A) Calculate the fraction of p-nitrophenol that would be in the ionized state an

Question

A) Calculate the fraction of p-nitrophenol that would be in the ionized state and the expected apparent coefficent at pH 7 and pH 8.

(8 pts.) In order to analyze quantitatively the data from an assay using p-nitrophenyl- phosphate, it is necessary to know the extinction coefficient of the product However the situation in this case is a bit tricky, because p-nitrophenol undergoes a deprotonation reaction, and it is only the ionized form that absorbs light at 405 nm. The ionization equilibrium is: pKa 7.0 OH N O H The extinction coefficient for the deprotonated form is 18, 300 cm. 1M-1, but the apparent extinction coefficient, E for a mixture of the conjugate acid and base depends on app H. The apparent extinction coefficient represents the observed absorbance, in a 1 cm cuvette, divided by the total concentration of the protonated and deprotonated forms.)

Explanation / Answer

First we need to know a little about absorbance.

Lambert-Beer-Bourger law says that the absorbance is directly and proportional to the concentration andn the extintion coefficient.

A=lEC

To calculate the concentration we need to know the other two variables.

Extintion coefficient and the cuvette large.

C=A/lE=405/1(18300)=0.022 mol/L

With this concentration value we can calculate the fraction ionized. Depending pon pka value we can say that at pH 7 the concentration of ionized molecule will be 50%. Because in this values the two species can coexist in equal fraction.

But at pH 8 only a little part of the non ionized moleculewill be ionized. less than 50%.

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