How to Identify a Strong-Acid Weak-Base Titration Curve

A burette contains hydrochloric acid, and the flask underneath contains ammonia solution. As acid is added, the pH falls sharply. But does the pH reach 7 at the equivalence point, where exactly enough acid has been added to react with the ammonia?

No. At equivalence, the ammonia has been converted into ammonium ions, which make the solution acidic.

To identify a strong-acid-weak-base titration curve, look for an equivalence point below pH 7. If the strong acid is added to the weak base, the curve starts at a basic pH and falls. If the weak base is added to the strong acid, the curve rises instead, but the equivalence point is still acidic.

The reliable method is to establish the reaction and equivalence volume first, then use the pH pattern to explain what is happening.

01Find equivalence before looking at pH

The most common mistake is to search the graph for pH 7 and call that equivalence.

Instead, start with the balanced chemical equation.

For hydrochloric acid reacting with ammonia:

\[ \ce{HCl(aq) + NH3(aq) -> NH4Cl(aq)} \]

The net ionic equation makes the proton transfer clearer:

\[ \ce{H3O+ + NH3 -> NH4+ + H2O} \]

Hydrochloric acid is a strong acid, so it ionises essentially completely in water. Ammonia is a weak base because it reacts only partially with water to produce hydroxide ions.

However, ammonia reacts essentially completely with the hydronium ions supplied by a strong acid during neutralisation.

The balanced equation tells us something more useful than either substance’s starting pH: one mole of hydronium ions reacts with one mole of ammonia.

At the equivalence point:

\[ n(\ce{H3O+})_{\text{added}} = n(\ce{NH3})_{\text{initial}} \]

Here, \(n\) represents amount in moles.

This is a stoichiometric condition. It says how much acid is needed, not what the final pH must be.

02What a pH probe would reveal

Imagine testing that prediction experimentally.

The apparatus contains a measured ammonia sample in the flask, hydrochloric acid in the burette, and a calibrated pH electrode immersed in the flask.

The experiment proceeds in a clear order:

1. Record the pH of the ammonia before adding acid.
2. Open the burette tap and add a measured volume of hydrochloric acid.
3. Stir the solution, allow the pH reading to stabilise, and record it.
4. Repeat, using smaller additions when the pH begins changing rapidly.

The burette measures the volume of acid delivered. The electrode measures pH. Plotting those two measurements produces the titration curve.

What should the original prediction show?

A student who assumes neutralisation always produces pH 7 might predict that the steepest part of the curve is centred on pH 7.

There is a better prediction.

Before equivalence, ammonia remains in the flask. Near equivalence, very little unreacted ammonia remains. At equivalence, the solution contains ammonium chloride, and the ammonium ions behave as weak acids.

So the steep fall should include an acidic equivalence point.

Illustrative pH data

For the 20.00 mL ammonia sample and \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid in the worked example, the following values are calculated predictions, not experimental measurements.

They assume \(25^\circ\mathrm{C}\), an ammonia base dissociation constant of \(K_b=1.8\times10^{-5}\), negligible activity effects, and additive solution volumes.

Notice the striking change around 30.00 mL.

At 29.80 mL, the pH is approximately 7.08. At 30.00 mL, it is 5.24. By 30.20 mL, it has fallen to 3.40.

The equivalence point is not where the pH happens to cross 7.

Why the measurement method needs refining

Suppose the first trial recorded pH only after every 5.00 mL of acid.

It would show the broad downward trend, but it might miss most of the steep region. The student could also mistake a reading near pH 7 for equivalence.

The revised method uses the balanced equation to estimate equivalence at 30.00 mL, then takes much smaller volume steps around that region.

A pH probe is useful because it records the curve without relying on an indicator’s colour change. Calibration, adequate stirring, and allowing the electrode reading to stabilise help distinguish a real pH change from measurement error.

The experimental limitation is important: a graph provides evidence about equivalence, but sparse or noisy readings may not locate it precisely. The balanced equation and measured reaction quantities still matter.

03Why the equivalence point is acidic

At equivalence, the strong acid has reacted with the ammonia. That does not mean the resulting solution contains chemically neutral particles.

Ammonium ions remain.

The reaction is:

\[ \ce{NH4+ + H2O <=> NH3 + H3O+} \]

Some ammonium ions donate protons to water, producing hydronium ions.

That makes the solution acidic.

The chloride ions from hydrochloric acid are spectators in this equilibrium. Under ordinary dilute aqueous conditions, they do not produce a significant change in pH.

A useful distinction: neutralisation versus neutrality

These words describe different things.

Neutralisation refers to an acid-base reaction. At equivalence, the acid and base have reacted in their required stoichiometric amounts.

Neutrality refers to the hydrogen ion balance in the resulting solution. At \(25^\circ\mathrm{C}\), a neutral aqueous solution has pH 7.

A strong acid reacting with a weak base can reach stoichiometric equivalence and still leave an acidic solution.

This is why finding equivalence must come before making a prediction about pH.

04How to recognise the curve without memorising its shape

A descending curve is not enough to prove that a strong acid is reacting with a weak base.

A strong acid added to a strong base also produces a descending curve.

The key difference is what remains at equivalence.

These descriptions assume ordinary dilute aqueous solutions, simple acid-base reactions, and no other significant acidic or basic species.

The important lesson is that the direction of the curve mainly tells you which reagent is being added.

The equivalence-point pH tells you more about the acid-base properties of the products.

What if the acid and base are reversed?

Suppose hydrochloric acid is in the flask, while ammonia is added from the burette.

The curve now rises because the initially acidic solution becomes less acidic as ammonia is added.

But at equivalence, the product is still ammonium chloride.

The equivalence-point pH remains below 7.

For the full calculations in this reversed arrangement, see Strong Acid-Weak Base Titration Curves for HSC Chemistry.

This also explains why a rising curve with an acidic equivalence point is not the same as a weak-acid-strong-base titration. The latter normally has a basic equivalence point because the weak acid’s conjugate base reacts with water.

That contrasting mechanism is explored in Weak Acid-Strong Base Titrations: HSC Chemistry Guide.

Three clues to combine

When given an unfamiliar graph, make these decisions in order.

First: What is reacting?

Identify the acid, the base, and the reaction ratio. A single pH value cannot reliably tell you whether a base is strong or weak, because concentration also affects pH.

Second: Where is equivalence?

Use the stoichiometry and the titre if concentrations are supplied. Otherwise, look for the equivalence region in the measured curve, where pH changes rapidly, and use the shape of the data carefully.

Do not substitute pH 7 for equivalence.

Third: What controls the pH there?

For a strong acid reacting with a weak base, the weak base’s conjugate acid remains and makes the equivalence solution acidic.

That combination gives a much stronger identification than the direction of the curve alone.

05The indicator provides a useful extra test

A pH probe can show the full curve. A colour indicator gives a much narrower piece of information: whether the pH has passed through the indicator’s transition range.

For the illustrative ammonia-hydrochloric acid titration, equivalence is around pH 5.24.

Methyl red changes colour approximately between pH 4.4 and 6.2. That interval lies within the steep equivalence region of this example.

Phenolphthalein changes colour approximately between pH 8.2 and 10.0. In this descending titration, it would change colour too early.

An indicator endpoint is the observed colour change. Equivalence is the stoichiometric point. They need to be close for an accurate titre, but they are not automatically identical.

06Questions and solutions

Question 1

A 15.00 mL sample of \(0.160\ \mathrm{mol\,L^{-1}}\) ammonia is titrated with \(0.120\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

a. Calculate the volume of hydrochloric acid required to reach equivalence.

b. State whether the equivalence-point pH is above, below, or equal to 7, and explain why.

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

Solution 1

The equivalence volume is 20.00 mL, and the equivalence-point pH is below 7 because ammonium ions form during neutralisation.

Part a: Calculate the volume.

The reaction is:

\[ \ce{HCl + NH3 -> NH4Cl} \]

The mole ratio is 1:1.

Calculate the initial ammonia amount:

\[ \begin{aligned} n(\ce{NH3}) &=(0.160)(0.01500)\\ &=2.400\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The same amount of hydrochloric acid is required.

\[ \begin{aligned} V(\ce{HCl}) &=\frac{2.400\times10^{-3}}{0.120}\\ &=0.02000\ \mathrm{L}\\ &=20.00\ \mathrm{mL} \end{aligned} \]

Part b: Explain the pH.

At equivalence, the ammonia has been converted into ammonium ions.

Those ions undergo weak-acid hydrolysis:

\[ \ce{NH4+ + H2O <=> NH3 + H3O+} \]

The resulting hydronium ions make the solution acidic.

The tempting prediction is pH 7 because the acid and base have neutralised each other. That confuses stoichiometric equivalence with a chemically neutral solution.

Question 2

Two unlabelled conical flasks, X and Y, each contain 20.00 mL of a base.

One contains ammonia, \(\ce{NH3}\), and the other contains sodium hydroxide, \(\ce{NaOH}\). Both are titrated separately with \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

Both require 30.00 mL of acid to reach equivalence.

The following pH readings are obtained:

Assume \(25^\circ\mathrm{C}\), \(K_b(\ce{NH3})=1.8\times10^{-5}\), and that the readings are consistent with ideal dilute-solution calculations.

a. Identify the base in each flask.

b. Calculate the initial concentration of each base.

c. Explain why both flasks require the same equivalence volume but have very different pH values at half-equivalence.

d. Explain how the equivalence-point readings confirm your identifications.

Solution 2

Flask X contains ammonia, flask Y contains sodium hydroxide, and both have an initial concentration of \(0.150\ \mathrm{mol\,L^{-1}}\). Their different pH values arise because ammonia forms a buffer with ammonium, while sodium hydroxide leaves excess hydroxide ions before equivalence.

Part a: Identify the bases.

Flask X has pH 9.26 at half-equivalence and an acidic equivalence point.

That matches ammonia reacting with strong acid.

Flask Y has a much higher pH at half-equivalence and reaches approximately pH 7 at equivalence.

That matches sodium hydroxide reacting with hydrochloric acid.

Part b: Calculate the concentrations.

Both bases react in a 1:1 mole ratio with hydrochloric acid.

At equivalence:

\[ \begin{aligned} n(\ce{HCl}) &=(0.100)(0.03000)\\ &=3.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Each flask initially contains the same amount of base.

\[ \begin{aligned} c&=\frac{n}{V}\\ &=\frac{3.00\times10^{-3}}{0.02000}\\ &=0.150\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Both bases have the same analytical concentration, but they behave differently in water.

Part c: Explain half-equivalence.

At 15.00 mL, half the base has been neutralised in each flask.

For flask X:

\[ \ce{NH3 + H3O+ -> NH4+ + H2O} \]

The solution contains approximately equal mole amounts of ammonia and ammonium.

Therefore:

\[ \begin{aligned} \mathrm{pH}&=\mathrm{p}K_a\\ &\approx9.26 \end{aligned} \]

For flask Y, sodium hydroxide is a strong base, so the remaining hydroxide controls the pH.

The hydroxide remaining is:

\[ \begin{aligned} n(\ce{OH-}) &=0.00300-0.00150\\ &=0.00150\ \mathrm{mol} \end{aligned} \]

The total solution volume is:

\[ 20.00+15.00=35.00\ \mathrm{mL} \]

Thus:

\[ \begin{aligned} [\ce{OH-}] &=\frac{0.00150}{0.03500}\\ &=0.0429\ \mathrm{mol\,L^{-1}} \end{aligned} \]

At \(25^\circ\mathrm{C}\), \(\mathrm{pH}+\mathrm{pOH}=14.00\), where \(\mathrm{pOH}=-\log_{10}[\ce{OH-}]\).

\[ \begin{aligned} \mathrm{pOH}&=-\log_{10}(0.0429)=1.37\\ \mathrm{pH}&=14.00-1.37=12.63 \end{aligned} \]

The difference is not caused by different amounts of base. It is caused by the different acid-base equilibria.

Part d: Explain equivalence.

Flask X contains ammonium chloride at equivalence. Ammonium is a weak acid, giving pH below 7.

Flask Y contains sodium chloride and water. Neither sodium nor chloride ions undergo significant acid-base hydrolysis, so the equivalence pH is approximately 7 at \(25^\circ\mathrm{C}\).

A common mistake is to assume that identical equivalence volumes imply identical curves. Equivalence volume reveals reaction amount; it does not establish the strength of the reacting base.

Question 3

Two flasks each initially contain 20.00 mL of \(0.120\ \mathrm{mol\,L^{-1}}\) ammonia.

Flask A is titrated directly with \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

Before titrating flask B, a student adds 60.00 mL of distilled water. Flask B is then titrated with the same hydrochloric acid.

Both titrations are carried out at \(25^\circ\mathrm{C}\).

Use:

\[ \begin{aligned} K_b(\ce{NH3})&=1.8\times10^{-5}\\ K_w&=1.0\times10^{-14} \end{aligned} \]

a. Calculate the equivalence volume for each flask.

b. Estimate the equivalence-point pH for each flask.

c. Predict whether the pH at half-equivalence will be different in the two flasks. Explain.

Assume the added water does not remove or chemically react with ammonia, and solution volumes are additive.

Solution 3

Both flasks reach equivalence after 24.00 mL of hydrochloric acid. Flask A has an approximate equivalence pH of 5.26, and diluted flask B has an approximate equivalence pH of 5.45. Both have approximately pH 9.26 at half-equivalence.

The important distinction is that adding water changes concentration, but not the number of moles of ammonia.

Part a: Equivalence volumes.

Both flasks initially contain:

\[ \begin{aligned} n(\ce{NH3}) &=(0.120)(0.02000)\\ &=2.40\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Since the reaction is 1:1:

\[ \begin{aligned} V(\ce{HCl}) &=\frac{2.40\times10^{-3}}{0.100}\\ &=0.02400\ \mathrm{L}\\ &=24.00\ \mathrm{mL} \end{aligned} \]

Both need the same amount of acid because both contain the same initial amount of ammonia.

Part b: Equivalence-point pH.

At equivalence, both solutions contain ammonium ions.

First calculate the ammonium acid dissociation constant:

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

For flask A, the total volume at equivalence is:

\[ 20.00+24.00=44.00\ \mathrm{mL} \]

Therefore:

\[ \begin{aligned} C_A &=\frac{2.40\times10^{-3}}{0.04400}\\ &=0.0545\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Using the weak-acid approximation:

\[ \begin{aligned} [\ce{H3O+}]_A &\approx\sqrt{K_aC_A}\\ &=5.51\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

So:

\[ \mathrm{pH}_A=5.26 \]

For flask B, the total volume at equivalence is:

\[ 20.00+60.00+24.00=104.00\ \mathrm{mL} \]

Therefore:

\[ \begin{aligned} C_B &=\frac{2.40\times10^{-3}}{0.10400}\\ &=0.0231\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Then:

\[ \begin{aligned} [\ce{H3O+}]_B &\approx\sqrt{K_aC_B}\\ &=3.58\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

So:

\[ \mathrm{pH}_B=5.45 \]

Flask B is less acidic at equivalence because its ammonium concentration is lower.

However, both remain acidic because ammonium ions still donate protons to water.

Part c: Half-equivalence.

Half-equivalence occurs after 12.00 mL of hydrochloric acid has been added to either flask.

Each then contains approximately:

\[ \begin{aligned} n(\ce{NH3})&=1.20\times10^{-3}\ \mathrm{mol}\\ n(\ce{NH4+})&=1.20\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The ammonia-to-ammonium mole ratio is 1:1 in both flasks.

Therefore:

\[ \mathrm{pH}\approx\mathrm{p}K_a=9.26 \]

The added water lowers both concentrations but does not change their ratio.

That is the key constraint. The buffer pH depends primarily on the ratio of the conjugate pair, while the equivalence-point pH depends on the concentration of the ammonium ions.

This conclusion uses the usual dilute-solution equilibrium approximations. Significant activity effects or extremely dilute solutions can introduce departures from these simple relationships.

Question 4

A student titrates 20.00 mL of \(0.150\ \mathrm{mol\,L^{-1}}\) ammonia with \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid.

Instead of using a pH probe to locate the equivalence region, the student chooses phenolphthalein.

In this hypothetical trial, the last visible pink colour disappears when the pH falls to 8.20. The student records that burette volume as the equivalence volume.

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

\[ K_b(\ce{NH3})=1.8\times10^{-5} \]

a. Estimate the hydrochloric acid volume added when the pH reaches 8.20.

b. Calculate the apparent ammonia concentration the student would obtain by treating that volume as equivalence.

c. State whether the result is too high or too low, and explain why.

d. Explain why methyl red, with a transition range of approximately pH 4.4-6.2, is more suitable for this titration. Would it guarantee an exactly correct equivalence volume?

Assume ideal buffer behaviour at pH 8.20.

Solution 4

The phenolphthalein colour change occurs after approximately 27.6 mL of acid, giving an apparent ammonia concentration of \(0.138\ \mathrm{mol\,L^{-1}}\). This is about 8% too low because the indicator changes colour before equivalence.

Methyl red is more suitable because its transition range lies within the steep acidic equivalence region, although its endpoint is still not mathematically identical to equivalence.

Part a: Find the volume at pH 8.20.

The initial ammonia amount is:

\[ \begin{aligned} n_0&=(0.150)(0.02000)\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

For ammonium:

\[ \mathrm{p}K_a\approx9.26 \]

Before equivalence, the solution contains both ammonia and ammonium.

Using the buffer relationship:

\[ \mathrm{pH} = \mathrm{p}K_a+ \log_{10}\frac{n(\ce{NH3})}{n(\ce{NH4+})} \]

Substitute pH 8.20:

\[ 8.20=9.26+\log_{10}r \]

where \(r=n(\ce{NH3})/n(\ce{NH4+})\).

Therefore:

\[ \begin{aligned} \log_{10}r&=-1.06\\ r&=10^{-1.06}\\ &=0.0871 \end{aligned} \]

Let \(n_a\) be the moles of hydrochloric acid added.

The acid converts ammonia into ammonium, so:

\[ n(\ce{NH3})=n_0-n_a \]

and:

\[ n(\ce{NH4+})\approx n_a \]

Thus:

\[ \frac{0.00300-n_a}{n_a}=0.0871 \]

Rearranging:

\[ \begin{aligned} 0.00300&=1.0871n_a\\ n_a&=2.76\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The acid volume is:

\[ \begin{aligned} V&=\frac{2.76\times10^{-3}}{0.100}\\ &=0.0276\ \mathrm{L}\\ &=27.6\ \mathrm{mL} \end{aligned} \]

Part b: Calculate the apparent concentration.

If the student treats 27.6 mL as equivalence, the calculated ammonia concentration becomes:

\[ \begin{aligned} c_\text{apparent} &=\frac{(0.100)(0.0276)}{0.02000}\\ &=0.138\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Part c: Determine the direction of error.

The true ammonia concentration is:

\[ 0.150\ \mathrm{mol\,L^{-1}} \]

The percentage error is approximately:

\[ \begin{aligned} \%\text{ error} &=\frac{0.138-0.150}{0.150}\times100\\ &=-8.0\% \end{aligned} \]

The result is too low.

This occurs because phenolphthalein becomes colourless while a measurable amount of ammonia remains unreacted.

The student has mistaken an early indicator endpoint for the true equivalence point.

Part d: Evaluate methyl red.

The actual equivalence volume is 30.00 mL, and the calculated equivalence pH is approximately 5.24.

Methyl red changes colour over approximately pH 4.4-6.2. That range includes the equivalence pH and lies within the rapid pH change around 30.00 mL.

It is therefore much more suitable than phenolphthalein.

However, it does not guarantee zero error.

An indicator changes colour over a range, and the observed endpoint depends on the selected colour threshold, mixing, lighting, and the amount of titrant added per drop.

A good indicator minimises the difference between endpoint and equivalence. It does not remove that difference by definition.

Question 5

Two identical conical flasks, A and B, initially contain 25.00 mL of \(0.100\ \mathrm{mol\,L^{-1}}\) ammonia.

Both are titrated with \(0.100\ \mathrm{mol\,L^{-1}}\) hydrochloric acid at \(25^\circ\mathrm{C}\).

After exactly 10.00 mL of acid has been added to each flask, both solutions are thoroughly mixed.

Flask A is left unchanged.

From flask B, the student withdraws exactly 7.00 mL of the well-mixed solution and replaces it with 7.00 mL of distilled water. No additional ammonia or hydrochloric acid is added during this operation.

The titrations then resume.

Use:

\[ \begin{aligned} K_b(\ce{NH3})&=1.8\times10^{-5}\\ K_w&=1.0\times10^{-14} \end{aligned} \]

Assume the withdrawn liquid has the same composition as the well-mixed flask, no ammonia escapes, the water adds no significant acid or base, and volumes are additive. Use the usual dilute-solution approximations.

a. Calculate the pH of flask A after the first 10.00 mL of acid. Determine the pH of flask B immediately after the withdrawal and replacement.

b. If another 5.00 mL of hydrochloric acid is added to each flask, which will have the lower pH? Calculate both approximate pH values.

c. Calculate the additional acid volume needed to reach equivalence in each flask, measured from the moment the titrations resume. State the total burette volume used in each titration.

d. Estimate the equivalence-point pH of each flask.

e. A student argues: “Both flasks have the same pH and the same total volume when the titrations resume. Therefore, they must need the same further volume of acid.”

Explain precisely why that conclusion is wrong.

Solution 5

Both flasks have approximately pH 9.43 when titration resumes, despite containing different amounts of ammonia. After another 5.00 mL of acid, flask B has the lower pH: approximately 8.99 compared with 9.08 in A.

Flask A then reaches equivalence at a cumulative acid volume of 25.00 mL, while flask B reaches it at 22.00 mL. Both equivalence solutions are acidic, with pH values of approximately 5.28 and 5.31, respectively.

The central idea is that pH can remain unchanged when the amounts of both members of a buffer are reduced in the same proportion, even though the amount of acid needed to finish neutralisation changes.

Part a: Determine the pH after the intervention.

Initially, each flask contains:

\[ \begin{aligned} n_0(\ce{NH3}) &=(0.100)(0.02500)\\ &=2.50\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The first 10.00 mL of hydrochloric acid supplies:

\[ \begin{aligned} n(\ce{HCl}) &=(0.100)(0.01000)\\ &=1.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

After neutralisation, each flask initially contains approximately:

\[ \begin{aligned} n(\ce{NH3})&=1.50\times10^{-3}\ \mathrm{mol}\\ n(\ce{NH4+})&=1.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

For flask A:

\[ \begin{aligned} \mathrm{pH} &=9.26+\log_{10}\left(\frac{1.50}{1.00}\right)\\ &=9.43 \end{aligned} \]

Now consider flask B.

Its volume before withdrawal is:

\[ 25.00+10.00=35.00\ \mathrm{mL} \]

Removing 7.00 mL removes one-fifth of the well-mixed solution.

Therefore, 80% of each dissolved species remains.

The ammonia remaining is:

\[ \begin{aligned} n_B(\ce{NH3}) &=0.80(1.50\times10^{-3})\\ &=1.20\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The ammonium remaining is:

\[ \begin{aligned} n_B(\ce{NH4+}) &=0.80(1.00\times10^{-3})\\ &=0.800\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Replacing the withdrawn liquid with distilled water restores the volume to 35.00 mL, but it does not restore the removed ammonia or ammonium.

The ratio remains:

\[ \frac{n_B(\ce{NH3})}{n_B(\ce{NH4+})} = \frac{1.20}{0.800}=1.50 \]

Hence:

\[ \mathrm{pH}_B=9.43 \]

The pH is approximately unchanged because the buffer ratio is unchanged.

That does not mean the chemical amounts are unchanged.

Part b: Add another 5.00 mL of acid.

The additional hydrochloric acid amount is:

\[ \begin{aligned} n(\ce{HCl}) &=(0.100)(0.00500)\\ &=0.500\times10^{-3}\ \mathrm{mol} \end{aligned} \]

For flask A, the new ammonia amount is:

\[ \begin{aligned} n_A(\ce{NH3}) &=(1.50-0.500)\times10^{-3}\\ &=1.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The ammonium amount becomes:

\[ n_A(\ce{NH4+})=1.50\times10^{-3}\ \mathrm{mol} \]

Therefore:

\[ \begin{aligned} \mathrm{pH}_A &=9.26+\log_{10}\left(\frac{1.00}{1.50}\right)\\ &=9.08 \end{aligned} \]

For flask B:

\[ \begin{aligned} n_B(\ce{NH3}) &=(1.20-0.500)\times10^{-3}\\ &=0.700\times10^{-3}\ \mathrm{mol} \end{aligned} \]

And:

\[ n_B(\ce{NH4+})=1.30\times10^{-3}\ \mathrm{mol} \]

So:

\[ \begin{aligned} \mathrm{pH}_B &=9.26+\log_{10}\left(\frac{0.700}{1.30}\right)\\ &=8.99 \end{aligned} \]

Flask B has the lower pH.

Both flasks received the same extra amount of acid, but that acid consumed a larger fraction of the remaining ammonia in flask B.

The buffer in B has a smaller capacity to resist the added acid because some buffer material was removed.

Part c: Calculate the equivalence volumes.

At the moment titration resumes, flask A contains \(1.50\times10^{-3}\) mol of unreacted ammonia.

It needs that amount of additional strong acid.

\[ \begin{aligned} V_A &=\frac{1.50\times10^{-3}}{0.100}\\ &=0.01500\ \mathrm{L}\\ &=15.00\ \mathrm{mL} \end{aligned} \]

The total acid volume dispensed from the burette is:

\[ 10.00+15.00=25.00\ \mathrm{mL} \]

Flask B contains only \(1.20\times10^{-3}\) mol of unreacted ammonia when titration resumes.

It therefore needs:

\[ \begin{aligned} V_B &=\frac{1.20\times10^{-3}}{0.100}\\ &=0.01200\ \mathrm{L}\\ &=12.00\ \mathrm{mL} \end{aligned} \]

Its cumulative burette volume is:

\[ 10.00+12.00=22.00\ \mathrm{mL} \]

The total burette reading is not simply the acid amount that remains inside the flask. Some acid has already reacted to form ammonium, and some of that ammonium was later withdrawn.

The correct calculation tracks the unreacted ammonia remaining in each flask.

Part d: Calculate the equivalence-point pH values.

For flask A, the total ammonium amount at equivalence is approximately:

\[ n_A(\ce{NH4+})=2.50\times10^{-3}\ \mathrm{mol} \]

Its final volume is:

\[ 25.00+25.00=50.00\ \mathrm{mL} \]

Therefore:

\[ C_A=\frac{0.00250}{0.05000} =0.0500\ \mathrm{mol\,L^{-1}} \]

Using \(K_a=5.56\times10^{-10}\):

\[ \begin{aligned} [\ce{H3O+}]_A &\approx\sqrt{K_aC_A}\\ &=5.27\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Hence:

\[ \mathrm{pH}_A=5.28 \]

For flask B, the final ammonium amount equals the ammonium retained after withdrawal plus that produced during the remaining neutralisation.

\[ \begin{aligned} n_B(\ce{NH4+}) &=(0.800+1.20)\times10^{-3}\\ &=2.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

The flask contains 35.00 mL when titration resumes, followed by another 12.00 mL of acid.

Final volume:

\[ 35.00+12.00=47.00\ \mathrm{mL} \]

Therefore:

\[ \begin{aligned} C_B &=\frac{0.00200}{0.04700}\\ &=0.0426\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The hydronium concentration is:

\[ \begin{aligned} [\ce{H3O+}]_B &\approx\sqrt{K_aC_B}\\ &=4.86\times10^{-6}\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Thus:

\[ \mathrm{pH}_B=5.31 \]

The equivalence-point pH values are surprisingly similar, even though the cumulative equivalence volumes differ by 3.00 mL.

Both solutions contain the same weak acid, ammonium, at fairly similar concentrations when equivalence is reached.

Part e: Explain the student’s error.

The student’s conclusion is wrong because pH does not directly measure the total amount of unreacted ammonia in the flask.

In this buffer system, pH depends mainly on the ammonia-to-ammonium ratio.

Withdrawing a well-mixed sample removes ammonia and ammonium in the same proportion. Replacing the lost volume with water restores the original total volume but not the original amounts.

The pH therefore remains approximately unchanged, while the amount of ammonia available to react with further acid decreases.

Flask B needs less additional hydrochloric acid to reach equivalence.

This is the deeper reason a titration curve cannot always be interpreted from pH alone: pH describes the current acid-base condition, while equivalence depends on the reacting amounts and how the sample has been treated.

07What to use this method for next

When identifying a strong-acid-weak-base titration curve, the most reliable order is to establish the reaction stoichiometry, locate equivalence, identify the chemical species present, and only then interpret the pH.

Once you can do that, you can make a more informed experimental decision: whether an indicator’s transition range is close enough to the acidic equivalence point to produce a useful titre.

That decision comes from the curve’s actual steep region, not from assuming that every neutralisation must finish at pH 7.