5.6.3 Titrations

Overview of Acid-Base Titrations

An acid-base titration involves adding a solution of known concentration (the titrant) to another solution of unknown concentration until the reaction between them is complete. This is indicated by reaching the equivalence point, where the moles of acid equal the moles of base, resulting in neutralization.

Performing Titration Calculations

To calculate the concentration of an unknown solution in a titration, use the formula:

\text{moles of acid} = \text{moles of base}

The steps are as follows:

Write the Balanced Equation: Ensure you know the stoichiometry of the acid-base reaction.
Calculate Moles of Titrant Added: Use the concentration and volume of the titrant to calculate the number of moles added:

\text{moles} = \text{concentration} \times \text{volume}

Determine Unknown Concentration: Rearrange to solve for the unknown concentration using the mole ratio from the balanced equation.

Example Calculation

infoNote

Example: Titrate 25.0 cm³ of hydrochloric acid ( $HCl$ ) with sodium hydroxide ( $NaOH$ ) of known concentration, 0.1 mol dm⁻³, and find the concentration of $HCl$ .

Step 1: Balanced Equation:

\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

Step 2: Calculate Moles of $NaOH$ Used

Volume of $NaOH = :highlight[20.0 cm³ = 0.020 dm³]$

Moles of $NaOH$

= 0.1 \, \text{mol dm}^{-3} \times 0.020 \, \text{dm}^3

= :highlight[0.002 \, \text{mol}]

Step 3: Determine Moles of $HCl$ :

Since the ratio is 1:1, moles of $HCl$ = moles of $NaOH$ = 0.002 mol

Step 4: Calculate Concentration of $HCl$ :

\text{Concentration of HCl} = \frac{0.002 \, \text{mol}}{0.025 \, \text{dm}^3}

= :success[0.08 \, \text{mol dm}^{-3}]

Titrations Simplified Revision Notes for A-Level AQA Chemistry

5.6.3 Titrations

Overview of Acid-Base Titrations

Performing Titration Calculations

Example Calculation

500K+ Students Use These Powerful Tools to Master Titrations For their A-Level Exams.

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