Question

Assume all of the ionized aspirin remains in the buffer solution and all of the un-ionized aspirin goes into the ethyl acetate, what is the molarity of the aspirin in the ethyl acetate at pH 2.0 and pH 8.0?

Is the calculated Molarity correct?

Ethyl acetate at pH =2:

pKa of aspirin = 3.49

pH = pKa + log [Base] / [Acid]

pH = pKa + log [Ionized] / [Unionized]

2 = 3.49 + log [Ionized] / [Unionized]

log [Ionized] / [Unionized] = -1.49

[Ionized] / [Unionized] = 0.0323 /1

Moles of aspirin in ethyl acetate = 1 + 0.0323 = 1.0323 mol of aspirin

Molarity = mol/L = 1.0323 mol / (2mL x 1L/1000mL) = 5.2 x 10² M aspirin

Ethyl acetate at pH = 8

pKa of aspirin = 3.49

pH = pKa + log [Base] / [Acid]

pH = pKa + log [Ionized] / [Unionized]

8 = 3.49 + log [Ionized] / [Unionized]

log [Ionized] / [Unionized] = 4.51

[Ionized] / [Unionized] = 32359.36

Moles of aspirin in ethyl acetate = 1 + 32359.36 = 32360.36 mol of aspirin

Molarity = mol/L = 32359.36 mol / (2mL x 1L/1000mL) = 1.6 x 10⁷ M aspirin

Answer 1

Upto the part   [Ionized] / [Unionized] = 0.323/ 1        is fine

this ratio inidcates that per 1 moles of Unionized aspririn we have 0.323 moles of aspirin in ionized form

so total moles of aspirin = 1+0.323 = 1.323

fraction of aspirin in unionized form = ( moles of aspirin / total moles of aspirin)

             = ( 1/ 1.323) = 0.756

Now based on initial aspirin moles we must find aspirin in unionized form

Aspririn moles in unionized form = ( fraction) x initial apsirin moles

                                             = 0.756 x ( initial apsirin) moles

( Initial aspirin moles not mentioned in this question , hence recheck that part)

Once we get aspriin in unionized form moles

then we do calculations as you did

Molarity= moles of aspirin in unionized form / ( total solution volume in L)

same calculations with pH = 8 ,

we first find fraction of aspirin in unionized form

and then find aspriin moles in unionized form by formula   fraction x initial aspirin moles