Assume all of the ionized aspirin remains in the buffer solutionand all of the un-ionized aspirin goes into the ethyl acetate, whatis the molarity of the aspirin in the ethyl acetate at pH 2.0 andpH 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 ofaspirin
Molarity = mol/L = 1.0323 mol / (2mL x 1L/1000mL) = 5.2 x102 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 molof aspirin
Molarity = mol/L = 32359.36 mol / (2mL x 1L/1000mL) = 1.6 x107 M aspirin
