Titration Notes

Titrations

Titration Calculations

• Volumetric Analysis: A process using the volume and concentration of a standard solution to determine the concentration of an unknown solution.

• Titration: The most common technique used in volumetric analysis.

• Equipment:

  • Volumetric or graduated pipette

  • Burette

• Standard Solution Preparation: Must be done before titration. Requires specific apparatus to ensure precise volume measurements.

Apparatus

Key pieces of apparatus used to prepare a volumetric solution and perform a simple titration:

1. Beaker

2. Burette

3. Volumetric Pipette

4. Conical Flask

5. Volumetric Flask

Making a Standard Solution

Chemists prepare solutions with precisely known concentrations for analysis. These are called volumetric solutions or standard solutions.

• Made as accurately and precisely as possible, using three decimal place balances and volumetric flasks to reduce measurement uncertainties.

• Steps:

  1. Weigh out a precise amount of the solid.

  2. Add to a small volume of water and pre-dissolve the solid.

  3. Rinse the beaker with distilled water and add the rinsings to the flask.

  4. Transfer to a volumetric flask using a funnel.

  5. Make up to the scratch mark with more water, ensuring the bottom of the meniscus aligns with the mark.

  6. Add stopper and mix the contents.

Performing the Titration

• Burette: The key piece of equipment used in the titration.

• Precision: Burettes are usually marked to a precision of $0.10 \text{ cm}^3$.

• Uncertainty: Recorded to half the smallest marking, i.e., $\pm 0.05 \text{ cm}^3$.

• End Point/Equivalence Point: Occurs when the two solutions have reacted completely, indicated by an indicator.

Steps in a Titration

1. Measure a known volume (usually 20 or 25 cm³) of one of the solutions with a volumetric pipette and place it into a conical flask.

2. Place the other solution in the burette.

3. Fill the burette to $0.00 \text{ cm}^3$ to start.

4. Add a few drops of the indicator to the solution in the conical flask.

5. Carefully open the tap on the burette and add the solution, portion by portion, to the conical flask until the indicator starts to change color.

6. Slow down the flow of the burette as you approach the end point, adding the solution dropwise.

7. Close the tap on the burette after one drop has caused the color change.

8. Place a white tile under the conical flask to make it easier to see the color change

9. Carry out multiple runs until concordant results are obtained (within $0.1 \text{ cm}^3$ of each other).

Recording and Processing Titration Results

• Record both the initial and final burette readings to a precision of $\pm 0.05 \text{ cm}^3$ (same as the uncertainty).

• The final digit is 0 or 5.

  • All results are recorded to 2 decimal places including zero readings.

• The rough result is usually far over the end-point.

• This result is discarded as it is too high.

• ✓ = concordant results. Used to calculate the average.

• The volume delivered (titre) is calculated as $V<em>{\text{final}} - V</em>{\text{initial}}$ and recorded to an uncertainty of $\pm 0.10 \text{ cm}^3$. The uncertainty is doubled because two burette readings are made.

• Concordant results are then averaged, and non-concordant results are discarded.

• Appropriate calculations are then done.

Volumes & Concentrations of Solutions

• Concentration of a Solution: The amount of solute dissolved in a solvent to make $1 \text{ dm}^3$ of solution.

• Solute: The substance that dissolves in a solvent to form a solution.

• Solvent: Often water.

$\text{Concentration (mol dm}^{-3}) = \frac{\text{number of moles of solute (mol)}}{\text{volume of solution (dm}^3)}$

• Concentrated Solution: A solution that has a high concentration of solute.

• Dilute Solution: A solution with a low concentration of solute.

When carrying out calculations involving concentrations in $\text{mol dm}^{-3}$, consider:

• Changing mass in grams to moles.

• Changing $\text{cm}^3$ to $\text{dm}^3$.

• To calculate the mass of a substance present in asolution of known concentration and volume:

$\text{number of moles (mol)} = \text{concentration (mol dm}^{-3}) \times \text{volume (dm}^3)$

$\text{mass of solute (g)} = \text{number of moles (mol)} \times \text{molar mass (g mol}^{-1})$

Worked Example

Neutralization calculation: $25.0 \text{ cm}^3$ of $0.050 \text{ mol dm}^{-3}$ sodium carbonate was completely neutralized by $20.00 \text{ cm}^3$ of dilute hydrochloric acid. Calculate the concentration in $\text{mol dm}^{-3}$ of hydrochloric acid.

Answer:

Step 1: Write the balanced symbol equation

$\text{Na}<em>2\text{CO}</em>3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}<em>2\text{O} + \text{CO}</em>2$

Step 2: Calculate the amount, in moles of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing the volume by 1000 to convert $\text{cm}^3$ to $\text{dm}^3$

$\text{Amount (Na}<em>2\text{CO}</em>3) = 0.025 \text{ dm}^3 \times 0.050 \text{ mol dm}^{-3} = 0.00125 \text{ mol}$

Step 3: Calculate the moles of hydrochloric acid required using the reaction's stoichiometry

• 1 mol of $\text{Na}<em>2\text{CO}</em>3$ reacts with 2 mol of HCl, so the molar ratio is 1:2

• Therefore 0.00125 moles of $\text{Na}<em>2\text{CO}</em>3$ react with 0.00250 moles of HCl

Step 4: Calculate the concentration, in $\text{mol dm}^{-3}$, of hydrochloric acid

$[\text{HCl}] = \frac{\text{amount (mol)}}{\text{volume (dm}^3)} = \frac{0.00250}{0.0200} = 0.125 \text{ mol dm}^{-3}$