Introduction: Welcome to Concentration and Acid–Base Titrations
Have you ever wondered how food scientists test the exact acidity of fruit juice, or how environmental chemists measure acid rain levels in rivers? The answer lies in volumetric analysis—specifically, acid–base titrations. Don't worry if quantitative chemistry feels a little intimidating at first. Once you break the calculations down into clear, repeatable steps, titrations become one of the most reliable scoring areas in OCR Chemistry B (Salters).
In this chapter, we will master how to prepare a standard solution, carry out a precise titration using the correct laboratory glassware, calculate unknown solution concentrations using the core relationship \(n = c V\), and evaluate experimental percentage uncertainties.
Cross-reference note: Fundamental mole conversions involving pure solids and masses (\(n = \frac{m}{M}\)) are detailed in Formulae, equations and the mole, while redox titrations are covered in Redox and iodine–thiosulfate titrations.
1. Understanding Solution Concentration
The concentration of a solution tells us how much solute is dissolved in a given volume of solvent (usually water). In chemistry, we express concentration in two main ways:
1. Molar concentration (molarity): Measured in \(\text{mol dm}^{-3}\). This tells us the amount of substance (in moles) present in \(1\text{ dm}^3\) of solution.
2. Mass concentration: Measured in \(\text{g dm}^{-3}\). This tells us the mass of solute (in grams) present in \(1\text{ dm}^3\) of solution.
Crucial Volume Conversions
Most laboratory apparatus (such as pipettes, burettes, and measuring cylinders) is calibrated in cubic centimetres (\(\text{cm}^3\)). However, concentration is defined per cubic decimetre (\(\text{dm}^3\)).
Remember: \(1\text{ dm}^3 = 1000\text{ cm}^3 = 1\text{ litre}\).
To convert from \(\text{cm}^3\) to \(\text{dm}^3\), always divide by \(1000\):
\(V\text{ in dm}^3 = \frac{V\text{ in cm}^3}{1000}\)
The Fundamental Formulas
The primary formula relating moles, concentration, and volume is:
\(n = c \times V\)
Where:
• \(n\) = amount of substance in moles (\(\text{mol}\))
• \(c\) = concentration in moles per cubic decimetre (\(\text{mol dm}^{-3}\))
• \(V\) = volume in cubic decimetres (\(\text{dm}^3\))
If your volume is in \(\text{cm}^3\), the equation becomes:
\(n = \frac{c \times V\text{ (in cm}^3)}{1000}\)
To convert between mass concentration (\(\text{g dm}^{-3}\)) and molar concentration (\(\text{mol dm}^{-3}\)), we use the molar mass (\(M\)) of the solute:
\(\text{concentration in g dm}^{-3} = \text{concentration in mol dm}^{-3} \times M\)
Memory Trick: Think of a triangle with \(n\) at the top, and \(c\) and \(V\) at the bottom. Cover the letter you want to find: to find \(c\), compute \(c = \frac{n}{V}\); to find \(V\), compute \(V = \frac{n}{c}\); to find \(n\), multiply \(c \times V\).
Key Takeaway: Always check your volume units before calculating. If the volume is given in \(\text{cm}^3\), divide by \(1000\) to convert it to \(\text{dm}^3\) before multiplying by concentration.
2. Preparing a Standard Solution
A standard solution is a solution of accurately known concentration. Preparing one is a fundamental practical technique assessed in written papers and the Practical Endorsement.
Step-by-Step Procedure:
Step 1: Accurate Weighing
Weigh a clean, dry weighing boat (or beaker) containing the solid solute on a balance reading to at least two decimal places. Tip the solid into a clean beaker. Re-weigh the empty boat to determine the precise mass transferred by difference:
\(\text{mass transferred} = \text{mass of boat + solid} - \text{mass of boat after emptying}\)
Step 2: Dissolving the Solute
Add a small volume of deionised (or distilled) water (e.g. \(50\text{ cm}^3\) to \(100\text{ cm}^3\)) to the beaker. Stir thoroughly with a glass rod until all the solid crystals have completely dissolved.
Step 3: Transferring the Solution
Pour the solution into a clean volumetric flask (e.g. \(250.0\text{ cm}^3\)) through a filter funnel. Rinse the beaker, glass rod, and funnel several times with small portions of deionised water from a wash bottle, transferring all washings directly into the volumetric flask. This ensures no solute is left behind on the apparatus.
Step 4: Making up to the Mark
Add deionised water to the volumetric flask until the liquid level approaches the graduation line. Use a dropping pipette to add the final drops slowly until the bottom of the meniscus sits exactly on the horizontal graduation mark at eye level.
Step 5: Inverting to Mix
Insert the stopper securely into the volumetric flask and invert it repeatedly (at least \(10\) to \(15\) times) to ensure thorough and uniform mixing. If you forget to invert the flask, the solution will be denser at the bottom and more dilute at the top, ruining subsequent titrations!
Common Mistake to Avoid: Filling past the graduation mark. If you overshoot the line, the total volume is unknown, the concentration is incorrect, and the entire solution must be discarded and remade from scratch.
Key Takeaway: Quantitatively transferring washings and inverting the flask after making up to the mark are essential practical steps to ensure accurate and uniform concentration.
3. The Acid–Base Titration Procedure
An acid–base titration involves reacting a standard solution of an acid or base with a known volume of a solution of unknown concentration until the reaction reaches its neutralisation point (the end-point), signalled by an acid–base indicator.
Apparatus and Technique:
• Volumetric Pipette: Used to measure and transfer an exact fixed volume (the aliquot, typically \(25.0\text{ cm}^3\)) of one solution into a conical flask. The pipette is rinsed first with deionised water, and then with the specific solution it will deliver.
• Conical Flask: Holds the reaction mixture. Its sloping sides allow vigorous swirling without spilling. It is rinsed only with deionised water before use—residual water droplets inside will not change the number of moles of solute transferred.
• Burette: Used to deliver variable, accurately measured volumes of the second solution. The burette is rinsed first with deionised water, and then with the solution it will contain to prevent dilution. Ensure the jet space below the tap is completely filled with liquid and contains no trapped air bubbles.
• White Tile: Placed under the conical flask to provide a clear background, making the indicator colour change easy to spot.
Running the Titration:
1. Record the initial burette reading to the nearest \(0.05\text{ cm}^3\). Read the bottom of the meniscus at eye level to prevent parallax error.
2. Add a few drops of a suitable acid–base indicator to the conical flask. (Never add too much indicator, as indicators are themselves weak acids or bases and adding excess alters the titre).
3. Run a quick trial (rough) titration to establish an approximate volume needed for neutralisation.
4. For accurate titrations, run the solution in rapidly until near the end-point, then add drop-by-drop with continuous swirling until a single drop causes an immediate, permanent colour change.
5. Record the final burette reading to \(0.05\text{ cm}^3\). Calculate the titre:
\(\text{titre} = \text{final reading} - \text{initial reading}\)
6. Repeat until you obtain concordant titres—results that agree within \(0.10\text{ cm}^3\) of each other.
Calculating the Mean Titre:
Only average the concordant titres. The rough (trial) titre and any non-concordant values must be discarded.
Example: If your titres are \(24.60\text{ cm}^3\) (rough), \(23.80\text{ cm}^3\), \(24.10\text{ cm}^3\), and \(23.90\text{ cm}^3\):
The concordant titres are \(23.80\text{ cm}^3\) and \(23.90\text{ cm}^3\) (difference = \(0.10\text{ cm}^3\)).
\(\text{mean titre} = \frac{23.80 + 23.90}{2} = 23.85\text{ cm}^3\)
Key Takeaway: Burette readings must be recorded to two decimal places, with the final digit being \(0\) or \(5\). Average only concordant titres (within \(0.10\text{ cm}^3\)).
4. Step-by-Step Titration Calculations
All acid–base titration calculations follow a structured four-step path. Memorise this four-step pathway:
Step 1: Calculate moles of the known substance (\(n_1\))
Use the mean titre volume and concentration of the solution whose details are fully known:
\(n_1 = c_1 \times V_1\)
Step 2: Use the stoichiometric ratio from the balanced chemical equation
Determine how many moles of the unknown substance (\(n_2\)) react with \(n_1\).
Step 3: Calculate the concentration or mass of the unknown substance in the sample
Use the mole value (\(n_2\)) and its sample volume (\(V_2\)):
\(c_2 = \frac{n_2}{V_2}\)
Step 4: Scale up to original solutions (if required)
If the sample tested was an aliquot taken from a larger volumetric flask (e.g. \(25.0\text{ cm}^3\) taken from \(250.0\text{ cm}^3\)), multiply the moles by the dilution factor (\(\frac{250.0}{25.0} = 10\)) to find the total moles in the original container.
Worked Example: Finding the Concentration of Hydrochloric Acid
A student titrates \(25.0\text{ cm}^3\) of an unknown concentration of \(\text{HCl}\) with a standard solution of \(0.100\text{ mol dm}^{-3}\) \(\text{Na}_2\text{CO}_3\). The mean concordant titre of \(\text{Na}_2\text{CO}_3\) is \(18.40\text{ cm}^3\). Find the concentration of the \(\text{HCl}\).
Balanced equation:
\(\text{Na}_2\text{CO}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}\)
Step 1: Calculate moles of \(\text{Na}_2\text{CO}_3\) used:
\(V = \frac{18.40}{1000} = 0.01840\text{ dm}^3\)
\(n(\text{Na}_2\text{CO}_3) = c \times V = 0.100\text{ mol dm}^{-3} \times 0.01840\text{ dm}^3 = 1.840 \times 10^{-3}\text{ mol}\)
Step 2: Use the mole ratio:
From the balanced equation, \(1\text{ mol of Na}_2\text{CO}_3\) reacts with \(2\text{ mol of HCl}\).
\(n(\text{HCl}) = 1.840 \times 10^{-3} \times 2 = 3.680 \times 10^{-3}\text{ mol}\)
Step 3: Calculate the concentration of \(\text{HCl}\):
The volume of \(\text{HCl}\) used was \(25.0\text{ cm}^3 = 0.0250\text{ dm}^3\).
\(c(\text{HCl}) = \frac{n}{V} = \frac{3.680 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.1472\text{ mol dm}^{-3}\)
Rounding to \(3\) significant figures (matching the data supplied):
\(c(\text{HCl}) = 0.147\text{ mol dm}^{-3}\)
Key Takeaway: Always check the stoichiometry of the chemical equation. In reactions between diprotic acids (like \(\text{H}_2\text{SO}_4\)) or dibasic salts (like \(\text{Na}_2\text{CO}_3\)), the mole ratio is not \(1:1\).
5. Experimental Uncertainties and Percentage Error
Every measuring instrument has an inherent degree of precision called its margin of error (or uncertainty). In the laboratory and written assessments, you must be able to calculate the percentage uncertainty for each measurement.
Percentage Uncertainty Formula:
\(\text{percentage uncertainty} = \frac{\text{uncertainty of apparatus}}{\text{measurement made}} \times 100\)
Important Guidelines for Different Apparatus:
• Pipette: Delivering a fixed volume involves a single measurement reading. For example, a \(25.0\text{ cm}^3\) volumetric pipette typically has an uncertainty of \(\pm 0.06\text{ cm}^3\):
\(\text{percentage uncertainty} = \frac{0.06}{25.0} \times 100 = 0.24\%\)
• Volumetric Flask: Preparing a standard solution in a \(250.0\text{ cm}^3\) flask typically has an uncertainty of \(\pm 0.20\text{ cm}^3\):
\(\text{percentage uncertainty} = \frac{0.20}{250.0} \times 100 = 0.08\%\)
• Burette (Crucial Exam Rule): A titre is calculated from two separate burette readings (an initial reading and a final reading). Therefore, the overall uncertainty is twice the reading uncertainty:
\(\text{total uncertainty} = 2 \times (\text{uncertainty of one reading})\)
If a burette has an uncertainty of \(\pm 0.05\text{ cm}^3\) per reading, the total uncertainty for a titre is \(2 \times 0.05 = \pm 0.10\text{ cm}^3\).
Example: For a mean titre of \(20.00\text{ cm}^3\):
\(\text{percentage uncertainty} = \frac{2 \times 0.05}{20.00} \times 100 = \frac{0.10}{20.00} \times 100 = 0.50\%\)
How to Minimise Percentage Uncertainty:
To reduce percentage uncertainty in a titration:
• Increase the titre volume: A larger titre volume (e.g. \(25\text{ cm}^3\) instead of \(10\text{ cm}^3\)) increases the denominator in the formula, significantly lowering the overall percentage error. This is achieved by either decreasing the concentration of the titrant in the burette or increasing the concentration/volume of the solution in the conical flask.
• Use a more precise balance: Using a balance reading to three decimal places (\(\pm 0.001\text{ g}\)) rather than two decimal places (\(\pm 0.01\text{ g}\)) reduces weighing uncertainty.
Key Takeaway: Titres always involve two burette readings, meaning the uncertainty is doubled when calculating the percentage error of the titre.
Quick Review: Summary of Key Terms
• Standard solution: A solution of accurately known concentration.
• Aliquot: An accurately measured volume of liquid taken from a larger sample, usually with a volumetric pipette.
• Concordant titres: Titre volumes agreeing within \(0.10\text{ cm}^3\) of each other.
• Meniscus: The curve in the upper surface of a liquid; readings are taken from the lowest point at eye level.
• End-point: The point during a titration when the indicator permanently changes colour, showing that stoichiometric neutralisation is complete.