One method for determination of p K a values of of purine and pyrimidine derivat
ID: 1023431 • Letter: O
Question
One method for determination of pKa values of of purine and pyrimidine derivatives relies of differences in the UV-Vis spectra of neutral and ionized species. Typically, the neutral species (acid) absorbs at a shorter wavelength while the anion (conjugate base) absorbs at longer wavelength. In a favorable situation, this difference can be employed to calculate the concentrations of the acid and the anion.
For example, the ionization of 9-methyluric acid occurs according to the scheme:
and the corresponding UV spectra of the neutral (blue) and anionic (purple) forms are shown below:
Note that the absorbance of the acid form is essentially zero at 304 nm while the molar absorptivity of the anion form at this wavelength is 9200 M-1cm-1. A solution for suitable measurements was prepared by dissolving 5.46 mg of 9-methyluric acid (Mw = 182.14 g/mol) in 145 mL water, adding NaOH drop-wise until pH reached 5.93, then bringing the volume up to 150.00 mL. This solution showed absorbance of 1.225 in 1 cm quartz cuvette when measured at 304 nm. The pKa of 9-methyluric acid is?
NExplanation / Answer
the mass of acid added = 5.46 mg
Mol wt = 182.14 g / ole
so moles of acid added = 5.46 / 182.14 = 0.03 millimoles
volume = 145mL
Concentration of acid = 0.03 / 145 = 2.07 X 10^-4 moles / L
now we know from Beer Lambert's law
Absorbance = absorptivity X concentration x length
1.225 = 9200 X concentration X 1
Concentration = 1.33 X 10^-4
The change in concentration of acid = (2.07 - 1.33 )X 10^-4 = 0.74 X 10^-4 molar
HA ---> H+ + A-
Initial 2.07 0 0
Change -x +x +x
Equil 2.07-x x x
Ka = [H+] [A-] / [HA]
Given ,x = 1.33 X 10^-4
Ka = 1.33 X 1.33 X 10^-4 X 10^-4 / (0.74 X 10^-4)
Ka = 2.39 X 10^-4
pKa = -logKa = 3.62
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