How to Choose an Indicator for an Acid-Base Titration

Imagine titrating 25.00 mL of ethanoic acid with sodium hydroxide. You calculate that equivalence occurs after adding 25.00 mL of sodium hydroxide. But which indicator should you use?

Bromothymol blue changes colour around pH 6.0-7.6. Phenolphthalein changes colour around pH 8.2-10.0. If you assume that neutralisation always happens at pH 7, bromothymol blue looks like the obvious choice.

It isn’t. For this titration, phenolphthalein is the better indicator because the solution is basic at equivalence, and the sharp pH rise passes through phenolphthalein’s transition range.

The rule is simple: choose an indicator whose transition range falls within the steep section of the titration curve around the equivalence point.

To do that reliably, calculate the equivalence volume first, establish what happens to the pH, and only then choose your indicator.

01What is the experiment actually measuring?

In an acid-base titration, you add a solution of known concentration (the titrant) to a measured volume of another solution (the analyte).

The usual apparatus consists of:

• A burette containing the titrant, such as sodium hydroxide.
• A conical flask containing the measured acid sample.
• A few drops of indicator in the conical flask.
• A clamp holding the burette vertically above the flask.

You slowly release the titrant into the flask while swirling the mixture. The burette measures how much titrant has been delivered.

As you add titrant, the pH changes. Eventually, the indicator changes colour.

There are two important points to distinguish:

• Equivalence point: the point at which the reactants have combined in exactly the stoichiometric mole ratio from the balanced chemical equation.
• Endpoint: the experimentally observed point at which the indicator changes colour sufficiently for you to recognise it.

Ideally, the endpoint occurs extremely close to the equivalence point.

An indicator does not create equivalence. It simply gives you a visible signal that the solution’s pH has entered a particular range.

For a more detailed explanation of this distinction, read Endpoint vs Equivalence Point: The Difference in Titrations.

02Step 1: Calculate the equivalence volume using stoichiometry

Before thinking about pH, establish how much titrant is required to react completely with the analyte.

This is where the balanced chemical equation matters.

For a general reaction:

\[ aA+bB\rightarrow \text{products} \]

At equivalence:

\[ \frac{n(A)}{a}=\frac{n(B)}{b} \]

Here, \(n\) represents the number of moles, while \(a\) and \(b\) are the stoichiometric coefficients.

Remember:

\[ n=CV \]

where:

• \(n\) = amount of substance in mol
• \(C\) = concentration in mol L\(^{-1}\)
• \(V\) = volume in L

For a 1:1 reaction, equal moles of acid and base are required. However, equal volumes are only required if their concentrations are also equal.

This distinction becomes particularly important when dealing with diprotic acids or reactions with different stoichiometric ratios.

03Step 2: Understand the indicator’s transition range

An acid-base indicator is a substance whose colour depends on the pH of its surroundings.

The indicator exists in different chemical forms, which have different colours. As the pH changes, the balance between these forms changes, producing a visible colour transition.

Importantly, this transition does not happen at one exact pH.

It occurs over a range of pH values.

For example, phenolphthalein does not suddenly switch from colourless to pink at precisely pH 9.0. Its visible transition occurs over approximately pH 8.2-10.0.

Think of the transition range as a strip on a ruler rather than a single marking. The indicator changes colour as the solution moves through that strip.

However, an indicator is not a precise pH-measuring instrument. The colour depends on the indicator concentration, lighting, and the observer’s judgement.

Common acid-base indicators

The following are approximate standard transition ranges.

The important part is not memorising which indicator changes colour at which pH.

It is understanding how those ranges relate to the titration curve.

04Step 3: Find the steep section of the titration curve

A titration curve plots:

• pH on the vertical axis
• Volume of titrant added on the horizontal axis

As titrant is added, the pH generally changes gradually at first.

Near the equivalence point, a relatively small addition of titrant can cause a large change in pH. This creates the steep, almost vertical section of the curve.

After equivalence, the pH generally changes more gradually again.

This is the section where you want the indicator to change colour.

Why?

Because if the indicator changes colour while the curve is relatively flat, a noticeable amount of extra titrant might be required before the reaction reaches equivalence.

That produces an endpoint error.

By contrast, when the indicator changes colour within the steep section, even a small volume change moves the solution through its transition range. The endpoint is much closer to the equivalence volume.

The important distinction: equivalence pH versus transition range

A common mistake is to choose an indicator simply because its transition range contains the pH at equivalence.

That is a useful starting point, but it is not the complete test.

You need to check whether the transition range fits inside the steep section of the curve.

For example, imagine an equivalence point at pH 7.0. An indicator with a transition range of pH 5.0-9.0 technically contains pH 7.0.

However, if much of that range extends into the relatively flat sections of the curve, it may produce a less accurate endpoint.

The best choice is an indicator whose entire transition range lies within the sharp pH change.

05Step 4: Recognise how the acid and base strengths affect your choice

The previous example involved a weak acid and a strong base.

Now consider the opposite arrangement: a weak base titrated with a strong acid.

The equivalence solution will generally be acidic because the conjugate acid of the weak base reacts with water to produce hydrogen ions.

This changes which indicator is appropriate.

06Step 5: Check the concentration before making your final choice

You might have heard that any indicator is suitable for a strong-acid/strong-base titration because the pH change near equivalence is so large.

This is a useful generalisation for many ordinary titrations, but it is not always true.

Diluting the solutions can make the steep section narrower in terms of pH.

Consider a strong acid and strong base. At higher concentrations, a very small addition of titrant can cause a large change in the concentration of excess hydrogen or hydroxide ions.

At lower concentrations, that same volume difference produces a smaller excess-ion concentration.

The pH jump is therefore less dramatic.

This matters because an indicator with a transition range extending beyond the steep section may no longer give an accurate endpoint.

The important lesson is that indicator selection depends on the actual concentrations and conditions, not just whether the acid and base are strong or weak.

07A reliable decision process

When you are given an acid-base titration question, use this sequence.

1. Balance the chemical equation.

Identify the mole ratio between the acid and base.

2. Calculate the equivalence volume.

Use \(n=CV\) and the stoichiometric coefficients. Do not assume a 1:1 ratio unless the equation supports it.

3. Establish the pH at and around equivalence.

Consider the strengths of the acid and base. If necessary, calculate the pH using the appropriate equilibrium or excess-reactant calculation.

4. Identify the steep section of the titration curve.

This is the region where a small addition of titrant produces a large change in pH.

5. Compare the indicator transition ranges with that section.

Choose an indicator whose full transition range lies within the steep region.

6. Consider the precision of the endpoint.

If multiple indicators fit, prefer one whose transition range is well positioned within the steep section and gives a clearly recognisable colour change.

If no suitable indicator exists, a pH probe can be used to construct a titration curve and identify equivalence more accurately.

Common indicator-selection mistakes

08Questions and solutions

The following questions progress from standard indicator selection to situations where you need to consider stoichiometry, endpoint error, and changes in concentration.

Question 1

A student titrates 25.00 mL of 0.1000 mol L\(^{-1}\) hydrochloric acid with 0.1000 mol L\(^{-1}\) sodium hydroxide.

The reaction is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

For the illustrative titration curve, the steep section extends from pH 4.0 to pH 10.0 as the volume of sodium hydroxide increases from 24.95 mL to 25.05 mL.

The available indicators have these transition ranges:

• Methyl orange: 3.1-4.4
• Bromothymol blue: 6.0-7.6
• Phenolphthalein: 8.2-10.0

(a) Calculate the equivalence volume.

(b) Identify the best-centred indicator and explain why it is suitable.

(c) Explain why methyl orange is a less suitable choice.

Solution 1

The equivalence volume is 25.00 mL, and bromothymol blue is the best-centred indicator.

(a) Calculate the equivalence volume.

First, calculate the initial moles of hydrochloric acid:

\[ n=CV \]

\[ n=0.1000\times0.02500 \]

\[ n=2.500\times10^{-3}\text{ mol} \]

The reaction has a 1:1 mole ratio, so the same number of moles of sodium hydroxide is required.

\[ V_{\mathrm{eq}}=\frac{2.500\times10^{-3}}{0.1000} \]

\[ \boxed{V_{\mathrm{eq}}=25.00\text{ mL}} \]

(b) Identify the best-centred indicator.

Bromothymol blue has a transition range of pH 6.0-7.6.

This range is entirely within the steep section of the supplied curve, which extends from pH 4.0 to 10.0.

It is also positioned around the equivalence pH of approximately 7.0, making it the best-centred choice.

Phenolphthalein would also be suitable according to the supplied transition interval, although its range sits towards the upper end of the steep section.

(c) Explain why methyl orange is less suitable.

Methyl orange changes colour between pH 3.1 and 4.4.

Part of this transition occurs below the supplied steep section. It could therefore begin changing colour before the reaction approaches equivalence closely enough.

This increases the potential endpoint error.

Question 2

A diprotic acid, represented by \(\ce{H2A}\), is titrated with sodium hydroxide.

A 25.00 mL sample of 0.0800 mol L\(^{-1}\) acid is titrated with 0.1000 mol L\(^{-1}\) sodium hydroxide.

The overall reaction is:

\[ \ce{H2A + 2OH^- -> A^2- + 2H2O} \]

An illustrative titration curve shows two distinct steep sections:

The objective is to identify the indicator for complete neutralisation of the diprotic acid.

Available indicators:

• Methyl orange: 3.1-4.4
• Bromothymol blue: 6.0-7.6
• Phenolphthalein: 8.2-10.0

(a) Calculate the first and second equivalence volumes.

(b) Select the most suitable indicator for complete neutralisation.

(c) Explain why choosing methyl orange because it changes colour during the first steep section would give the wrong endpoint.

Solution 2

The first equivalence volume is 20.00 mL, the second is 40.00 mL, and phenolphthalein is suitable for detecting complete neutralisation.

(a) Calculate the equivalence volumes.

First, calculate the initial moles of diprotic acid:

\[ n(\ce{H2A})=CV \]

\[ =0.0800\times0.02500 \]

\[ =2.000\times10^{-3}\text{ mol} \]

At the first equivalence point, one mole of hydroxide ions has reacted per mole of diprotic acid.

Therefore:

\[ n(\mathrm{OH}^{-})=2.000\times10^{-3}\text{ mol} \]

\[ V_1=\frac{2.000\times10^{-3}}{0.1000} \]

\[ \boxed{V_1=20.00\text{ mL}} \]

At the second equivalence point, two moles of hydroxide ions have reacted per mole of diprotic acid.

Therefore:

\[ n(\mathrm{OH}^{-})=2(2.000\times10^{-3}) \]

\[ =4.000\times10^{-3}\text{ mol} \]

\[ V_2=\frac{4.000\times10^{-3}}{0.1000} \]

\[ \boxed{V_2=40.00\text{ mL}} \]

(b) Select the indicator for complete neutralisation.

Phenolphthalein is the correct choice.

Its transition range of pH 8.2-10.0 lies entirely within the second steep section, which extends from pH 8.0 to 10.4.

It therefore provides a suitable endpoint for the second equivalence point.

(c) Explain why methyl orange is unsuitable for complete neutralisation.

Methyl orange has a transition range of pH 3.1-4.4, which corresponds to the first steep section rather than the second.

At the first equivalence point, only one proton per molecule of diprotic acid has been neutralised.

The hydrogen-containing ion still has another acidic proton available to react.

Consequently, using methyl orange to identify complete neutralisation would stop the titration too early.

The key is to identify which equivalence point the experiment is designed to detect before selecting an indicator.

Question 3

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

The true equivalence volume is 25.00 mL.

An illustrative titration curve gives the following volume intervals for two indicators:

For this question, assume that the student identifies the endpoint at the midpoint of the indicator’s visible transition interval.

(a) Calculate the endpoint volume for each indicator.

(b) Calculate the percentage titre error for each indicator using:

\[ \%\text{ error} = \frac{V_{\mathrm{endpoint}}-V_{\mathrm{eq}}} {V_{\mathrm{eq}}}\times100 \]

(c) Determine which indicator produces the more accurate endpoint.

(d) Explain whether using each endpoint would cause the calculated acid concentration to be overestimated or underestimated.

Solution 3

Phenolphthalein gives the more accurate endpoint, with a smaller absolute titre error of approximately 0.10%, compared with 0.84% for methyl orange.

(a) Calculate the endpoint volumes.

For methyl orange:

\[ V_{\mathrm{endpoint}} =\frac{24.60+24.98}{2} \]

\[ \boxed{V_{\mathrm{endpoint}}=24.79\text{ mL}} \]

For phenolphthalein:

\[ V_{\mathrm{endpoint}} =\frac{25.001+25.050}{2} \]

\[ \boxed{V_{\mathrm{endpoint}}=25.0255\text{ mL}} \]

(b) Calculate the percentage titre errors.

For methyl orange:

\[ \%\text{ error} = \frac{24.79-25.00}{25.00}\times100 \]

\[ \boxed{-0.84\%} \]

For phenolphthalein:

\[ \%\text{ error} = \frac{25.0255-25.00}{25.00}\times100 \]

\[ \boxed{+0.102\%} \]

The negative sign indicates an endpoint before equivalence.

The positive sign indicates an endpoint after equivalence.

(c) Determine the more accurate indicator.

Phenolphthalein is more accurate because its endpoint is much closer to the true equivalence volume.

Its absolute percentage error is approximately 0.10%, compared with 0.84% for methyl orange.

(d) Explain the effect on the calculated acid concentration.

For a 1:1 reaction:

\[ C_{\mathrm{acid}} = \frac{C_{\mathrm{base}}V_{\mathrm{base}}} {V_{\mathrm{acid}}} \]

The calculated acid concentration is directly proportional to the measured titre.

Therefore:

• Methyl orange gives an endpoint that is too early. The measured titre is too small, so the calculated acid concentration is underestimated by approximately 0.84%.
• Phenolphthalein gives an endpoint that is slightly too late. The measured titre is too large, so the calculated acid concentration is overestimated by approximately 0.10%.

This question demonstrates why an indicator’s transition range matters even when it eventually produces a visible colour change.

The important consideration is how close that colour change occurs to the true equivalence volume.

The midpoint assumption is a calculation convention for this question. Actual visual endpoints depend on the observer and experimental conditions.

Question 4

A student titrates 25.00 mL of 0.00100 mol L\(^{-1}\) hydrochloric acid with 0.00100 mol L\(^{-1}\) sodium hydroxide.

The reaction is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The student argues that phenolphthalein must be suitable because this is a strong-acid/strong-base titration.

However, calculations show the following:

• At 24.90 mL of sodium hydroxide added, the solution contains excess hydrogen ions.
• At 25.10 mL of sodium hydroxide added, the solution contains excess hydroxide ions.

Available indicators:

• Methyl orange: 3.1-4.4
• Bromothymol blue: 6.0-7.6
• Phenolphthalein: 8.2-10.0

Assume complete dissociation and ideal solution behaviour at 25°C.

(a) Calculate the equivalence volume.

(b) Calculate the pH at 24.90 mL.

(c) Calculate the pH at 25.10 mL.

(d) Choose the most suitable indicator and explain why the student’s reasoning is incorrect.

Solution 4

Bromothymol blue is the most suitable indicator because its entire transition range lies within the narrow pH jump. Phenolphthalein is unsuitable for this particular dilute titration.

(a) Calculate the equivalence volume.

Initial moles of hydrochloric acid:

\[ n=0.00100\times0.02500 \]

\[ =2.500\times10^{-5}\text{ mol} \]

The reaction is 1:1, so the same number of moles of sodium hydroxide is required.

\[ V_{\mathrm{eq}} =\frac{2.500\times10^{-5}}{0.00100} \]

\[ \boxed{V_{\mathrm{eq}}=25.00\text{ mL}} \]

(b) Calculate the pH at 24.90 mL.

Moles of sodium hydroxide added:

\[ n(\ce{NaOH})=0.00100\times0.02490 \]

\[ =2.490\times10^{-5}\text{ mol} \]

The excess hydrogen ion amount is:

\[ n_{\mathrm{excess}} =2.500\times10^{-5}-2.490\times10^{-5} \]

\[ =1.00\times10^{-7}\text{ mol} \]

Total volume:

\[ V=25.00+24.90=49.90\text{ mL} \]

\[ =0.04990\text{ L} \]

Therefore:

\[ [\mathrm{H}^{+}] =\frac{1.00\times10^{-7}}{0.04990} \]

\[ =2.00\times10^{-6}\text{ mol L}^{-1} \]

The pH is:

\[ pH=-\log(2.00\times10^{-6}) \]

\[ \boxed{pH=5.70} \]

(c) Calculate the pH at 25.10 mL.

The excess hydroxide ion amount is:

\[ n_{\mathrm{excess}} =2.510\times10^{-5}-2.500\times10^{-5} \]

\[ =1.00\times10^{-7}\text{ mol} \]

Total volume:

\[ V=25.00+25.10 \]

\[ =50.10\text{ mL}=0.05010\text{ L} \]

Therefore:

\[ [\mathrm{OH}^{-}] =\frac{1.00\times10^{-7}}{0.05010} \]

\[ =2.00\times10^{-6}\text{ mol L}^{-1} \]

\[ pOH=-\log(2.00\times10^{-6}) \]

\[ pOH=5.70 \]

\[ pH=14.00-5.70 \]

\[ \boxed{pH=8.30} \]

(d) Choose the indicator.

The pH changes from approximately 5.70 to 8.30 across the steep section.

Compare the indicators:

• Methyl orange: 3.1-4.4, too acidic.
• Bromothymol blue: 6.0-7.6, entirely within the steep section.
• Phenolphthalein: 8.2-10.0, mostly outside the steep section.

Bromothymol blue is therefore the best choice.

The student’s mistake is assuming that all strong-acid/strong-base titrations have equally large pH jumps.

The acid and base are both strong, but their concentrations are very low. This makes the pH change around equivalence much narrower than in a more concentrated titration.

Indicator selection must account for concentration as well as acid and base strength.

Question 5

A student performs two titrations using identical acid and base solutions.

Each experiment starts with:

• 25.00 mL of 0.1000 mol L\(^{-1}\) hydrochloric acid.
• 0.1000 mol L\(^{-1}\) sodium hydroxide in the burette.

In Experiment A, the acid is titrated without additional water.

In Experiment B, the student adds 950.0 mL of distilled water to the acid in the conical flask before beginning the titration.

Assume that the water contains no acid or base, volumes are additive, and both solutions behave ideally at 25°C.

A student makes two claims:

1. Adding water changes the equivalence volume because it changes the acid concentration.
2. Adding water cannot affect indicator selection because the number of moles of acid remains unchanged.

Evaluate both claims.

(a) Calculate the equivalence volume in both experiments.

(b) Calculate the pH at 24.90 mL and 25.10 mL of sodium hydroxide added in Experiment A.

(c) Repeat the pH calculations for Experiment B.

(d) Using the following transition ranges, identify all suitable indicators for each experiment. Assume the pH interval between the two calculated volumes represents the steep section for the purpose of this question.

• Methyl orange: 3.1-4.4
• Bromothymol blue: 6.0-7.6
• Phenolphthalein: 8.2-10.0

(e) Which single indicator would be suitable for both experiments? Explain why the two student claims are not both correct.

Solution 5

Both experiments have an equivalence volume of 25.00 mL. However, their pH curves differ, and bromothymol blue is the only indicator suitable for both under the stated criterion.

This is the important distinction: dilution changes concentration and pH, but it does not change the number of moles of acid initially present.

(a) Calculate the equivalence volume in both experiments.

First, calculate the initial moles of hydrochloric acid:

\[ n(\ce{HCl})=CV \]

\[ =0.1000\times0.02500 \]

\[ =2.500\times10^{-3}\text{ mol} \]

The balanced reaction is:

\[ \ce{HCl + NaOH -> NaCl + H2O} \]

The mole ratio is 1:1.

Therefore, both experiments require:

\[ n(\ce{NaOH})=2.500\times10^{-3}\text{ mol} \]

The required volume is:

\[ V_{\mathrm{eq}} =\frac{2.500\times10^{-3}}{0.1000} \]

\[ \boxed{V_{\mathrm{eq}}=25.00\text{ mL}} \]

Adding water to the conical flask does not change the initial amount of hydrochloric acid.

The equivalence volume therefore remains unchanged.

(b) Calculate the pH values for Experiment A.

At 24.90 mL, the amount of sodium hydroxide added is:

\[ n(\ce{NaOH})=0.1000\times0.02490 \]

\[ =2.490\times10^{-3}\text{ mol} \]

Excess hydrochloric acid:

\[ n_{\mathrm{excess}} =2.500\times10^{-3}-2.490\times10^{-3} \]

\[ =1.00\times10^{-5}\text{ mol} \]

Total volume:

\[ V=25.00+24.90=49.90\text{ mL} \]

\[ =0.04990\text{ L} \]

Therefore:

\[ [\mathrm{H}^{+}] =\frac{1.00\times10^{-5}}{0.04990} \]

\[ =2.00\times10^{-4}\text{ mol L}^{-1} \]

\[ \boxed{pH=3.70} \]

At 25.10 mL, there is excess sodium hydroxide:

\[ n_{\mathrm{excess}} =2.510\times10^{-3}-2.500\times10^{-3} \]

\[ =1.00\times10^{-5}\text{ mol} \]

The total volume is:

\[ V=0.05010\text{ L} \]

Therefore:

\[ [\mathrm{OH}^{-}] =\frac{1.00\times10^{-5}}{0.05010} \]

\[ =2.00\times10^{-4}\text{ mol L}^{-1} \]

\[ pOH=3.70 \]

\[ \boxed{pH=10.30} \]

The pH changes from approximately 3.70 to 10.30.

(c) Calculate the pH values for Experiment B.

This is where the extra water matters.

At 24.90 mL, the excess hydrogen ion amount is still:

\[ 1.00\times10^{-5}\text{ mol} \]

However, the total volume is now:

\[ 25.00+950.0+24.90 \]

\[ =999.90\text{ mL} \]

\[ =0.99990\text{ L} \]

Therefore:

\[ [\mathrm{H}^{+}] =\frac{1.00\times10^{-5}}{0.99990} \]

\[ =1.00\times10^{-5}\text{ mol L}^{-1} \]

\[ \boxed{pH=5.00} \]

At 25.10 mL, the excess hydroxide ion amount remains:

\[ 1.00\times10^{-5}\text{ mol} \]

The total volume is:

\[ 25.00+950.0+25.10 \]

\[ =1000.10\text{ mL} \]

\[ =1.00010\text{ L} \]

Therefore:

\[ [\mathrm{OH}^{-}] =\frac{1.00\times10^{-5}}{1.00010} \]

\[ \approx1.00\times10^{-5}\text{ mol L}^{-1} \]

\[ pOH=5.00 \]

\[ \boxed{pH=9.00} \]

The pH now changes from approximately 5.00 to 9.00.

Notice that the equivalence volume has not moved, but the pH jump has become narrower.

(d) Identify suitable indicators for each experiment.

For Experiment A, the steep section spans approximately pH 3.70-10.30.

• Methyl orange: 3.1-4.4. Not entirely within the steep section.
• Bromothymol blue: 6.0-7.6. Suitable.
• Phenolphthalein: 8.2-10.0. Suitable.

Therefore, bromothymol blue and phenolphthalein both satisfy the stated criterion for Experiment A.

For Experiment B, the steep section spans approximately pH 5.00-9.00.

• Methyl orange: 3.1-4.4. Outside the steep section.
• Bromothymol blue: 6.0-7.6. Suitable.
• Phenolphthalein: 8.2-10.0. Its upper transition range extends beyond the steep section.

Therefore, bromothymol blue is the only suitable choice for Experiment B.

(e) Identify the indicator suitable for both experiments and evaluate the claims.

Bromothymol blue is suitable for both experiments because its full transition range of pH 6.0-7.6 falls within the steep section of each curve.

The first claim is incorrect.

Adding water does not change the number of moles of hydrochloric acid, so the stoichiometric amount of sodium hydroxide required remains unchanged. Both experiments reach equivalence at 25.00 mL.

The second claim is also incorrect.

Although the number of acid moles is unchanged, the additional water increases the total solution volume. This reduces the concentration of excess hydrogen or hydroxide ions at a given volume before or after equivalence.

As a result, the pH curve changes and some indicators that worked in Experiment A no longer fit the steep section in Experiment B.

This is the key distinction between stoichiometry and concentration:

• Stoichiometry determines how much titrant is required to reach equivalence.
• Concentration determines how the pH changes as titrant is added.

They are related, but they are not interchangeable.

09Where to go next

Once you can choose an indicator, the next useful skill is calculating the pH on either side of equivalence.

This allows you to predict the shape of a titration curve rather than relying on memorised indicator pairings.

For more practice with those calculations, continue with Calculating pH Before and After Equivalence in Strong-Acid Titrations.

The central decision remains the same: calculate the stoichiometric equivalence first, understand the pH change around it, and choose an indicator whose transition range fits the actual curve.