Final reading / $cm3\text{cm}^3$

21.50

41.80

22.30

Initial reading / $cm3\text{cm}^3$

0.00

21.50

2.10

A student uses the reaction of aspirin with cold $NaOH(aq)\text{NaOH(aq)}$ to determine the mass of aspirin in one tablet.

In this reaction, $mol1\text{ mol}$ of aspirin reacts with $mol1\text{ mol}$ of cold $NaOH(aq)\text{NaOH(aq)}$ .

The student’s method is outlined below.

Step 1: The student reacts two aspirin tablets with $cm350.0\text{ cm}^3$ of $NaOH(aq)0.500\text{ mol dm}^{-3}\ \text{NaOH(aq)}$ . The $NaOH\text{NaOH}$ is in excess. A colourless solution forms.
Step 2: The colourless solution from Step 1 is made up to $cm3250.0\text{ cm}^3$ with distilled water.
Step 3: A $cm325.00\text{ cm}^3$ sample of the diluted solution from Step 2 is titrated with $HCl(aq)0.100\text{ mol dm}^{-3}\ \text{HCl(aq)}$ in the burette.

The $HCl(aq)\text{HCl(aq)}$ reacts with excess $NaOH(aq)\text{NaOH(aq)}$ that remains in Step 1:

\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}

The student repeats the titration to obtain concordant (consistent) titres.

Titration results The trial titre (Titre 1) is not used because it is not concordant. The burette readings have been read to the nearest $cm30.05\text{ cm}^3$ :

	1	2	3
Final reading / $cm3\text{cm}^3$	21.50	41.80	22.30
Initial reading / $cm3\text{cm}^3$	0.00	21.50	2.10

Determine the mass, in $mg\text{mg}$ , of aspirin in one aspirin tablet. (Molar mass of aspirin, $mol−1M_r = 180\text{ g mol}^{-1}$ ).

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