Question

1.250 g mass of standard aspirin is dissolved in approximately 20 mL of 1 M NaOH. This solution is added to a 250-mL volumetric flask and diluted to the mark with DI water. How many grams per mL of aspirin are in the resulting solution. (Note a 250-mL volumetric flask has a precision of 0.01 mL and the amount of NaOH added has no bearing on the significant figures in the final concentration).

Answer 1

Answer – We are given, mass of standard aspirin = 1.250 g

Volume of 20 mL of 1.0 M NaOH

We know aspirin is the monoprotic acid

Molarity of NaOH at 250 mL –

M2 = 1 M * 20 mL / 250 mL = 0.08 M

So moles of NaOH = 0.08 M * 0.250 L

                               = 0.020 moles

So, moles of NaOH = moles of aspirin = 0.020 moles

Moles of aspirin = 1.250 g / 180.16 g/mol

                           = 0.00694 moles

So in the NaOH all aspirin gets reacted with NaOH and convert to the sodium salt of aspirin. If aspirin not reacted with NaOH, then in the resulting solution = 1.250 g / 250 mL

                       = 0.005 g/mL