0.10}\right)\\
&=4.74+\log_{10}(19)\\
&=6.02
\end{aligned}
\]

The estimated pH is 6.02.

Notice that the solution has only \(0.10\ \text{mmol}\) of weak acid remaining. Although the equation gives a numerical pH, the solution is no longer an especially effective buffer. Its remaining ability to absorb additional hydroxide is limited.

**Step 5**: Calculate the pH at equivalence using the conjugate base equilibrium.

At equivalence, all \(2.00\ \text{mmol}\) of weak acid has been converted into \(\ce{A-}\).

The total volume is:

\[
25.00+20.00=45.00\ \text{mL}
\]

Therefore:

\[
[\ce{A-}]
=\frac{0.00200}{0.04500}
=0.04444\ \text{mol L}^{-1}
\]

The conjugate base reacts with water:

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

At \(25^\circ\text{C}\), assuming \(K_w=1.0\times10^{-14}\):

\[
\begin{aligned}
K_b&=\frac{K_w}{K_a}\\
&=\frac{1.0\times10^{-14}}{1.82\times10^{-5}}\\
&=5.49\times10^{-10}
\end{aligned}
\]

Using the usual small-ionisation approximation:

\[
K_b\approx\frac{[\ce{OH-}]^2}{[\ce{A-}]}
\]

Rearrange and substitute:

\[
\begin{aligned}
[\ce{OH-}]
&\approx\sqrt{K_b[\ce{A-}]}\\
&=\sqrt{(5.49\times10^{-10})(0.04444)}\\
&=4.94\times10^{-6}\ \text{mol L}^{-1}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
\text{pOH}&=-\log_{10}(4.94\times10^{-6})\\
&=5.31\\
\text{pH}&=14.00-5.31\\
&=8.69
\end{aligned}
\]

At equivalence, the pH is approximately 8.69.

The Henderson-Hasselbalch approximation is no longer suitable because there is essentially no stoichiometric weak acid remaining. The conjugate base's reaction with water now determines the pH.

This is why the curve rises sharply near equivalence. The solution has lost much of its ability to absorb additional hydroxide without a substantial change in pH.

## Where the buffer approximation fails

The most important distinction is between having a buffer and having enough buffer capacity to resist a meaningful addition of acid or base.

A solution can contain both conjugate partners but still have very little capacity against one particular addition.

Consider the stages of a weak-acid/strong-base titration:

| Stage | What is happening? | How to interpret the pH |
|---|---|---|
| Beginning | Mostly \(\ce{HA}\), with a small equilibrium amount of \(\ce{A-}\) | Initial weak-acid equilibrium determines pH |
| Early titration | Increasing amounts of \(\ce{A-}\) form | Buffer behaviour develops |
| Middle of buffer region | Significant amounts of both species | pH changes gradually |
| Half-equivalence | Approximately equal amounts of \(\ce{HA}\) and \(\ce{A-}\) | pH approximately equals pKa |
| Near equivalence | Very little \(\ce{HA}\) remains | Buffer capacity against added base becomes weak |
| Equivalence | Almost all acid has become \(\ce{A-}\) | Conjugate-base hydrolysis determines pH |
| After equivalence | Excess \(\ce{OH-}\) is present | Excess strong base determines pH |

There are three situations where students commonly apply the buffer model incorrectly.

**At the beginning of the titration**

The weak acid has partially ionised, so some conjugate base exists. However, its amount is usually too small to provide the substantial reserve needed for effective buffering.

Do not assume that every weak acid solution is automatically a useful buffer.

**Very close to equivalence**

The weak acid is nearly exhausted. A small additional amount of hydroxide can consume a large fraction of what remains.

The ratio of conjugate base to weak acid becomes very large, and the pH begins changing rapidly.

**At and after equivalence**

At equivalence, the conjugate base's reaction with water determines pH.

After equivalence, excess hydroxide from the titrant dominates.

Neither situation should be treated as an ordinary buffer calculation.

A useful diagnostic question is:

*Are there still meaningful amounts of both components, and is the amount of acid or base being added small compared with the available component that must react with it?*

If not, stop applying the buffer approximation and return to stoichiometry and the appropriate equilibrium calculation.

## How the experiment affects the curve

The curve is only as useful as the measurements used to construct it.

During a titration, the pH electrode must be calibrated and allowed to stabilise after each addition. The solution should be stirred consistently so the probe measures a representative pH rather than a locally concentrated pocket of titrant.

The volume increments also matter.

If sodium hydroxide is added in large \(5.00\ \text{mL}\) portions throughout the experiment, the recorded data might miss much of the steep rise near equivalence. The curve could appear to have a less precise equivalence volume than it actually does.

A better approach is to:
- Use reasonably spaced additions through the broad buffer region.
- Reduce the volume increments as the steep region approaches.
- Allow the pH reading to stabilise after each addition.
- Keep temperature consistent, since \(K_a\), \(K_w\), and pH equilibria depend on temperature.

[IMAGE: A comparison of two experimentally sampled weak-acid/strong-base titration curves. Show one dataset with widely spaced volume measurements that misses the steep equivalence rise, and another with smaller additions near equivalence that clearly resolves the inflection point. Label the measured data points and the resulting smooth curves.]

This is an important distinction when interpreting experimental data: the chemical system has a continuous response, but the student only observes the values recorded at particular volumes.

A missing data point does not mean the buffer region or equivalence point did not exist. It means the experiment did not measure it closely enough.

For a broader explanation of how equivalence points, curve shapes, and different acid-base combinations are identified, see [Titration Curves and Conductivity Graphs for HSC Chemistry](https://tutorgum.com/learn/resources/hsc-chemistry-titration-curves-and-conductivity-graphs/).

## Questions and solutions

### Question 1

A \(25.00\ \text{mL}\) sample of a monoprotic acid is titrated with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide.

The following illustrative data are recorded:

| Volume of NaOH added (mL) | Measured pH |
|---:|---:|
| 0.00 | 2.92 |
| 5.00 | 4.26 |
| 10.00 | 4.74 |
| 15.00 | 5.22 |
| 19.00 | 6.02 |
| 20.00 | 8.69 |
| 21.00 | 11.34 |

Answer the following:

a) Identify the approximate buffer region.

b) Determine the half-equivalence volume and estimate the acid's \(\text{p}K_a\).

c) Explain why the equivalence pH is above 7.

### Solution 1

The buffer region is approximately between \(5.00\) and \(15.00\ \text{mL}\), the half-equivalence volume is \(10.00\ \text{mL}\), and the acid has a \(\text{p}K_a\) of approximately 4.74.

**a) Identify the buffer region.**

The pH rises gradually from 4.26 at \(5.00\ \text{mL}\) to 5.22 at \(15.00\ \text{mL}\).

This is characteristic of a buffer region, where appreciable quantities of weak acid and conjugate base coexist.

The data at \(19.00\ \text{mL}\) show a larger increase in pH. The solution is approaching equivalence, and its ability to resist further additions of hydroxide is weakening.

**b) Determine the half-equivalence volume and pKa.**

The equivalence point occurs at approximately \(20.00\ \text{mL}\).

Therefore:

\[
V_{\frac{1}{2}\text{eq}}=\frac{20.00}{2}
=10.00\ \text{mL}
\]

At half-equivalence:

\[
\text{pH}\approx\text{p}K_a
\]

The measured pH at \(10.00\ \text{mL}\) is 4.74.

Therefore:

\[
\boxed{\text{p}K_a\approx4.74}
\]

**c) Explain why the equivalence pH is above 7.**

The original weak acid has been converted into its conjugate base, \(\ce{A-}\).

This conjugate base reacts with water to produce hydroxide ions:

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

The increase in hydroxide concentration makes the solution basic.

Therefore, the equivalence pH is greater than 7.

### Question 2

A \(25.00\ \text{mL}\) sample of a monoprotic weak acid is titrated with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide.

The equivalence point occurs at \(30.00\ \text{mL}\). When \(12.00\ \text{mL}\) of sodium hydroxide has been added, the measured pH is 5.10.

Assuming ideal buffer behaviour:

a) Calculate the ratio of conjugate base to weak acid at \(12.00\ \text{mL}\).

b) Determine the acid's \(\text{p}K_a\) and \(K_a\).

c) Estimate the pH at half-equivalence.

### Solution 2

The acid has \(\text{p}K_a\approx5.28\), \(K_a\approx5.30\times10^{-6}\), and a half-equivalence pH of approximately 5.28.

**a) Calculate the conjugate-base-to-acid ratio.**

The original acid amount corresponds to \(30.00\ \text{mL}\) of the sodium hydroxide titrant.

At \(12.00\ \text{mL}\), the amount of conjugate base formed is proportional to 12.00 mL.

The remaining weak acid is proportional to:

\[
30.00-12.00=18.00\ \text{mL}
\]

Therefore:

\[
\frac{n(\ce{A-})}{n(\ce{HA})}
=\frac{12.00}{18.00}
=0.667
\]

**b) Calculate pKa and Ka.**

Using the Henderson-Hasselbalch equation:

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

Rearranging:

\[
\begin{aligned}
\text{p}K_a
&=5.10-\log_{10}(0.667)\\
&=5.10+0.176\\
&=5.276
\end{aligned}
\]

Now calculate \(K_a\):

\[
\begin{aligned}
K_a&=10^{-5.276}\\
&=5.30\times10^{-6}
\end{aligned}
\]

**c) Estimate the half-equivalence pH.**

Half-equivalence occurs at:

\[
V_{\frac{1}{2}\text{eq}}=
\frac{30.00}{2}
=15.00\ \text{mL}
\]

At this point, the amounts of \(\ce{HA}\) and \(\ce{A-}\) are approximately equal.

Therefore:

\[
\text{pH}\approx\text{p}K_a
\]

\[
\boxed{\text{pH}\approx5.28}
\]

The important distinction is that \(12.00\ \text{mL}\) is within the buffer region but is not the half-equivalence point. The pH only equals the \(\text{p}K_a\) approximately when the two buffer components are present in equal amounts.

### Question 3

A buffer solution contains:

- \(1.00\ \text{mmol}\) of weak acid, \(\ce{HA}\).
- \(1.50\ \text{mmol}\) of its conjugate base, \(\ce{A-}\).

The acid has a \(\text{p}K_a\) of 4.80.

A student adds \(0.60\ \text{mmol}\) of hydrochloric acid. Assume the volume change is negligible.

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

b) Estimate the new pH.

c) Determine how much additional strong acid could be neutralised stoichiometrically before the original conjugate-base reserve is exhausted.

d) Explain why the pH does not remain constant.

### Solution 3

After the addition, the solution contains \(2.10\ \text{mmol}\) of \(\ce{HA}\) and \(0.40\ \text{mmol}\) of \(\ce{A-}\), giving an approximate pH of 4.08.

**a) Calculate the new amounts.**

Hydrochloric acid is a strong acid, so it supplies hydronium ions.

These react with the conjugate base:

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

The \(0.60\ \text{mmol}\) of added acid consumes the same amount of \(\ce{A-}\).

Therefore:

\[
n(\ce{A-})=1.50-0.60
=0.90\ \text{mmol}
\]

\[
n(\ce{HA})=1.00+0.60
=1.60\ \text{mmol}
\]

**b) Calculate the new pH.**

Apply the Henderson-Hasselbalch approximation:

\[
\begin{aligned}
\text{pH}
&=4.80+\log_{10}\left(\frac{0.90}{1.60}\right)\\
&=4.80+\log_{10}(0.5625)\\
&=4.80-0.250\\
&=4.55
\end{aligned}
\]

The new pH is approximately 4.55.

**c) Determine the remaining acid-neutralising capacity.**

The remaining conjugate base is \(0.90\ \text{mmol}\).

Therefore, a further \(0.90\ \text{mmol}\) of strong acid could be neutralised stoichiometrically before the conjugate-base reserve is exhausted.

In practice, the solution's useful buffering ability would become weak before reaching complete exhaustion.

**d) Explain why the pH changes.**

The buffer does not remove added acid without changing its composition.

Instead, \(\ce{A-}\) reacts with the added hydronium ions and becomes \(\ce{HA}\).

The ratio of conjugate base to weak acid decreases from:

\[
\frac{1.50}{1.00}=1.50
\]

to:

\[
\frac{0.90}{1.60}=0.5625
\]

The lower ratio produces a lower pH.

The buffer still functions because both components remain present in appreciable amounts, but its composition and pH have changed.

### Question 4

A student has \(20.00\ \text{mL}\) of \(0.1000\ \text{mol L}^{-1}\) weak monoprotic acid with:

\[
K_a=1.8\times10^{-5}
\]

The student adds \(80.00\ \text{mL}\) of distilled water before titrating the sample with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide.

The student makes three predictions:

1. The equivalence volume will increase because the acid has been diluted.
2. The pH at half-equivalence will decrease because the solution is more dilute.
3. The buffer will have a lower capacity per unit volume, even though the total number of acid and conjugate-base moles available in the flask has not changed.

Assess all three predictions.

Calculate the equivalence volume, the pH at half-equivalence, and the approximate pH at equivalence.

Assume \(25^\circ\text{C}\), ideal dilute-solution behaviour, and \(K_w=1.0\times10^{-14}\).

### Solution 4

The first two predictions are incorrect, while the third is correct with an important qualification: the total mole reserve in the flask is unchanged, but its concentration and capacity per unit volume are lower.

**Step 1**: Calculate the original amount of weak acid.

\[
\begin{aligned}
n(\ce{HA})
&=cV\\
&=0.1000\times0.02000\\
&=0.002000\ \text{mol}
\end{aligned}
\]

Adding water changes the concentration but does not change the number of moles of acid.

**Step 2**: Determine the equivalence volume.

At equivalence:

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

Therefore:

\[
\begin{aligned}
V(\ce{NaOH})
&=\frac{n}{c}\\
&=\frac{0.002000}{0.1000}\\
&=0.02000\ \text{L}\\
&=20.00\ \text{mL}
\end{aligned}
\]

The equivalence volume is unchanged.

The tempting mistake is to assume that dilution removes some of the acid. It doesn't. The same number of acid moles must still be neutralised.

**Step 3**: Calculate the pH at half-equivalence.

Half-equivalence occurs at:

\[
V_{\frac{1}{2}\text{eq}}=10.00\ \text{mL}
\]

At this point:

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

Therefore:

\[
\begin{aligned}
\text{pH}&\approx\text{p}K_a\\
&=-\log_{10}(1.8\times10^{-5})\\
&=4.74
\end{aligned}
\]

The pH at half-equivalence is approximately 4.74, just as it would be without the added water.

This is because dilution lowers both buffer-component concentrations by the same factor. Their ratio remains approximately one.

**Step 4**: Calculate the conjugate-base concentration at equivalence.

At equivalence, all \(2.000\ \text{mmol}\) of acid has become \(\ce{A-}\).

The total volume is:

\[
20.00+80.00+20.00=120.00\ \text{mL}
\]

Therefore:

\[
[\ce{A-}]
=\frac{0.002000}{0.12000}
=0.01667\ \text{mol L}^{-1}
\]

**Step 5**: Calculate the pH at equivalence.

First calculate \(K_b\):

\[
\begin{aligned}
K_b&=\frac{K_w}{K_a}\\
&=\frac{1.0\times10^{-14}}{1.8\times10^{-5}}\\
&=5.56\times10^{-10}
\end{aligned}
\]

Using the small-ionisation approximation:

\[
\begin{aligned}
[\ce{OH-}]
&\approx\sqrt{K_b[\ce{A-}]}\\
&=\sqrt{(5.56\times10^{-10})(0.01667)}\\
&=3.04\times10^{-6}\ \text{mol L}^{-1}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
\text{pOH}&=-\log_{10}(3.04\times10^{-6})\\
&=5.52\\
\text{pH}&=14.00-5.52\\
&=8.48
\end{aligned}
\]

The equivalence pH is approximately 8.48.

It is lower than the equivalence pH without the added water because the conjugate-base concentration is lower, so its hydrolysis produces fewer hydroxide ions.

**Step 6**: Assess the buffer capacity.

The original amount of weak acid has not changed. At half-equivalence, the flask still contains equal amounts of \(\ce{HA}\) and \(\ce{A-}\), each with \(1.000\ \text{mmol}\).

Therefore, the total number of moles available to react with added acid or base is unchanged.

However, the solution is more dilute. Its buffer capacity per unit volume is lower.

This distinction matters:

- Comparing the entire flask, the total stoichiometric reserve is essentially unchanged.
- Comparing equal-volume samples taken from each solution, the diluted sample contains fewer moles of buffer components and has a lower capacity.

The deeper lesson is that concentration, total amount, and pH are related but are not interchangeable.

### Question 5

A \(20.00\ \text{mL}\) sample contains a mixture of hydrochloric acid, \(\ce{HCl}\), and a weak monoprotic acid, \(\ce{HA}\).

The mixture is titrated with \(0.1000\ \text{mol L}^{-1}\) sodium hydroxide.

An idealised titration curve shows two distinct equivalence features:

- The first occurs at \(6.00\ \text{mL}\).
- The second occurs at \(26.00\ \text{mL}\).

Between these points, the curve has a broad buffer-like region.

At \(16.00\ \text{mL}\), the measured pH is 4.65.

Assume the first equivalence feature represents completion of the strong-acid component, and the second represents completion of both acids. Treat the neutralisation of the strong acid and weak acid as two distinguishable stages.

[IMAGE: A two-stage titration curve for a mixture of strong HCl and weak monoprotic HA titrated with NaOH. Show a first equivalence feature at 6.00 mL, a broad weak-acid buffer region extending from 6.00 to 26.00 mL, the weak-acid half-equivalence point at 16.00 mL where pH = 4.65, and the final equivalence point at 26.00 mL. Clearly distinguish the first strong-acid neutralisation stage from the subsequent weak-acid neutralisation stage.]

Answer the following:

a) Calculate the original concentrations of \(\ce{HCl}\) and \(\ce{HA}\).

b) Determine the \(\text{p}K_a\) and \(K_a\) of the weak acid.

c) A student claims that \(13.00\ \text{mL}\) is the half-equivalence point because it is half of the final equivalence volume. Explain why this is incorrect and calculate the approximate pH at \(13.00\ \text{mL}\).

d) Calculate the approximate pH at \(21.00\ \text{mL}\).

e) At \(21.00\ \text{mL}\), a further \(1.00\ \text{mL}\) of \(0.1000\ \text{mol L}^{-1}\) hydrochloric acid is added. Calculate the new pH and explain why the solution still behaves as a buffer.

### Solution 5

The original concentrations are \(0.0300\ \text{mol L}^{-1}\) for \(\ce{HCl}\) and \(0.1000\ \text{mol L}^{-1}\) for \(\ce{HA}\). The weak acid has \(\text{p}K_a=4.65\) and \(K_a\approx2.24\times10^{-5}\).

The critical point is that the weak acid's half-equivalence volume is measured from the completion of the strong-acid neutralisation, not from zero titrant volume.

**a) Calculate the original concentrations.**

The first equivalence feature occurs at \(6.00\ \text{mL}\).

This represents the amount of sodium hydroxide required to neutralise the hydrochloric acid.

\[
\begin{aligned}
n(\ce{HCl})
&=cV\\
&=0.1000\times0.00600\\
&=0.000600\ \text{mol}
\end{aligned}
\]

The original sample contained \(0.600\ \text{mmol}\) of \(\ce{HCl}\).

Its concentration was:

\[
\begin{aligned}
c(\ce{HCl})
&=\frac{0.000600}{0.02000}\\
&=0.0300\ \text{mol L}^{-1}
\end{aligned}
\]

The second equivalence feature occurs at \(26.00\ \text{mL}\).

The total amount of hydroxide required to neutralise both acids is:

\[
\begin{aligned}
n(\ce{OH-})
&=0.1000\times0.02600\\
&=0.002600\ \text{mol}
\end{aligned}
\]

Subtract the amount used to neutralise \(\ce{HCl}\):

\[
\begin{aligned}
n(\ce{HA})
&=2.600-0.600\\
&=2.000\ \text{mmol}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
c(\ce{HA})
&=\frac{0.002000}{0.02000}\\
&=0.1000\ \text{mol L}^{-1}
\end{aligned}
\]

**b) Calculate pKa and Ka.**

The weak acid's neutralisation occurs between the two equivalence features.

Its half-equivalence point is halfway through this interval:

\[
\begin{aligned}
V_{\frac{1}{2}\text{eq}}
&=6.00+\frac{26.00-6.00}{2}\\
&=16.00\ \text{mL}
\end{aligned}
\]

The measured pH at this volume is 4.65.

Therefore:

\[
\text{p}K_a\approx4.65
\]

And:

\[
\begin{aligned}
K_a&=10^{-4.65}\\
&=2.24\times10^{-5}
\end{aligned}
\]

**c) Explain why \(13.00\ \text{mL}\) is not the weak acid's half-equivalence point.**

The \(13.00\ \text{mL}\) value is half of the total volume required to neutralise both acids.

However, the first \(6.00\ \text{mL}\) is used to neutralise \(\ce{HCl}\). It does not represent neutralisation of the weak acid.

At \(13.00\ \text{mL}\), the amount of sodium hydroxide used to neutralise \(\ce{HA}\) is equivalent to:

\[
13.00-6.00=7.00\ \text{mL}
\]

The weak acid requires \(20.00\ \text{mL}\) of sodium hydroxide for complete neutralisation.

Therefore, the amounts are proportional to:

\[
n(\ce{A-})=7.00
\]

\[
n(\ce{HA})=20.00-7.00=13.00
\]

The ratio is:

\[
\frac{n(\ce{A-})}{n(\ce{HA})}
=\frac{7}{13}
=0.538
\]

Using Henderson-Hasselbalch:

\[
\begin{aligned}
\text{pH}
&=4.65+\log_{10}\left(\frac{7}{13}\right)\\
&=4.65-0.269\\
&=4.38
\end{aligned}
\]

The pH at \(13.00\ \text{mL}\) is approximately 4.38, not 4.65.

The tempting shortcut of halving the final equivalence volume ignores the first neutralisation reaction. The weak acid's buffer region only begins after the strong-acid component has been neutralised.

**d) Calculate the pH at \(21.00\ \text{mL}\).**

The amount of sodium hydroxide used to neutralise the weak acid is:

\[
21.00-6.00=15.00\ \text{mL}
\]

This corresponds to:

\[
n(\ce{A-})=0.1000\times0.01500
=1.500\ \text{mmol}
\]

The original weak acid amount was \(2.000\ \text{mmol}\).

Therefore:

\[
n(\ce{HA})=2.000-1.500
=0.500\ \text{mmol}
\]

Apply the Henderson-Hasselbalch approximation:

\[
\begin{aligned}
\text{pH}
&=4.65+\log_{10}\left(\frac{1.500}{0.500}\right)\\
&=4.65+\log_{10}(3)\\
&=4.65+0.477\\
&=5.13
\end{aligned}
\]

The pH is approximately 5.13.

The ratio of conjugate base to weak acid is \(3:1\), so both components remain in appreciable amounts. The solution is still within the usual useful buffer range.

**e) Calculate the pH after adding hydrochloric acid.**

The added hydrochloric acid amount is:

\[
\begin{aligned}
n(\ce{HCl})
&=0.1000\times0.00100\\
&=0.000100\ \text{mol}\\
&=0.100\ \text{mmol}
\end{aligned}
\]

Hydrochloric acid reacts with the conjugate base:

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

Therefore, the new amounts are:

\[
n(\ce{A-})=1.500-0.100
=1.400\ \text{mmol}
\]

\[
n(\ce{HA})=0.500+0.100
=0.600\ \text{mmol}
\]

The new pH is:

\[
\begin{aligned}
\text{pH}
&=4.65+\log_{10}\left(\frac{1.400}{0.600}\right)\\
&=4.65+\log_{10}(2.333)\\
&=4.65+0.368\\
&=5.02
\end{aligned}
\]

The new pH is approximately 5.02.

The solution still behaves as a buffer because both \(\ce{HA}\) and \(\ce{A-}\) remain present in substantial amounts. The added acid has converted some conjugate base into weak acid, changing the ratio rather than overwhelming the available buffer components.

The central lesson is that the buffer region belongs to a particular acid-base pair, not necessarily to the entire titration from its starting volume. When more than one acid is present, identify which component is reacting before locating half-equivalence or applying the buffer approximation.

## The next step: buffer capacity

Reading a buffer region tells you where the solution resists pH change. The next question is how much acid or base it can actually absorb before that resistance becomes weak.

That depends on the amounts of both buffer components, not just the pH.

For a closer look at calculating how the composition changes after adding strong acid or base, see [How to Calculate Buffer Composition After Adding Strong Acid or Base](https://tutorgum.com/learn/resources/hsc-chemistry-how-to-calculate-buffer-composition-after-adding-strong-acid-or-strong-base/).

The distinction to carry into your next titration question is simple: **the buffer region shows resistance to pH change, but the finite amounts of \(\ce{HA}\) and \(\ce{A-}\) determine how long that resistance lasts.**

Correction to Question 3: The starting amounts in the question are \(1.00\ \text{mmol}\) of HA and \(1.50\ \text{mmol}\) of A−. The corrected solution is:

### Solution 3

After adding hydrochloric acid, the solution contains \(1.60\ \text{mmol}\) of \(\ce{HA}\) and \(0.90\ \text{mmol}\) of \(\ce{A-}\), giving an approximate pH of 4.55.

**a) Calculate the new amounts.**

The reaction is:

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

Therefore:

\[
n(\ce{A-})=1.50-0.60=0.90\ \text{mmol}
\]

\[
n(\ce{HA})=1.00+0.60=1.60\ \text{mmol}
\]

**b) Calculate the new pH.**

\[
\begin{aligned}
\text{pH}
&=4.80+\log_{10}\left(\frac{0.90}{1.60}\right)\\
&=4.80+\log_{10}(0.5625)\\
&=4.80-0.250\\
&=4.55
\end{aligned}
\]

**c) Determine the remaining acid-neutralising capacity.**

The remaining conjugate base is \(0.90\ \text{mmol}\).

Therefore, a further \(0.90\ \text{mmol}\) of strong acid could be neutralised stoichiometrically before the conjugate-base reserve is exhausted.

**d) Explain why the pH changes.**

The ratio of conjugate base to weak acid decreases from:

\[
\frac{1.50}{1.00}=1.50
\]

to:

\[
\frac{0.90}{1.60}=0.5625
\]

This decreases the pH from approximately 4.98 to 4.55.

The solution still functions as a buffer because both components remain in appreciable amounts. However, its composition has changed, and it has less capacity to absorb further added acid.