Question

In titration 1, to determine M_ASA, you weigh out 269.1 mg of pure aspirin (ASA) on the analytical balance. After dissolving this mass in ethanol and water, you titrate the solution with the KOH to the phenolpthaline equilvalence point. Subtracting the initial buret reading from the final one, you find that the amount of KOH consumed is equal to 24.58 mL.

Based on this titration, what is the value of M_ASA   (mg / mmol ) ?

Answer 1

Answer –

We are given, mass of aspirin = 216.1 g

Volume of KOH = 24.58 mL

First we need to calculate the moles of aspirin

We know molar mass of aspirin is 180.157 g.mol^-1

Now we need to convert given mass mg to g

So, 1 mg = 0.001g

So, 269.1 mg = ?

= 0.2691 g

Moles of aspirin = 0.2691 g / 180.157 g.mol¹

                           = 0.001494 moles of aspirin

We know mole ratio between aspirin and KOH is 1:1,

So moles of aspirin = moles of KOH

Moles of KOH = 0.001494 moles

Moles of MASA = moles of aspirin

So, moles of MASA = 0.001494 moles

So molar mass of MASA = g / 0.001494 mole

Mass of MASA formed = mass of aspirin + mass of KOH

Mass of KOH = 0.001494 mole * 39.083 g/mol¹

                       = 0.0584 g

So, Mass of MASA formed = 0.2691+0.0584

                                               = 0.3274 g

                                                = 327.4 mg

Mmole of MASA = 0.001491 *1000

                     = 1.491 mmol

So, value of MASA = 237.4 mg / 1.491 mmol

                                  = 219.58 mg/mmol