Titrations

Choosing the Right Indicator

Universal indicator produces a beautiful rainbow of colours across the pH scale, but this is exactly why it is useless for a titration. Titrations require an indicator with a sharp colour change to pinpoint the exact moment of neutralisation. Universal indicator has a gradual colour change, making it impossible to identify a precise endpoint.

For AQA GCSE, you must use one of two specific indicators, depending on the acid and alkali being used. Phenolphthalein is pink in an alkali and becomes colourless in acid or neutral solutions. Methyl orange is yellow in an alkali, red in an acid, and reaches a distinct orange colour at the endpoint.

Concordant Results and Calculating the Mean

How do scientists know their results are reliable? By repeating the experiment until the data becomes highly consistent. In a titration, you must repeat the procedure until you achieve concordant results. These are titre volumes that are within $0.10\text{ cm}^3$ of each other.

Results must be recorded to two decimal places, ending in a '0' or '5' (e.g., $23.45\text{ cm}^3$). Obtaining concordant results increases the precision of your data, which improves the accuracy of the final calculation. When calculating your final mean, you must exclude your rough titration and any anomalies.

Titration Calculations (Higher Tier Only)

Finding the volume is only half the battle; the real goal is using that volume to unlock the mystery concentration. Concentration can be expressed as a molar concentration ($mol/dm^3$) or a mass concentration ($g/dm^3$). You must always convert your volume from $cm^3$ to $dm^3$ by dividing by 1000.

The core formulas required are:

(Where $n$ is moles, $c$ is concentration in $mol/dm^3$, $V$ is volume in $dm^3$, and $M_r$ is the relative formula mass)

Worked Example:

$25.0\text{ cm}^3$ of $0.10\text{ mol/dm}^3$ $NaOH$ reacts with $20.0\text{ cm}^3$ of $H_2SO_4$. Calculate the concentration of $H_2SO_4$ in $g/dm^3$. (The balanced equation is $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ and the of is )

Step 1: Calculate the moles of the known substance ($NaOH$).

• $V = 25.0 / 1000 = 0.025\text{ dm}^3$
• $n = 0.10 \times 0.025 = 0.0025\text{ mol}$

Step 2: Use the mole ratio from the balanced equation to find the moles of the unknown substance.

• The ratio of $H_2SO_4 : NaOH$ is $1:2$.
• Moles of $H_2SO_4 = 0.0025 / 2 = 0.00125\text{ mol}$

Step 3: Calculate the molar concentration ($mol/dm^3$) of the unknown.

• $V = 20.0 / 1000 = 0.020\text{ dm}^3$
• $c = 0.00125 / 0.020 = 0.0625\text{ mol/dm}^3$

Step 4: Convert the concentration to $g/dm^3$.

• Concentration in $g/dm^3 = 0.0625 \times 98 = 6.13\text{ g/dm}^3$ (to 3 sig figs).