# Acid-base Titrations ## Volumetric analysis - Volumetric analysis determines the concentration of an unknown solution by reacting it with a solution of known concentration (a **standard solution**) - The most common method is a **titration** - This uses precise volume measurements - The two main pieces of equipment used are a **volumetric pipette** (or graduated pipette) and a **burette** - A standard solution must first be prepared carefully before the titration can be performed - Accurate apparatus is essential throughout the process to minimise uncertainty

A **standard solution** is a solution with a precisely known concentration

## Key apparatus - Some key pieces of apparatus used in preparing a standard solution and performing a titration include: | Number | Apparatus | Purpose | |--------|-----------|---------| | 1 | Beaker | Dissolving solute in water | | 2 | Burette | Delivering variable volumes of solution with precision | | 3 | Volumetric pipette | Measuring fixed, accurate volumes of solution | | 4 | Conical flask | Containing the solution being titrated | | 5 | Volumetric flask | Preparing standard solutions to exact volumes | ## Making a standard solution - Chemists often need to prepare solutions with accurately known concentrations - These are called **standard solutions** or **volumetric solutions** - To ensure precision, a **3 d.p. balance** and a **volumetric flask** are used - This helps reduce uncertainty in measurement ### Method 1. **Weigh out the solute** - Use a 3 d.p. balance to accurately weigh the required mass of solid 2. **Dissolve the solute** - Transfer the solid into a beaker containing a small volume of distilled water - Stir using a glass rod until completely dissolved 3. **Transfer to a volumetric flask** - Pour the solution into a clean volumetric flask using a funnel 4. **Rinse the beaker** - Rinse the beaker and glass rod with distilled water - Add the rinsings to the volumetric flask to ensure no solute is lost 5. **Make up to the mark** - Add distilled water until the bottom of the meniscus sits exactly on the scratch mark - Stopper the flask and invert to mix thoroughly

Always rinse apparatus with distilled water and add the rinsings to avoid losing solute

## Performing the titration - The key piece of equipment used in the titration is the **burette** - **Burettes** are usually marked to a precision of $0.10 \text{ cm}^3$ - Since they are analogue instruments, the uncertainty is recorded to **half** the smallest marking, i.e. $\pm 0.05 \text{ cm}^3$ - The **end point** or **equivalence point** occurs when the two solutions have reacted completely and is shown with the use of an **indicator** ### Method 1. Measure a known volume (typically $25 \text{ cm}^3$) of one solution using a volumetric pipette 2. Transfer it into a conical flask 3. Add a few drops of indicator to the solution in the conical flask 4. Place the conical flask on a white tile - The white tile makes it easier to observe the colour change 5. Fill the burette with the second solution and record the initial volume 6. Slowly add the solution from the burette to the conical flask while swirling 7. As the end point approaches, add the solution drop by drop until the colour changes 8. Record the final burette volume 9. Repeat the titration until you obtain **concordant titres** - Concordant results are within $0.10 \text{ cm}^3$ of each other

The uncertainty of a burette reading is $\pm 0.05 \text{ cm}^3$, but the titre uncertainty is $\pm 0.10 \text{ cm}^3$ because two readings are used

## Recording and processing titration results - Record **initial and final burette readings** to **2 decimal places**, ending in **.00** or **.05** - This is because the precision/uncertainty of the burette is $\pm 0.05 \text{ cm}^3$ - Calculate the **titre** by subtracting the initial from the final volume - The uncertainty of the titre is $\pm 0.10 \text{ cm}^3$ - This is because two readings with $0.05 \text{ cm}^3$ uncertainty are used to determine the titre - Discard any rough or inconsistent results - The **rough titre** is usually far over the end point and should be discarded - Calculate an **average titre** using only concordant values ### Example titre table | | Rough | Run 1 | Run 2 | Run 3 | |---|---|---|---|---| | Initial ($\text{cm}^3$) | 0.00 | 23.15 | 0.25 | 23.00 | | Final ($\text{cm}^3$) | 23.75 | 45.95 | 23.00 | 46.10 | | Titre ($\text{cm}^3$) | 23.75 | 22.80 | 22.75 | 23.10 | | Concordant? | ✗ | ✓ | ✓ | ✗ | - ✓ = Concordant results used to calculate the average titre - Run 3 is discarded as an outlier since it is outside the concordant range - The average titre is calculated from Run 1 and Run 2 only: $$\frac{22.80 + 22.75}{2} = 22.775 \rightarrow 22.78 \text{ cm}^3$$ ## Percentage uncertainties - Percentage uncertainty helps assess how significant an error is compared to the measured value - The percentage uncertainty formula is: $$\text{percentage uncertainty} = \frac{\text{uncertainty}}{\text{measured value}} \times 100$$ ### Adding or subtracting measurements - When adding or subtracting quantities, **add the absolute uncertainties** - This applies to: - Mass measurements (initial and final) - Temperature changes - Burette readings (initial and final) - Each instrument is read twice, so uncertainties are doubled

For burette readings, the uncertainty in the titre is $\pm 0.10 \text{ cm}^3$ because you read it twice (initial and final)

## Acid-base titration calculations ### Volumes & concentrations of solutions - The **concentration** of a solution is the amount of **solute** in $1 \text{ dm}^3$ of solution - The **solute** is the substance being dissolved - The **solvent** is the liquid doing the dissolving (usually water) - A **concentrated** solution contains a high amount of solute - A **dilute** solution contains a small amount of solute - Concentration can be expressed as: - Moles per $\text{dm}^3$ - Mass per $\text{dm}^3$ - Parts per million (ppm), often used for trace substances in environmental chemistry ### Using concentration in calculations - To calculate values involving concentration, use: $$\text{concentration} \, (\text{mol dm}^{-3}) = \frac{\text{number of moles of solute (mol)}}{\text{volume of solution (dm}^3)}$$ - This rearranges to: $$\text{number of moles (mol)} = \text{concentration (mol dm}^{-3}) \times \text{volume (dm}^3)$$ - To convert between mass and moles: $$\text{mass (g)} = \text{amount (mol)} \times \text{molar mass (g mol}^{-1})$$

Convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by 1000

### Worked Example: Titration calculation A student uses $25.0 \text{ cm}^3$ of $0.050 \text{ mol dm}^{-3}$ sodium hydroxide to neutralise $19.6 \text{ cm}^3$ of dilute nitric acid in a titration. Calculate the concentration of the nitric acid. **Answer:** 1. Write the balanced symbol equation $$\text{NaOH} + \text{HNO}_3 \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}$$ 2. Calculate the amount, in moles, of sodium hydroxide: - Rearrange the equation for amount of substance (mol) - Divide the volume by 1000 to convert $\text{cm}^3$ to $\text{dm}^3$ $$\text{amount (NaOH)} = 0.025 \text{ dm}^3 \times 0.050 \text{ mol dm}^{-3} = 0.00125 \text{ mol}$$ 3. Calculate the moles of nitric acid required using the reaction's stoichiometry: - 1 mol of NaOH reacts with 1 mol of $\text{HNO}_3$ - The molar ratio is 1:1 - So, 0.00125 moles of NaOH react with 0.00125 moles of $\text{HNO}_3$ 4. Calculate the concentration, in $\text{mol dm}^{-3}$, of nitric acid: $$\text{concentration of HNO}_3 = \frac{\text{amount (mol)}}{\text{volume (dm}^3)}$$ $$\text{concentration of HNO}_3 = \frac{0.00125}{0.0196}$$ $$\text{concentration of HNO}_3 = 0.064 \text{ mol dm}^{-3}$$

Always use the balanced equation to determine the molar ratio between reactants before calculating concentrations