Acid-base titrations

Chemistry 1110 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to Acid-Base Titrations!

Have you ever wondered how food scientists check the exact acidity of orange juice, or how pharmaceutical companies make sure an aspirin tablet contains the precise dose of active medicine? The answer is volumetric analysis—specifically, acid-base titrations.

Titration is one of the most fundamental laboratory techniques in Chemistry. At first, all the glassware, colour changes, and calculations might feel a little overwhelming, but don't worry! By breaking it down into simple, step-by-step stages, you will master both the practical techniques and the maths behind them.

1. Standard Solutions: The Starting Point

Before you can find the unknown concentration of an acid or alkali, you need a reference point. This reference is called a standard solution.

A standard solution is defined as a solution of accurately known concentration.

What Makes a Good Primary Standard?

To make a reliable standard solution directly by weighing, a compound (known as a primary standard) must meet specific criteria:

• High purity: It must be completely free from impurities so that the mass you weigh is 100% the intended substance.
• Stability: It must not react with air, decompose, or gain/lose moisture.
• High molar mass (\(M_r\)): A high molar mass reduces the percentage weighing error on the balance.
• Readily soluble in water: It must dissolve easily at room temperature.

Did you know? Solid sodium hydroxide (\(\text{NaOH}\)) is not a primary standard! It is hygroscopic (it absorbs water vapour and carbon dioxide from the air), so you cannot weigh it accurately to make a direct standard solution.

Step-by-Step: Preparing a Standard Solution

Examiners love asking for this exact experimental procedure. Here is the foolproof 6-step method:

Step 1: Weigh accurately
Weigh the solid in a weighing boat using a balance (preferably reading to at least 2 decimal places). Use the weighing by difference technique (mass of boat + solid, tip solid out, reweigh empty boat) to find the exact mass transferred.

Step 2: Dissolve the solid
Transfer the solid into a clean beaker. Add a small volume of deionised water (usually \(50\text{ cm}^3\) to \(100\text{ cm}^3\)) and stir thoroughly with a glass rod until all the solid dissolves completely.

Step 3: Transfer quantitatively
Pour the solution through a filter funnel into a clean volumetric flask (e.g., \(250.0\text{ cm}^3\)).

Step 4: Rinse everything (Quantitative Washings)
Rinse the beaker, glass stirring rod, and funnel with deionised water, and pour all the washings into the volumetric flask. This ensures no moles of solute are left behind!

Step 5: Fill to the mark
Add deionised water until the bottom of the meniscus sits exactly on the calibration line at eye level. Near the mark, add water dropwise using a teat pipette so you do not overshoot.

Step 6: Invert to mix
Stopper the flask securely and invert it repeatedly (at least 10 to 15 times) to ensure thorough mixing and a uniform concentration throughout.

Key Takeaway: A standard solution is a solution of known concentration. Preparing it requires complete dissolution, careful transfer of washings, filling to the bottom of the meniscus, and thorough inversion.

2. The Titration Technique: Apparatus & Practical Skills

In an acid-base titration, a solution of known concentration is delivered from a burette into a known volume of another solution measured using a volumetric pipette inside a conical flask until the reaction is complete.

How to Rinse Your Glassware Correctly

A classic exam question asks what each piece of glassware should be rinsed with before starting:

• Volumetric Pipette: Rinse first with deionised water, then with the solution it is going to measure. (Rinsing only with water leaves droplets that dilute the solution!).
• Burette: Rinse first with deionised water, then with the solution it is going to contain.
• Conical Flask: Rinse ONLY with deionised water! Never rinse it with the solution being placed in it, as this would add extra unmeasured moles of reactant.
• Volumetric Flask: Rinse ONLY with deionised water.

Carrying Out the Titration

1. Use a pipette filler to draw the solution into the volumetric pipette until the bottom of the meniscus touches the line at eye level.
2. Discharge the pipette into a clean conical flask. Touch the tip of the pipette against the side of the flask (do not blow out the tiny drop remaining in the tip; the pipette is calibrated to account for this).
3. Fill the burette using a funnel, ensuring the jet space below the tap is completely filled with liquid and contains no air bubbles. Remove the funnel before taking any readings!
4. Add 2–3 drops of the appropriate indicator to the conical flask.
5. Place the conical flask on a white tile under the burette (the white tile makes it much easier to see the exact colour change at the endpoint).
6. Perform a rough "trial" titration first to get an approximate volume, swirling the flask continuously.
7. Perform subsequent accurate titrations: add solution rapidly until near the endpoint, then add drop-by-drop while swirling until a single drop causes a permanent colour change.
8. Read the burette at eye level to the bottom of the meniscus, recording all burette readings to \(2\text{ decimal places}\) (ending in \(.00\) or \(.05\text{ cm}^3\)).

Concordant Titres

Concordant results are titre values that are within \(0.10\text{ cm}^3\) of each other.

• Always ignore your rough trial titration when calculating the mean.
• Calculate the average (mean) titre using only the concordant values.
• Example: If your titres are \(24.80\text{ cm}^3\) (Rough), \(23.40\text{ cm}^3\), \(23.50\text{ cm}^3\), and \(23.85\text{ cm}^3\), you only average \(23.40\text{ cm}^3\) and \(23.50\text{ cm}^3\):
\(\text{Mean Titre} = \frac{23.40 + 23.50}{2} = 23.45\text{ cm}^3\)

Key Takeaway: Pipettes and burettes are rinsed with their respective solutions; conical flasks are rinsed only with water. Always calculate the mean titre from concordant results within \(0.10\text{ cm}^3\).

3. Acid-Base Indicators

An indicator is a weak acid (or base) that changes colour depending on the \( \text{pH} \) of the solution. Because different acid-base reactions neutralise at different \( \text{pH} \) levels, choosing the right indicator is vital.

The Two Main Indicators in AS Chemistry

1. Methyl Orange:
• Colour in acid: Red
• Colour in alkali: Yellow
• Colour at endpoint (neutral point): Orange
• \( \text{pH} \) range of change: \(3.1 - 4.4\)

2. Phenolphthalein:
• Colour in acid: Colourless
• Colour in alkali: Pink (or magenta)
• Colour at endpoint: Faint pale pink (when titrating acid with alkali) or Colourless (when titrating alkali with acid)
• \( \text{pH} \) range of change: \(8.3 - 10.0\)

Choosing the Correct Indicator

• Strong Acid + Strong Base: (e.g., \( \text{HCl} \) and \( \text{NaOH} \))
Either methyl orange or phenolphthalein can be used because the \( \text{pH} \) change at the equivalence point is very steep (from around \( \text{pH } 3 \) to \( 11 \)).

• Strong Acid + Weak Base: (e.g., \( \text{HCl} \) and \( \text{NH}_3 \))
Use Methyl Orange (equivalence point lies below \( \text{pH } 7 \)).

• Weak Acid + Strong Base: (e.g., \( \text{CH}_3\text{COOH} \) and \( \text{NaOH} \))
Use Phenolphthalein (equivalence point lies above \( \text{pH } 7 \)).

• Weak Acid + Weak Base: (e.g., \( \text{CH}_3\text{COOH} \) and \( \text{NH}_3 \))
Neither indicator works because there is no sharp, vertical \( \text{pH} \) change at equivalence. A \( \text{pH} \) meter must be used instead.

Memory Aid for Indicator Colours:
"Methyl Orange is Red in acid, Yellow in alkali" \(\rightarrow\) M-O-R-Y.
"Phenolphthalein is Pink in alkali" \(\rightarrow\) P-P.

Key Takeaway: Match the indicator to the strength of the acid and base. Strong acid/weak base = Methyl orange; Weak acid/strong base = Phenolphthalein.

4. Titration Calculations: Step-by-Step

Don't be intimidated by titration maths! Every single problem follows the exact same 4-step logical flow:

1. Calculate moles of the known substance: \(n = c \times \frac{V}{1000}\)
2. Use the balanced equation: Apply the mole ratio to find moles of the unknown substance.
3. Scale up (if necessary): If a portion/aliquot (e.g., \(25.0\text{ cm}^3\)) was taken from a larger volumetric flask (e.g., \(250.0\text{ cm}^3\)), multiply by the dilution factor (\(\times 10\)).
4. Calculate the required quantity: Find concentration (\(c = \frac{n \times 1000}{V}\)), molar mass (\(M_r = \frac{m}{n}\)), or mass (\(m = n \times M_r\)).

Worked Example 1: Finding an Unknown Concentration

A \(25.0\text{ cm}^3\) sample of \( \text{NaOH} \) solution requires \(21.50\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) \( \text{H}_2\text{SO}_4 \) for complete neutralisation. Calculate the concentration of the \( \text{NaOH} \) solution in \( \text{mol dm}^{-3} \).

Step 1: Write the balanced chemical equation
\(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)

Step 2: Find moles of the known substance (\( \text{H}_2\text{SO}_4 \))
\(n(\text{H}_2\text{SO}_4) = c \times V = 0.100\text{ mol dm}^{-3} \times \frac{21.50}{1000}\text{ dm}^3 = 0.00215\text{ mol}\)

Step 3: Use the mole ratio to find moles of unknown (\( \text{NaOH} \))
From the equation: \(1\text{ mole of } \text{H}_2\text{SO}_4 : 2\text{ moles of } \text{NaOH}\)
\(n(\text{NaOH}) = 0.00215 \times 2 = 0.00430\text{ mol}\)

Step 4: Calculate the concentration of \( \text{NaOH} \)
\(c(\text{NaOH}) = \frac{n}{V} = \frac{0.00430\text{ mol}}{\frac{25.0}{1000}\text{ dm}^3} = 0.172\text{ mol dm}^{-3}\)

Worked Example 2: Determining Water of Crystallisation

A student dissolves \(2.86\text{ g}\) of hydrated sodium carbonate crystals (\( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \)) in deionised water and makes the volume up to \(250.0\text{ cm}^3\) in a volumetric flask. A \(25.0\text{ cm}^3\) portion of this solution neutralises \(20.00\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) \( \text{HCl} \). Determine the value of \(x\).

Step 1: Equation
\(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\)

Step 2: Moles of known substance (\( \text{HCl} \))
\(n(\text{HCl}) = 0.100 \times \frac{20.00}{1000} = 0.00200\text{ mol}\)

Step 3: Moles of \( \text{Na}_2\text{CO}_3 \) in the \(25.0\text{ cm}^3\) sample
Ratio is \(2\text{ HCl} : 1\text{ Na}_2\text{CO}_3\)
\(n(\text{Na}_2\text{CO}_3\text{ in } 25\text{ cm}^3) = \frac{0.00200}{2} = 0.00100\text{ mol}\)

Step 4: Scale up to the total \(250.0\text{ cm}^3\) volumetric flask
\(n(\text{Na}_2\text{CO}_3\text{ in } 250\text{ cm}^3) = 0.00100 \times \left(\frac{250.0}{25.0}\right) = 0.0100\text{ mol}\)

Step 5: Find the molar mass (\(M_r\)) of \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \)
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{2.86\text{ g}}{0.0100\text{ mol}} = 286\text{ g mol}^{-1}\)

Step 6: Find \(x\)
Molar mass of anhydrous \( \text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1}\)
Mass of water of crystallisation \(= 286 - 106 = 180\text{ g mol}^{-1}\)
Since \(M_r(\text{H}_2\text{O}) = 18.0\text{ g mol}^{-1}\):
\(x = \frac{180}{18.0} = 10\)
Formula: \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \)

Key Takeaway: Always follow the 4-step sequence: find moles of known \(\rightarrow\) apply mole ratio \(\rightarrow\) scale up for dilution \(\rightarrow\) calculate target value.

5. Common Exam Pitfalls to Avoid

• Forgetting to convert \(\text{cm}^3\) to \(\text{dm}^3\): Always divide volumes in \(\text{cm}^3\) by \(1000\) before multiplying by concentration!
• Including the rough titre in the mean: Never average the rough titre. Only use concordant values within \(0.10\text{ cm}^3\).
• Leaving the funnel in the burette: Drops of solution can drip from the funnel during titration, giving a falsely small titre volume.
• Air bubble in the burette jet: If the bubble dislodges during titration, the titre volume recorded will be larger than the true volume delivered.
• Missing the scaling factor: Always check whether the titration used the whole original solution or just an aliquot (e.g., \(25.0\text{ cm}^3\) taken from \(250.0\text{ cm}^3\)).

Quick Review Checklist

• Primary standard: pure, stable, high \(M_r\), soluble.
• Pipette & Burette: rinse with water then reagent solution.
• Conical Flask: rinse with deionised water only.
• Concordancy: within \(0.10\text{ cm}^3\).
• Methyl orange: Red (acid) \(\rightarrow\) Orange (endpoint) \(\rightarrow\) Yellow (alkali).
• Phenolphthalein: Colourless (acid) \(\rightarrow\) Pale pink (endpoint) \(\rightarrow\) Pink (alkali).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is a standard solution?

A standard solution is a solution of accurately known concentration.

What key properties must a compound have to be suitable as a primary standard?

It must have high purity, high stability, a high molar mass ((M_r)), and be readily soluble in water.

What are the key steps required to prepare a standard solution in the laboratory?

Weigh accurately by difference, dissolve in deionised water in a beaker, transfer quantitatively with rinsings into a volumetric flask, fill until the meniscus touches the line, and invert repeatedly to mix.

How should volumetric glassware (pipette, burette, and conical flask) be rinsed before a titration?

Rinse the volumetric pipette and burette with deionised water followed by the solution they will contain; rinse the conical flask with deionised water only.

What are concordant titres and how are they used to calculate the mean titre?

Concordant titres are results within \(0.10\text{ cm}^3\) of each other, and only these concordant values (excluding the rough trial) are averaged to find the mean titre.

What are the colours and suitable applications of methyl orange and phenolphthalein indicators?

Methyl orange is red in acid and yellow in alkali (used for strong acid-weak base); phenolphthalein is colourless in acid and pink in alkali (used for weak acid-strong base).

What is the general 4-step sequence used to solve titration calculation problems?

1. Find moles of known (\(n = c \times V\)); 2. Use the stoichiometric mole ratio; 3. Scale up for aliquots/dilutions; 4. Calculate the required final quantity (\(c\), \(M_r\), or mass).

How do an air bubble in the burette jet and leaving the funnel in the burette affect titration results?

An air bubble leads to a falsely high recorded titre, whereas a funnel left in place can drip liquid into the burette, leading to a falsely low recorded titre.

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